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INF3707: Database implementation and design

The purpose of this course is to enable a systems analyst and programmer to apply the steps of the database life cycle to effectively implement database structures to solve client problems according to specific standards.

Lesson 01: Installing oracle XE

Instructions: In this lesson, you are required to make use of the prescribed study manual, Oracle Database 10g: SQL Fundamentals 1, Introduction and Lesson 1.

Lesson Objectives

After completing this lesson, you should be able to do the following:

To install and configure Oracle XE database software List the features of Oracle 10g. Discuss the theoretical and physical aspects of a relational database. Describe the Oracle implementation of RDBMS and ORDBMS. Describe the main SQL statements.

In this lesson, it is important that you understand the major SQL statements and how they are used in carrying database operations. Here is a list of statements that you should be familiar with:

oo SELECToo INSERToo UPDATEoo DELETEoo MERGEoo CREATEoo ALTERoo DROPoo RENAMEo

https://my.unisa.ac.za/portal/tool/dc3552c8-1aea-4abc-8717-3080cbd4924a/print_module.jsf?printModuleId=135725080

o TRUNCATEoo COMMENToo GRANToo REVOKEoo COMMIToo ROLLBACKoo SAVEPOINTo

Resources:

1.2. Oracle Study Mananual, Fundamentals 13.4. https://www.youtube.com/watch?v=CWCDDuJnJR05.6. http://docs.oracle.com/cd/B25329_01/doc/admin.102/b25107/toc.htm

Practical exercise 1: Installing Oracle XE

Download the file Installing Oracle XE from myUnisa under additional resources and follow the instructions on how to install and configure Oracle database.

Lesson 2: Retrieving Data Using the SQL SELECT Statements

Instructions: This lesson teaches you how to extract data from a database using SQL SELECT statement. You are referred to the prescribed study manual lesson 1.

Lesson objectives

List the capabilities of SQL SELECT statement Execute a basic SELECT statement

This lesson explores the concepts of extracting or retrieving data stored in relational tables using the SELECT statement. The statement is introduced in its basic form and is progressively built on to extend its core functionality. As you learn the rules governing this statement, an important point to remember is that the SELECT statement never alters information stored in the database. Instead, it provides a read-only method of extracting information. When you ask a question using SQL SELECT, you are guaranteed to extract results, even from the difficult questions.

Exercise 2.1

How do you describe the Human Resources Schema? The HR schema contains seven tables representing a data model of a fictitious Human Resources department. Follow the steps given below to view all the tables in the human resources database and to view the structure of the tables

Answer:

Follow the given steps to describe the human resources schema.

1. Launch SQL Developer. From the start menu choose All Programs, Oracle Database 10g Expression edition, Go to database home page, Choose SQL, SQL commands. The SQL developer editor will be launched.

2. Now view all the tables that are in the Human Resources Schema.a. Enter the statement: Select * from tab; and run the command.b. Enter the statement: Select * from tabs; and run the command.c. What is the difference between these two statements?

3. Now you need to view the structure of each and every table that is in the Human Resources Schema.

a. Type in the command: DESCRIBE JOBS. Terminating this command with a semicolon is optional. Run the command.

4. Repeat steps 4 to describe the remaining JOB_HISTORY, LOCATIONS, COUNTRIES, and REGIONS tables.

Exercise 2.2

How many unique departments are there in the database that has employees currently working in them?

Answer: Follow the steps given below to answer the above questions.

1. Launch SQL developer and connect to the HR schema.

2. You may initially be tempted to find the answer in the DEPARTMENTS table. A careful examination reveals that the question asks for information about employees. This information is contained in the EMPLOYEES table.

3. The word “unique” should guide you to use the DISTINCT keyword.

4. Combining steps 2 and 3, you can construct the following SQL statement:

select distinct department_idfrom employees;

5. This query returns 12 rows. Notice that the third row is empty. This is a null value in the DEPARTMENT_ID column.

6. What is the answer to question 1?

Exercise 2.2

How many countries are there in the Europe region?

Answer: follow the steps given below to answer the above questions.

1. This question comprises of two parts. Consider the REGIONS table, which contains four regions each uniquely identified by a REGION_ID value, and the COUNTRIES table, which has a REGION_ID column indicating which region a country belongs to.

2. The first query needs to identify the REGION_ID of the Europe region. This is accomplished by the SQL statement:

select * from regions;

3. The results show that the Europe region has a REGION_ID value of 1:

4. To identify which countries have 1 as their REGION_ID, you need to execute the following SQL query

select region_id, country_name from countries;

5. Manually counting the country rows with a REGION_ID of 1 in from the results helps in answering the second question:

6. The answer to the second question is therefore: There are eight countries in the Europe region as far as the HR data model is concerned.

Exercise 2.3

Answer the following MCQ questions. Discuss the solution of these questions with other students through the discussion forums. No solutions to these questions will be provided by the lecturer.

1. Which query creates a projection of the DEPARTMENT_NAME and LOCATION_ID columns from the DEPARTMENTS table? (Choose the best answer.)

a. SELECT DISTINCT DEPARTMENT_NAME, LOCATION_ID

FROM DEPARTMENTS;

b. SELECT DEPARTMENT_NAME, LOCATION_ID

FROM DEPARTMENTS;

c. SELECT DEPT_NAME, LOC_ID

FROM DEPT;

d. SELECT DEPARTMENT_NAME AS “LOCATION_ID”

FROM DEPARTMENTS;

2. After describing the EMPLOYEES table, you discover that the SALARY column has a data type of NUMBER(8,2). Which SALARY value(s) will not be permitted in this column?

(Choose all that apply.)

a. SALARY=12345678

b. SALARY=123456.78

c. SALARY=12345.678

d. SALARY=123456

e. SALARY=12.34

3. After describing the JOB_HISTORY table, you discover that the START_DATE and END_DATE columns have a data type of DATE. Consider the expression END_DATE-START_DATE. (Choose two correct statements.)

a. A value of DATE data type is returned.

b. A value of type NUMBER is returned.

c. A value of type VARCHAR2 is returned.

d. The expression is invalid since arithmetic cannot be performed on columns with DATE data types.

e. The expression represents the days between the END_DATE and START_DATE less one day.

4. The DEPARTMENTS table contains a DEPARTMENT_NAME column with data type VARCHAR2(30). (Choose two true statements about this column.)

a. This column can store character data up to a maximum of 30 characters.

b. This column must store character data that is at least 30 characters long.

c. The VARCHAR2 data type is replaced by the CHAR data type.

d. This column can store data in a column with data type VARCHAR2(50) provided that the contents are at most 30 characters long.

5. Which statement reports on unique JOB_ID values from the EMPLOYEES table? (Choose all that apply.)

a. SELECT JOB_ID FROM EMPLOYEES;

b. SELECT UNIQUE JOB_ID FROM EMPLOYEES;

c. SELECT DISTINCT JOB_ID, EMPLOYEE_ID FROM EMPLOYEES;

d. SELECT DISTINCT JOB_ID FROM EMPLOYEES;

Lesson 3: Using single raw functions to sort data

Instructions: In this lesson you are required to read through Fundamentals 1, lesson 3 of the prescribed study manual.

Objectives

Describe various types of functions that are available in SQL Use character, number and date functions in SELECT statements Describe the use of conversion functions

A function is a program written to optionally accept input parameters, perform an operation, or return a single value. A function returns only one value per execution. For information on functions, you are required to read lesson 3 of Fundamentals 1. After reading lesson 3, please do the following practical exercises.

Exercise 3.1: Using the Case Conversion Functions

Retrieve a list of all FIRST_NAME and LAST_NAME values from the EMPLOYEES table where FIRST_NAME contains the character string “li.”


1. Start SQL Developer and connect to the HR schema.The SELECT clause is

1. SELECT FIRST_NAME, LAST_NAME2. The FROM clause is3. FROM EMPLOYEES4. The WHERE clause must compare the FIRST_NAME column values with a pattern of

characters containing all possible case combinations of the string “li.” Therefore, if the FIRST_NAME contains the character strings “LI,” “Li,” “lI,” or “li,” that row must be retrieved.

5. The LIKE operator is used for character matching, and four combinations can be extracted with four WHERE clauses separated by the OR keyword. However, the case conversion functions can simplify the condition. If the LOWER function is used on the FIRST_NAME column, the comparison can be done with one WHERE clause condition. The UPPER or INITCAP functions could also be used.

6. The WHERE clause is

WHERE LOWER(FIRST_NAME) LIKE '%li%'

Execute the above statements.

Exercise 3.2: Using the Case Manipulation Functions

Envelope printing restricts the addressee field to 16 characters. Ideally, the addressee field contains employees’ FIRST_NAME and LAST_NAME values separated by a single space. When the combined length of an employee’s FIRST_NAME and LAST_NAME exceeds 15 characters, the addressee field should contain their formal name. An employee’s formal name is made up of the first letter of their FIRST_NAME and the first 14 characters of their LAST_NAME.

You are required to retrieve a list of FIRST_NAME and LAST_NAME values and formal names for employees where the combined length of FIRST_NAME and LAST_NAME exceeds 15 characters.

Answer: Follow the steps given below to answer the above questions.

1. Start SQL developer and connect to the HR schema.2. The formal name is constructed by concatenating the first character in the

FIRST_NAME field with a space and the first 14 characters of the LAST_NAME field to return a string that is 16 characters long. The SUBSTR function is used to extract the initial and surname portions.

3. The SELECT clause is4. SELECT FIRST_NAME, LAST_NAME, SUBSTR (FIRST_NAME,1,1)||' '||

SUBSTR(LAST_NAME,1,14) FORMAL_NAME5. The FROM clause is

6. FROM EMPLOYEES7. The WHERE clause must limit the records returned to only those where the combined

lengths of their FIRST_NAME and LAST_NAME exceeds 15 characters.8. The WHERE clause is

WHERE LENGTH (FIRST_NAME) + LENGTH (LAST_NAME) > 15

7. Execute the statements.

Exercise 3.3: Using the Date Functions

You are required to obtain a list of EMPLOYEE_ID, LAST_NAME, and HIRE_DATE values for the employees who have worked more than 100 months between the date they were hired and 01-JAN-2000.


1. Start SQL Developer and connect to the HR schema.2. The SELECT clause is: SELECT EMPLOYEE_ID, LAST_NAME, HIRE_DATE3. The FROM clause is: FROM EMPLOYEES4. The WHERE clause must compare the months between the given date literal and the

HIRE_DATE value with the numeric literal 100.5. The MONTHS_BETWEEN function may be used in the WHERE clause.6. The WHERE clause is WHERE MONTHS_BETWEEN('01-JAN-2000', HIRE_DATE)

> 1007. Execute the above statements

Exercise 3.4: MCQ

Do the following multiple choice questions. Solutions to these questions are not provided to students. You are required share and discuss your solutions with other students through myUnisa discussion forums.

Describe Various Types of Functions Available in SQL

1. Which statements regarding single-row functions are true? (Choose all that apply.)A. They may return more than one result.B. They execute once for each record processed.C. They may have zero or more input parameters.D. They must have at least one mandatory parameter.

2. Which of these are single-row character-case conversion functions? (Choose all that apply.)A. LOWERB. SMALLERC. INITCASED. INITCAP

Use Character, Number, and Date Functions in SELECT Statements

3. What value is returned after executing the following statement:

SELECT LENGTH('How_long_is_a_piece_of_string?') FROM DUAL; (Choose the best answer.)

A. 29B. 30C. 24D. None of the above

4. What value is returned after executing the following statement:

SELECT SUBSTR('How_long_is_a_piece_of_string?', 5,4) FROM DUAL; (Choose the best answer.)

A. longB. _longC. string?D. None of the above

5. What value is returned after executing the following statement?

SELECT INSTR('How_long_is_a_piece_of_string?','_',5,3) FROM DUAL; (Choose the best answer.)

A. 4B. 14C. 12D. None of the above

Lesson 4: Reporting aggregated data using the group functions

Instructions: In this lesson, you are required to make use of the prescribed study manual, Oracle Database 10g: SQL Fundamentals 1, Lesson 4.

Objectives

Identify the available group functions Describe the use of group functions Group data by using the GROUP BY clause Include or exclude grouped rows by using the HAVING clause

Group functions operate on aggregated data and return a single result per group. These groups usually consist of zero or more rows of data. Single-row functions are defined with the formula: F(x, y, z, …) = result, where x,y,z… are input parameters. The function F executes on one row of the data set at a time and returns a result for each row. For information on functions, you are

required to read lesson 4 of Fundamentals 1. After reading lesson 4, please do the following practical exercises.

Exercise 4.1

The COUNTRIES table stores a list of COUNTRY_NAME values. You are required to calculate the average length of all the country names. Any fractional components must be rounded to the nearest whole number.


1. Start SQL developer and connect to the HR schema2. The length of the country name value for each row is calculated using the LENGTH

function. The average length may be determined using the AVG function. It may be rounded to the nearest whole number using the ROUND function.

3. The SELECT clause using alias AVERAGE_COUNTRY_NAME_LENGTH is: SELECT ROUND(AVG( LENGTH(COUNTRY_NAME))) AVERAGE_COUNTRY_NAME_LENGTH

4. The FROM clause is: FROM COUNTRIES5. Executing this statement returns a single row representing the average length of all the

country names in the COUNTRIES table.

Exercise 4.2

Analysis of staff turnover is a common reporting requirement. You are required to create a report containing the number of employees who left their jobs, grouped by the year in which they left. The jobs they performed are also required. The results must be sorted in descending order based on the number of employees in each group. The report must list the year, the JOB_ID, and the number of employees who left a particular job in that year.


1. Start SQL Developer and connect to the HR schema.2. The JOB_HISTORY table contains the END_DATE and JOB_ID columns, which

constitute the source data for this report.3. The year component is extracted using the TO_CHAR function. The number of

employees who quit a particular job in each year is obtained using the COUNT(*) function.

4. The SELECT clause is: TO_CHAR(END_DATE,‘YYYY’) “Quitting Year”, JOB_ID, COUNT(*) “Number of Employees”

5. The FROM clause is : FROM EMPLOYEES6. There is no WHERE clause.7. Since the report requires employees to be listed by year and JOB_ID, these two items

must appear in the GROUP BY clause, which is GROUP BY

TO_CHAR(END_DATE,‘YYYY’), JOB_ID8. The sorting is performed with ORDER BY COUNT(*) DESC9. Executing this statement returns the staff turnover report requested.

Exercise 4.3

The company is planning a recruitment drive and wants to identify the days of the week on which 20 or more staff members were hired. Your report must list the days and the number of employees hired on each of them.


1. Start Developer and connect to the HR schema.2. EMPLOYEES records must be divided into groups based on the day component of the

HIRE_DATE column. The number of employees per group may be obtained using the COUNT function.

3. The SELECT clause is: SELECT TO_CHAR(HIRE_DATE,'DAY'), COUNT(*)4. No WHERE clause is required since all the physical rows of the EMPLOYEES table are

considered.5. The GROUP BY clause is: GROUP BY TO_CHAR(HIRE_DATE,'DAY')6. This GROUP BY clause potentially creates seven group-level rows, one for each day of

the week.7. The COUNT function in the SELECT clause then lists the number of staff members

employed on each day. The HAVING clause must be used to restrict these seven rows to only those where the count is greater than or equal to 20.

8. The HAVING clause is: HAVING COUNT(*) >= 209. The FROM clause is: FROM EMPLOYEES10. Executing this statement returns the days of the week on which 20 or more employees

were hired.

lesson 5:Displaying data from multiple tables


Objectives

Write SELECT statements to access data from more than one table using equijoins and no equijoins Join a table to itself by using a self-join View data that generally does not meet a join condition by using outer join Generate a Cartesian product of all rows from two or more tables

Tables may be joined in several ways. The most common technique is called an equijoin. A row is associated with one or more rows in another table based on the equality of column values or

expressions. Tables may also be joined using a nonequijoin. In this case, a row is associated with one or more rows in another table if its column values fall into a range determined by inequality operators. A less common technique is to associate rows with other rows in the same table. This association is based on columns with logical and usually hierarchical relationships with each other. This is called a self-join. Rows with null or differing entries in common join columns are excluded when equijoins and nonequijoins are performed. An outer join is available to fetch these one-legged or orphaned rows if necessary.

A cross join or Cartesian product is formed when every row from one table is joined to all rows in another. This join is often the result of missing or inadequate join conditions but is occasionally intentional. To learn more about this, you are now requested to read Lesson 5 in Fundamentals 1. After reading through the lesson, do the following practical exercises.

Exercise 5.1

The JOB_HISTORY table shares three identically named columns with the EMPLOYEES table: EMPLOYEE_ID, JOB_ID, and DEPARTMENT_ID. You are required to describe the tables and fetch the EMPLOYEE_ID, JOB_ID, DEPARTMENT_ID, LAST_NAME, HIRE_DATE, and END_DATE values for all rows retrieved using a pure natural join. Alias the EMPLOYEES table as EMP and the JOB_HISTORY table as JH and use dot notation where possible.


1. Start SQL developer and connect to the HR schema.2. The tables are described using the commands DESC EMPLOYEES and DESC

JOB_HISTORY, and the columns with identical names and their data types may be examined.

3. The SELECT clause is SELECT EMPLOYEE_ID, JOB_ID, DEPARTMENT_ID, EMP.LAST_NAME, EMP.HIRE_DATE, JH.END_DATE

4. The FROM clause is FROM JOB_HISTORY JH5. The JOIN clause is NATURAL JOIN EMPLOYEES EMP6. Executing this statement returns a single row with the same EMPLOYEE_ID, JOB_ID,

and DEPARTMENT_ID values in both tables.

Exercise 5.2

Each record in the DEPARTMENTS table has a MANAGER_ID column matching an EMPLOYEE_ID value in the EMPLOYEES table. You are required to produce a report with one column aliased as Managers. Each row must contain a sentence of the format FIRST_NAME LAST_NAME is manager of the DEPARTMENT_NAME department. Alias the EMPLOYEES table as E and the DEPARTMENTS table as D and use dot notation where possible.


1. Start SQL Developer and connect to the HR schema.2. The Managers column may be constructed by concatenating the required items and

separating them with spaces.3. The SELECT clause is SELECT E.FIRST_NAME||' '||E.LAST_NAME||' is manager of

the '||D.DEPARTMENT_NAME||' department.' "Managers"4. The FROM clause is FROM EMPLOYEES E5. The JOIN…ON clause is JOIN DEPARTMENTS D ON (E.

EMPLOYEE_ID=D.MANAGER_ID).6. Executing this statement returns 11 rows describing the managers of each department.

Exercise 5.3

There is a hierarchical relationship between employees and their managers. For each row in the EMPLOYEES table the MANAGER_ID column stores the EMPLOYEE_ID of every employee’s manager. Using a self-join on the EMPLOYEES table, you are required to retrieve the employee’s LAST_NAME, EMPLOYEE_ID, manager’s LAST_NAME, and employee’s DEPARTMENT_ID for the rows with DEPARMENT_ID values of 10, 20, or 30. Alias the EMPLOYEES table as E and the second instance of the EMPLOYEES table as M. Sort the results based on the DEPARTMENT_ID column.


1. Start SQL Developer and connect to the HR schema.2. The SELECT clause is SELECT E.LAST_NAME EMPLOYEE, E.EMPLOYEE_ID,

E.MANAGER_ID, M.LAST_NAME MANAGER, E.DEPARTMENT_ID.3. The FROM clause with source table and alias is FROM EMPLOYEES E.4. The JOIN…ON clause with aliased target table is JOIN EMPLOYEES M ON

(E.MANAGER_ID=M.EMPLOYEE_ID).5. The WHERE clause is WHERE E.DEPARTMENT_ID IN (10,20,30).6. The ORDER BY clause is ORDER BY E.DEPARTMENT_ID.

Exercise 5.4

The DEPARTMENTS table contains details of all departments in the organization. You are required to retrieve the DEPARTMENT_NAME and DEPARTMENT_ID values for those departments to which no employees are currently assigned.


1. Start SQL developer and connect to the HR schema.2. The SELECT clause is SELECT D.DEPARTMENT_NAME, D.DEPARTMENT_ID.3. The FROM clause with source table and alias is FROM DEPARTMENTS D.4. The LEFT OUTER JOIN clause with aliased target table is LEFT OUTER JOIN

EMPLOYEES E ON E.DEPARTMENT_ID=D.DEPARTMENT_ID.5. The WHERE clause is WHERE E.DEPARTMENT_ID IS NULL.

6. Execute the statements.

Exercise 5.5

You are required to obtain the number of rows in the EMPLOYEES and DEPARTMENTS table as well as the number of records that would be created by a Cartesian product of these two tables. Confirm your results by explicitly counting and multiplying the number of rows present in each of these tables.


1. Start SQL developer and connect to the HR schema.2. The SELECT clause to find the number of rows in the Cartesian product is: SELECT

COUNT(*).3. The FROM clause is: FROM EMPLOYEES.4. The Cartesian product is performed using CROSS JOIN DEPARTMENTS.5. Explicit counts of the rows present in the source tables are performed using: SELECT

COUNT(*) FROM EMPLOYEES; SELECT COUNT(*) FROM DEPARTMENTS;6. Explicit multiplication of the values resulting from the previous queries may be

performed by querying the DUAL table.7. Execute these statements.

Lesson 6: Using subqueries to solve queries


Objectives Define subqueries Describe the types of problems that subqueries can solve List the types of subqueries Write single row and multiple row subqueries

A subquery is a query that is nested inside a SELECT, INSERT, UPDATE, or DELETE statement or inside another subquery. A subquery can return a set of rows or just one row to its parent query. A scalar subquery is a query that returns exactly one value: a single row, with a single column. Scalar subqueries can be used in most places in a SQL statement where you could use an expression or a literal value. To learn more about subqueries, you now requested to go and read lesson 6 in Fundamentals 1. After this lesson, attempt to do the practical exercises below.

Exercise 6.1

In this exercise, you will write code that demonstrates the places where subqueries can be used.

All the queries should be run when connected to the HR schema.

1. 1. Log on to your database as user HR.

2. 2. Write a query that uses subqueries in the column projection list. The query will report on the current numbers of departments and staff:

select sysdate Today,select count(*) from departments) Dept_count,select count(*) from employees) Emp_countfrom dual;

3. 3. Write a query to identify all the employees who are managers. This will require using a subquery in the WHERE clause to select all the employees whose EMPLOYEE_ID appears as a MANAGER_ID:

select last_name from employees where(employee_id in (select manager_id from employees));

4. 4. Write a query to identify the highest salary paid in each country. This will require using a subquery in the FROM clause:

select max(salary),country_id from(select salary,department_id,location_id,country_id fromemployees natural join departments natural join locations)group by country_id;

Exercise 6.2

In this exercise, you will write some more complicated subqueries. Use SQL Developer. All the queries should be run when connected to the HR schema.

1. Log on to your database as user HR.

2. Write a query that will identify all employees who work in departments located in the United Kingdom. This will require three levels of nested

subqueries:select last_name from employees where department_id in(select department_id from departmentswhere location_id in(select location_id from locationswhere country_id =(select country_id from countrieswhere country_name=’United Kingdom’))

);

3. Check that the result from step 2 is correct by running the subqueries independently.

First, find the COUNTRY_ID for the United Kingdom:

select country_id from countries where country_name=’United Kingdom’;

The result will be UK. Then find the corresponding locations:

select location_id from locations where country_id = ’UK’;

The LOCATION_IDs returned will be 2400, 2500, and 2600. Then find the

DEPARTMENT_IDs of department in these locations:

select department_id from departments where location_id in (2400,2500,2600)

The result will be two departments, 40 and 80. Finally, find the relevant employees:

select last_name from employees where department_id in (40,80);

4. Write a query to identify all the employees who earn more than the average and who work in any of the IT departments. This will require two subqueries, not nested:

select last_name from employeeswhere department_id in(select department_id from departments where department_name like ’IT%’)and salary > (select avg(salary) from employees);

Exercise 6.3

In this exercise, you will demonstrate problems that can occur with different types of subqueries. Use SQL Developer. All the queries should be run when connected to the HR schema: it is assumed that the EMPLOYEES table has the standard sets of rows.

1. Log on to your database as user HR.

2. Write a query to determine who earns more than Mr. Tobias:

select last_name from employees where

salary > (select salary from employees where last_name=’Tobias’)

order by last_name;

This will return 86 names, in alphabetical order.

3. Write a query to determine who earns more than Mr. Taylor:


salary > (select salary from employees where last_name=’Taylor’)

order by last_name;

This will fail with the error “ORA-01427: single-row subquery returns more than one row.” The following illustration shows the last few lines of the output from step 2 followed by step 3 and the error.

4. Determine why the query in step 2 succeeded but failed in step 3. The answer lies in the state of the data:

select count(last_name) from employees where last_name=’Tobias’;

select count(last_name) from employees where last_name=’Taylor’;

The use of the “greater than” operator in the queries for steps 2 and 3 requires a single-row subquery, but the subquery used may return any number of rows, depending on the search predicate used.

5. Fix the code in steps 2 and 3 so that the statements will succeed no matter what LAST_NAME is used. There are two possible solutions: one uses a different comparison operator that can handle a multiple-row subquery; the other uses a subquery that will always be single-row.

The first solution:


salary > all (select salary from employees where last_name=’Taylor’)

order by last_name;

The second solution:


salary > (select max(salary) from employees where last_name=’Taylor’)

order by last_name;

Lesson 7: Objectives


Objectives

After completing this lesson you should be able to do the following:

Describe set operators Use a set operator to combine multiple queries into a single query Control the order of rows returned

Any SQL query can be described mathematically (relational algebra is the term used for this), but the use of set operators is what recalls school mathematics to many programmers. As your math curriculum may not have included set theory, following is a short summary.

All SELECT statements return a set of rows. The set operators take as their input the results of two or more SELECT statements and from these generate a single result set. This is known as a compound query.

Oracle provides three set operators:

UNION, INTERSECT, and MINUS. UNION can be qualified with ALL. There is a significant deviation from the ISO standard for SQL here, in that ISO SQL uses EXCEPT where Oracle uses MINUS, but the functionality is identical. The difference can be important when porting applications (or skills) developed for a third-party database to the Oracle environment. To learn more about set operators, you now requested to go and read lesson 7 in Fundamentals 1. After this lesson, attempt to do the practical exercises below.

Exercise 7.1

In this exercise, you will see the effect of the set operators. SQL Developer can be used.

1. Connect to your database as user HR.

2. Run this query:select region_name from regions;

Note the result, in particular the order of the rows. If the table is as originally created, there will be four rows returned. The order will be Europe, America, Asia, Middle East.

3. Query the Regions table twice, using UNION:select region_name from regions union select region_name fromregions;

The rows returned will be as for step 1 but sorted alphabetically.

4. This time, use UNION ALL:select region_name from regions union all select region_namefrom regions;

There will be double the number of rows, and they will not be sorted.

5. An intersection will retrieve rows common to two queries:select region_name from regions intersect select region_namefrom regions;

All four rows are common, and the result is sorted.

6. A MINUS will remove common rows:

select region_name from regions minus select region_name from

regions;

The second query will remove all the rows in the first query. Result: no rows left.

Exercise 7.2

In this exercise, you will run more complex compound queries.


2. Run this query to count the employees in three departments:select department_id, count(1) from employeeswhere department_id in (20,30,40)group by department_id;

3. Obtain the same result with a compound query:select 20,count(1) from employees where department_id=20union allselect 30,count(1) from employees where department_id=30union allselect 40,count(1) from employees where department_id=40;

4. Find out if any managers manage staff in both departments 20 and 30, and exclude any managers with staff in department 40:

select manager_id from employees where department_id=20intersectselect manager_id from employees where department_id=30

minusselect manager_id from employees where department_id=40;

5. Use a compound query to report salaries subtotaled by department, by manager, and the overall total:

select department_id,to_number(null),sum(salary) from employeesgroup by department_idunionselect to_number(null),manager_id,sum(salary) from employeesgroup by manager_idunion allselect to_number(null),to_number(null),sum(salary) fromemployees;

Exercise 7.3

In this exercise, you will tidy up the result of the final step in Exercise 2. This step produced a listing of salaries totalled by department and then by manager, but the results were not very well formatted.


2. Generate better column headings for the query:select department_id dept,to_number(null) mgr,sum(salary)from employeesgroup by department_idunion allselect to_number(null),manager_id,sum(salary) from employeesgroup by manager_idunion allselect to_number(null),to_number(null),sum(salary) fromemployees;

3. Attempt to sort the results of the queries that subtotal by using UNION instead of UNION ALL:

select department_id dept,to_number(null) mgr,sum(salary)from employeesgroup by department_idunionselect to_number(null),manager_id,sum(salary) from employeesgroup by manager_idunion allselect to_number(null),to_number(null),sum(salary) fromemployees;

This would be fine, except that the subtotals for staff without a department or a manager are

placed at the bottom of the output above the grand total, not within the sections for departments and managers.

4. Generate a value to replace the NULLs for department and manager codes and for the overall total:

select 20,count(1) from employees where department_id=20union allselect 30,count(1) from employees where department_id=30union allselect 40,count(1) from employees where department_id=40;

5. Find out if any managers manage staff in both departments 20 and 30, and exclude any managers with staff in department 40:

select manager_id from employees where department_id=20intersectselect manager_id from employees where department_id=30minusselect manager_id from employees where department_id=40;

6. Use a compound query to report salaries subtotaled by department, by manager, and the overall total:

select department_id dept,to_number(null) mgr,sum(salary)from employeesgroup by department_idunionselect to_number(null),manager_id,sum(salary) from employeesgroup by manager_idunion allselect to_number(null),to_number(null),sum(salary) from

Lesson 8: Manipulating Data


Objectives

After completing this lesson, you should be able to do the following: Describe each data manipulation language(DML) statement Insert rows into a table Update rows in a table Delete rows in a table Control transactions

This chapter is all about the commands that change data. Syntactically, the DML commands are

much simpler than the SELECT command that has been studied so far. However, there is a large amount of relational database theory that comes along with DML. As well as understanding the commands that make the changes, it is essential to understand the theory of transaction management, which is part of the relational database paradigm. The transaction management mechanisms provided by the Oracle database guarantee conformance to the relational standards for transactions: they implement what is commonly referred to as the ACID (atomicity, consistency, isolation, and durability) test.

Following detail on the DML commands, this chapter includes a full treatment of the necessary theory: transactional integrity, record locking, and read consistency.

Strictly speaking, there are five DML commands: SELECT INSERT UPDATE DELETE MERGE

To learn more about Data Manipulation Language, you now requested to go and read lesson 8 in Fundamentals 1. After this lesson, attempt to do the practical exercises below.

Exercise 8.1

In this exercise, use various techniques to insert rows into a table.

1. Connect to the HR schema, with SQL Developer.

2. Query the REGIONS table, to check what values are already in use for the

REGION_ID column:


This exercise assumes that values above 100 are not in use. If they are, adjust the values suggested below to avoid primary key conflicts.

3. Insert a row into the REGIONS table, providing the values in line:insert into regions values (101,'Great Britain');

4. Insert a row into the REGIONS table, providing the values as substitution variables:

insert into regions values (&Region_number,'&Region_name');

When prompted, give the values 102 for the number, Australasia for the name. Note the use of quotes around the string.

5. Insert a row into the REGIONS table, calculating the REGION_ID to be one higher than the current high value. This will need a scalar subquery:

insert into regions values ((select max(region_id)+1from regions), 'Oceania');

6. Confirm the insertion of the rows:select * from regions;

7. Commit the insertions:commit;

Exercise 8.2

In this exercise, use various techniques to update rows in a table.

1. Connect to the HR schema using SQL Developer.

2. Update a single row, identified by primary key:

update regions set region_name='Scandinavia' where region_id=101;

This statement should return the message “1 row updated.”

3. Update a set of rows, using a nonequality predicate:

update regions set region_name='Iberia' where region_id > 100;

This statement should return the message “3 rows updated.”

4. Update a set of rows, using subqueries to select the rows and to provide values:update regionsset region_id=(region_id+(select max(region_id) from regions))where region_id in (select region_id from regions whereregion_id > 100);

This statement should return the message “3 rows updated.”

5. Confirm the state of the rows:


6. Commit the changes made:

commit;

Exercise 8.3

In this exercise, use various techniques to delete rows in a table.

1. Connect to the HR schema using SQL Developer.

2. Remove one row, using the equality predicate on the primary key:

delete from regions where region_id=204;

This should return the message “1 row deleted.”

3. Attempt to remove every row in the table by omitting a WHERE clause:

delete from regions

This will fail, due to a constraint violation.

4. Remove rows with the row selection based on a subquery:delete from regions whereregion_id in (select region_id from regions where region_name='Iberia');

This will return the message “2 rows deleted.”

5. Confirm that the REGIONS table now contains just the original four rows:


6. Commit the deletions:

Commit

Lesson 9: Using DDL statements to create and manage tables


Objectives

After completing this lesson, you should be able to do the following: Categorize the main database objects Review the table structure

List the data types that are available for columns Create a simple table Explain how constraints are created at the time of table creation Describe how schema objects work

There are several types of data objects in a database that can be addressed by users with SQL. The most commonly used type of object is the table. Tables come in various forms, but SQL is not aware of this. A table may also be associated with other objects such as indexes or LOBs (a large object—a structure designed for storing big items of information, such as video recordings) that are addressed implicitly. The statement will address only the table with which the other objects are associated. This chapter details table creation; Chapter 12 will cover some other object types.

When creating a table, there are certain rules that must be followed regarding the table’s structure: its columns may only be of certain data types. There are also rules that can be defined for the individual rows; these are known as constraints. The structural rules and the constraint rules together restrict the data that can be inserted into the table. To learn more about Using DDL Statements to Create and Manage Tables, you now requested to go and read lesson 9 in Fundamentals 1. After this lesson, attempt to do the practical exercises below.

Exercise 9.1

In this exercise, query various data dictionary views as user HR to determine what objects are in the HR schema and what objects are in other schemas HR has access to.

1. Connect to the database with SQL Developer as user HR.

2. Determine how many objects of each type are in the HR schema:

select object_type,count(*) from user_objects group by object_type;

The USER_OBJECTS view lists all objects owned by the schema to which the current session is connected, in this case HR.

3. Determine how many objects in total HR has permissions on:

select object_type,count(*) from all_objects group by object_type;

The ALL_OBJECTS view lists all objects to which the user has some sort of access.

4. Determine who owns the objects HR can see:

select distinct owner from all_objects;

Exercise 9.2

In this exercise, query various data dictionary views as user HR to determine the structure of a table.

1. Connect to the database with SQL Developer as user HR.

2. Determine the names and types of tables that exist in the HR schema:select table_name,cluster_name,iot_type from user_tables;

Clustered tables and index organized tables (IOTs) are advanced table structures. In the HR schema, all tables are standard heap tables except for COUNTRIES which is an IOT.

3. Use the DESCRIBE command to display the structure of a table

describe regions;

4. Retrieve similar information by querying a data dictionary view:select column_name,data_type,nullable from user_tab_columnswhere table_name='REGIONS';

Exercise 9.3

In this exercise, find out what data types are used in the tables in the HR schema, using two techniques.

1. Connect to the database as user HR with SQL Developer.

2. Use the DESCRIBE command to show the data types in some tables:describe employees;describe departments;

3. Use a query against a data dictionary view to show what columns make up the EMPLOYEES table, as the DESCRIBE command would:

select column_name,data_type,nullable,data_length,data_precision,data_scale from user_tab_columns where table_name='EMPLOYEES';

The view USER_TAB_COLUMNS shows the detail of every column in every table in the current user’s schema.

Exercise 9.4

Use SQL Developer to create tables, add constraints, and demonstrate their use.

1. Connect to the database as user HR.

2. Create a table EMP as a copy of some columns from EMPLOYEES:create table emp asselect employee_id empno, last_name ename, department_id deptnofrom employees;

3. Create a table DEPT as a copy of some columns from DEPARTMENTS:create table dept asselect department_id deptno, department_name dname from departments;

4. Use DESCRIBE to describe the structure of the new tables. Note that the not null constraint on ENAME and DNAME has been carried over from the source tables.

5. Add a primary key constraint to EMP and to DEPT and a foreign key constraint linking the tables:

alter table emp add constraint emp_pk primary key (empno);alter table dept add constraint dept_pk primary key (deptno);alter table emp add constraintdept_fk foreign key (deptno) references dept on delete set null;

The preceding last constraint does not specify which column of DEPT to reference; this will default to the primary key column.

6. Demonstrate the effectiveness of the constraints by trying to insert data that will violate them:

insert into dept values(10,'New Department');insert into emp values(9999,'New emp',99);truncate table dept;

Lesson 10: Creating other schema objects


Objectives

After completing this lesson, you should be able to do the following: Create a simple and complex views Retrieve data from views Create, maintain and use sequences Create and maintain indexes Create private and public synonyms

So far, most of the SQL statements used in this book have addressed tables. This seems reasonable: SQL commands interact with data, and data is stored in tables. In practice, most users never issue SQL that addresses tables: they issue SQL that addresses views or synonyms.

Views and synonyms do not themselves store data; they provide a layer of abstraction between the users and the data. As tables, views, and synonyms share the same namespace, users need never be aware of which they are addressing.

Sequences are a mechanism for issuing unique numbers. In many databases, primary key values are defined as a unique number. A sequence can issue such numbers on demand, without programmers needing to worry about whether they really are unique. The structure of a sequence means that hundreds of numbers can, if necessary, be issued every second without any performance issues. Index management is sometimes said to lie in the database administration domain, not the SQL developer’s domain. For this reason, the treatment here is very brief. But whether or not index management is part of the developer’s job, all developers must understand the purpose and design of indexes. In many cases the indexing strategy is crucial to adequate performance: get it right, and the benefits will be enormous; get it wrong, and the results may be disastrous. Use of indexes is transparent, but if the programmers are aware of them they can ensure that their code will be able to make the best use of what indexes are available and advise on what indexes should be created.

Exercise 10.1

In this exercise, you will create some simple and complex views, using data in the HR schema. SQL developer can be used.


2. Create views on the EMPLOYEES and DEPARTMENT tables that remove all personal information. Run the two pieces of code separately.

create view emp_anon_v asselect hire_date, job_id,salary,commission_pct,department_id fromemployees;

create view dept_anon_v asselect department_id,department_name,location_id from departments;

3. Create a complex view that will join and aggregate the two simple views. Note that there is no reason not to have views of views.

create view dep_sum_v as

select e.department_id,count(1) staff, sum(e.salary) salaries,

d.department_name from emp_anon_v e join dept_anon_v d

on e.department_id=d.department_id

group by e.department_id,d.department_name;

4. Confirm that the view works by querying it

Exercise 10.2

In this exercise, you will use the views created in Exercise 10.1 for SELECT and DML statements. SQL developer can be used.


2. Insert a new department through the DEPT_ANON_V view and attempt to insert an employee through EMP_ANON_V:

insert into DEPT_ANON_V values(99,'Temp Dept',1800);insert into emp_anon_v values(sysdate,'AC_MGR',10000,0,99);

The insert into EMP_ANON_V will fail because of missing mandatory values. You will, however, be able to do this update through it:

update emp_anon_v set salary=salary*1.1;

Then roll back the changes:rollback;

3. Find out the average salary of the department with the highest average salary, by querying the EMPLOYEES table:

select max(avg_sal) from(select AVG(SALARY) avg_sal from employees group by department_id);

and find the same information from the DEP_SUM_V view, which is a much simpler query:

select max(salaries / staff) from dep_sum_v;

Exercise 10.3

Do questions 1, 2, and 3 in fundamentals 1 lesson 10. Discuss your solutions with the rest of the class on discussion forum on myUnisa.

Lesson 11: Managing Objects with Data Dictionary Views


ObjectiveAfter completing this lesson, you should be able to do the following: Use data dictionary views to research data on your objects Query various data dictionary views

This lesson describes a valuable tool that all Oracle professionals should understand. The data dictionary is Oracle’s built-in real-time reference source for all information about the applications that you build in your database. Armed with the full capabilities of the data dictionary, you can obtain information on the database objects that have been created within your database, who created them, when, and much more.

The data dictionary is a collection of database tables and views. It is automatically built and populated by the Oracle database. The information stored in the data dictionary includes the full description of all the database objects you create as part of your application: tables, views, indexes, constraints, synonyms, sequences, and more. In other words, the result of each DDL statement you’ve studied in this book is recorded in the dictionary, and the information is automatically maintained by the Oracle system in real time as you change database objects and their structures. To learn more about Data Dictionary, read lesson 11 in Fundamentals 1 and do the following exercises.

Exercise 11.1

In practice 11 do questions 1 and 2. Discuss your solutions on the discussion forum.

Lesson 12: Controlling User Access


Objectives

After completing this lesson, you should be able to do the following: Differentiate system privilege Grant privileges on tables View privileges in the data dictionary Grant roles Distinguish between privileges and roles

This chapter explores the subject of user access and the privileges associated with performing actions in the database. Every action performed by any user account requires a corresponding privilege or set of privileges to perform that action. There are two categories of privileges—system privileges are required to perform a task in the database; object privileges are required to use those system privileges on any given database object in particular. Privileges may be granted to a user account, or to another database object called a role. A role, in turn, can be granted to a user account, which effectively grants the set of privileges collected within the role. Once granted, privileges and roles may later be revoked. Together, privileges and roles are the mechanism for managing and controlling access to the database by user accounts. This chapter looks at how to create and manage privileges and roles.

A word of warning about the sample code contained in this chapter: Some of it has the ability to

change your database permanently with results that may be undesirable. Some of our code samples will look at SQL code that uses the SYSTEM user account, a very important account that should be controlled by experienced database administrators in any production database. You should always check with your DBA before trying any code samples from any book, but this chapter in particular includes code that you should not execute in a professional installation without first checking with your DBA.

To learn more about Controlling User Access, read lesson 1 in Fundamentals 2 and do the following exercises.

Exercise 12.1

In Fundamentals 11, lesson 1, practice 1, do questions 1 to 10. Discuss your solutions on the discussion forum.

Lesson 13: Managing Schema Objects


Objectives

After completing this lesson, you should be able to do the following: Add constraints Create index Create indexes using the CREATE TABLE statement Creating function based indexes Drop columns and set column UNUSED Perform FLASHBACK operations Create and use external tables

This lesson looks at a variety of tasks, features, and SQL statements, all designed to support the management of database objects and the data contained within them using constraints and indexes in new ways, in new statements and new clauses and in new combinations. This chapter also introduces new features like FLASHBACK and external tables.

To learn more on Managing Schema Objects, read lesson 2 and do the following exercises.

Exercise 13.1

In Fundamentals 11, lesson 2, practice 2, do questions 1, 2, 3, and 4. Discuss your solutions on the discussion forum.

Lesson 14: Generating Reports by Grouping Data


Objectives

After completing this lesson, you should be able to do the following: Use the ROLLUP operation to produce subtotal values Use the CUBE operation to produce cross-tabulation values Use the GROUPING function to identify the row values created by ROLLUP or

CUBE Use GROUPING SETS to produce a single results set

This chapter is about a number of extensions to the capabilities of the GROUP BY clause of the SELECT statement, CUBE, ROLLUP, and GROUPING SETS. We’ll also examine a function named GROUPING that supports the GROUP BY operations discussed in this chapter. Each is addressed separately in the pages that follow, starting with ROLLUP.

To learn more about Generating Reports by Grouping Data, read lesson 4 and do the following exercises.

Exercise 14.1

In Fundamentals 11, lesson 4, practice 4, do questions 1, 2, 3, and 4. Discuss your solutions on the discussion forum.

Lesson 15: Regular Expression Support


Objectives

After completing this lesson, you should be able to use regular expression support in SQL to: Search Match replace strings

in terms of regular expressions.

This chapter explores the use of regular expressions in Oracle SQL. The term “regular expressions” refers to a specific language for working with text. You can use regular expressions to specify some complex character patterns and perform complex searches in a succinct and efficient manner.

Regular expressions have an identity that goes beyond SQL and Oracle. They are used in Unix,

C, and elsewhere. This chapter will look at the functionality of regular expressions and Oracle SQL’s capabilities for working with them.

To learn more about Regular Expression Support, read lesson 8 and do the following exercises.

Exercise 15.1

In Fundamentals 11, lesson 8, practice 8, do questions 1, 2, and 3. Discuss your solutions on the discussion forum.

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