1. MS SQL Server 2005
Lab # 4 :
 Practicing Queries, part #2
Complex Queries
Adavnced Database Programming 1

2. 2. Complex Queries
 WHERE Clause Form:
USE database_name
SELECT Column_List
[ INTO New_table ]
From table1[table2,….]
[ WHERE conditions ]
[ GROUP BY group_by_expression ]
[ HAVING search_conditions ]
[ ORDER BY order_expression[ASC|DESC] ]
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3. Subqueries
 An Outer query contains an Inner query,
which can be included in the WHERE Clause
or in the FROM Clause
 Outer Query:
SELECT
FROM
WHERE (Inner Query)
 Outer Query:
SELECT
FROM (Inner Query)
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4. Subqueries And Comparison Operators
 Example :
Q: View all Employees First & Last name who work in
Innovation Department
A:
Use Company
Select FName, LName
From Employee
where DNO=( Select DNumber
From Department
Where DName=‘Innovation’)
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5. Subqueries And Comparison Operators
 Example :
Q: View all Employees Last name who work on project Building
A:
Use Company
SelectLName
From Employee
where SSN IN( Select ESSN
From Works_On
Where PNO IN(SELECT PNumber
From Project
WHERE PName=‘Building’))
 Can we Substitute the IN clause with = ?
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6. Subqueries (Inner Query included in FROM
Clause)
 Example :
Q: View all Employees Last name who their SSN
is greater than 1000
A:
Use Company
Select LName Should I give
the inner query
From (Select * result a name?
From Employee
Where SSN>1000) AS NewEmp
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7. Subqueries ( ANY & ALL Operators )
 ANY: Returns true if the result of an inner query
contains at least one row that satisfies the
comparison
 ALL: Evaluates true if the evaluation of the table
column in the inner query returns all values of that
column
 Syntax :
Outer_query column operator [ANY | ALL] Inner_query
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8. Subqueries ( ANY & ALL Operators )
 Example :
Q: View the oldest Employees SSN who their
BDate is less than all other employees
A:
Use Company
Select SSN
From Employee
where Bdate>=ALL (Select BDate
From Employee)
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9. EXIST Function
 Checks the inner query of a subquery and
evaluates to true if its result contains at least
one row
 Syntax:
[NOT] EXISTS (query)
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10. EXIST Function
Q: View all employees info only if there is an
employee who’s salary is 5000
A:
Use Company
Select *
From Employee
where Exists (Select *
From Employee
Where salary=5000)
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11. GROUP BY Clause
 Defines one or more columns as a group
such that all rows within any group have the
same values for those columns
 ALL columns specified in the SELECT clause
must appear in the GROUP BY clause
 If an aggregate function appears in the
SELECT clause, then you can’t include a
simple column in it unless if its used in the
Group By clause
Adavnced Database Programming 11

12. GROUP BY Clause
Q: For each Project View all employees SSN
with hours working on more than 20 hours
A:
Use Company
Select PNO,Hours,ESSN
From Works_On
where hours>20
Group by PNO,Hours,ESSN
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13. GROUP BY Clause
 Example :
Q: How many Employees works in each
project?
A:
Use Company
Select PNumber ,COUNT(*) emp_count
From Works_IN
Group by PNumber
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14. HAVING Clause
 Defines conditions that is then applied to groups of
rows (Such as WHERE Clause condition applied on
rows) and its Syntax: HAVING condition
Q: Get Project numbers for all projects employing less
than four persons
A:
Use Company
Select PNO
From Works_On
Group by PNO
Having Count(*)<4
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15. HAVING Clause
 Example :
Q: View Projects names that their name starts
with S
A:
Use Company
Select PName
From Project
Group by PName
HAVING PName LIKE ‘S%’
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16. ORDER BY Clause
 Defines the order of the rows in the result and its
Syntax:
ORDER BY [col_name | col_no [ASC | DESC]] , …
Q: Get employees all info ordered by their SSN
A:
Use Company
Select *
From Employee
Order BY SSN
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17. ORDER BY Clause
Q: Get employees first & last names ordered by
their last name then their first name
A:
Use Company
Select FName, LName
From Employee
Order BY LName, FName
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18. ORDER BY Clause
Q: Get employees first & last names ordered by
their last name Ascending and then their first
name Descending
A:
Use Company
Select FName, LName
From Employee
Order BY 2 Asc, 1 DESC
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19. INTO Clause
It performs two parts:
1. Create a new table with columns as same the
columns specified in the select clause
2. Insert the rows that matches the query into the
new table
Basic INTO form:
Use database_name
Select columns_List
INTO NewTableName
From table(s)
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20. INTO Clause
Q: Create a separate table for all employees
who aren’t supervised
A:
Use Company
Select *
INTO Head_Supervisors
From Employee
Where MGRSSN IS NULL
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21. Set Operators
 UPDATE table_name SET column_name =
newValue
WHERE condition
 EX.
UPDATE employee SET Salary=32400
WHERE SSN=2341
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22. Set Operators
 Union
is the set of all elements appearing in either
or both of tables
Form: Query_1 UNION [ALL]Query_2 …
ALL option includes all resulting rows
including duplicates are to be displayed
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23. Set Operators ( UNION)
Q: View all employees numbers who either belong to department 5
or work on their project more than 20 hours in ascending order of
employee number
A:
Use Company
Select SSN
From Employee
Where DNO=5
UNION
Select ESSN
From Works_On
Where Hours>20
Order By 1
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24. Set Operators
 INTERSECTION
is the set of all elements belongs to both of
the tables
Form: Query_1 INTERSECT Query_2 …
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25. Set Operators ( INTERSECT)
Q: View all employees numbers who belong to department 5 and
work on their project more than 20 hours in ascending order of
employee number
A:
Use Company
Select SSN
From Employee
Where DNO=5
INTERSECT
Select ESSN
From Works_On
Where Hours>20
Order By 1
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26. Set Operators
 DIFFERENCE
is the set of all elements belongs to the first
table but doesn’t belong to the second table
Form: Query_1 EXCEPT Query_2 …
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27. Set Operators (EXCEPT)
Q: View all employees numbers who belongs to department 5
except those who work on their project more than 20 hours in
ascending order of employee number
A:
Use Company
Select SSN
From Employee
Where DNO=5
EXCEPT
Select ESSN
From Works_On
Where Hours>20
Order By 1
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28. CASE Expression
 CASE Basic form:
CASE expression_1
{WHEN exp_2 THEN result_1}
{ELSE result_n}
END
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29. CASE Expression
Q: View all employees SSN,FNAME,Lname and Grade as following:
Salary  Employee Grade:
0-6000  1 , 6000-15000  2 , 15000-22000  3 ,
>22000 4
A:
Use Company
Select SSN,FNAME,LNAME,
CASE
when Salary>0 and Salary<=6000 Then 1
when Salary>6000 and Salary<=15000 Then 2
when Salary>15000 and Salary<=22000 Then 3
ELSE 4
END Employee_Grade Column name after END
From Employee
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30. COMPUTE Clause
 Uses aggregate function(Min, Max,…) to calculate
summary values that appear as additional rows in
the result of query
Basic COMPUTE form:
Use database_name
Select columns_List
From table(s)
Where condition
COMPUTE aggre_Func(column_name)
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31. COMPUTE Clause
Q: View all employees numbers, project their working
on, and working hours for employees work on
project 3 or 10. And view the minimum working
hours
A:
Use Company
Select *
From Works_on
Where PNO=3 OR PNO=10
Compute MIN(Hours)
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32. COMPUTE Clause (Using BY option)
Q: View all employees numbers, project their working
on, and working hours for employees work on
project 3 or 10. And view the minimum working
hours for each project
A:
Use Company
Select *
From Works_on
Where PNO=3 OR PNO=10
Order By PNO
Compute MIN(Hours) BY PNO
Adavnced Database Programming 32

33. COMPUTE Clause(Multiple aggregate
function)
Q: View all employees numbers, project their working on,
and working hours for employees work on project 3 or
10. And view the minimum, maximum, avrage working
hours
A:
Use Company
Select *
From Works_on
Where PNO=3 OR PNO=10
Compute MIN(Hours), MAX(Hours), AVG(Hours)
Adavnced Database Programming 33

34. COMPUTE Clause
 SELCET INTO is not allowed (because the
result of the COMPUTE clause is not a
table)
 All columns in the COMPUTE clause must
appear in the SELECT list
 The name of each column in the COMPUTE
BY clause must appear in the ORDER BY
clause
 The order of the columns in the COMPUTE
BY & ORDER BY clauses must be identical
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