1. SQL: Structured Query Language
Part -2
Instructor: Mohamed Eltabakh

2. Aggregation + GroupBy

3. Restrictions of GROUP BY
If you group by A1, A2, …An, then any other column projected in SELECT clause must be inside an aggregation function
SELECT pNumber, address, count(sName) AS CNT, sum(sNumber) AS SUM
FROM Student
WHERE sNumber > 1
GROUP BY pNumber, address;
SELECT pNumber, address, sName, sum(sNumber) AS SUM
FROM Student
WHERE sNumber > 1
GROUP BY pNumber, address; X
SELECT pNumber, count(sName) AS CNT, sum(sNumber) AS SUM
FROM Student
WHERE sNumber > 1
GROUP BY pNumber, address;

4. HAVING Clause: Putting Condition on Groups
How to add conditions on each group?
Select only the groups where the COUNT > 5
These conditions are after you build the groups (not before)
Remember: WHERE conditions are executed before the groups are formed
New optional clause called “HAVING”, added after the GROUP BY clause
SELECT pNumber, COUNT (sName)
FROM Student
GROUP BY pNumber
HAVING SUM(sNumber) > 2;
Can reference aggregation inside HAVING

5. HAVING Clause: Example
Student
sNumber
sName
address
pNumber 1
Dave
320FL 2
Greg 3
Matt 4
Jan
500MA
 (SUM> 3) (pNumber,address, CNT  count(sName), SUM  sum(sNumber) ( (sNumber > 1) (Student)))
SELECT pNumber,address, count(sName) AS CNT, sum(sNumber) AS SUM
FROM Student
WHERE sNumber > 1
GROUP BY pNumber, address
HAVING sum(sNumber) > 3;
Applied before the grouping
Applied after the grouping on each group
pNumber
address
CNT
SUM 2
500MA 1 4

6. Example Queries
Find customers having loans with sum > 20,000. Report the customer names and the total loans amount
SELECT customer_name, sum(amount) AS SUM
FROM Loan L, Borrower B
WHERE L.loan_number = B. loan_number
GROUP BY customer_name
HAVING sum(amount) > 20,000;

7. Example Queries Find the cities with more than 3 branches
SELECT branch_city
FROM Branch
GROUP BY branch_city
HAVING count(branch_name) > 3;

8. Example Queries
Find the customer names who have accounts with balance > 100,000 in more than two branches. Report the customer names & number of branches
SELECT customer_name, count(branch_name) As CNT
FROM Account A, Depositor D
WHERE A.account_number = D.account_number
AND A.balance > 100,000
GROUP BY customer_name
HAVING count(branch_name) > 2;

9. Example Queries
Report the largest and smallest loans given by each branch in NY city along with the branch name.
SELECT branch_name, Max(amount) As MaxLoan, Min(amount) As MinLoan
FROM Loan L, Branch B
Where L.branch_name = B.branch_name
AND branch_city = ‘NY’
GROUP BY branch_name;

10. SELECT Statement Clauses
SELECT <projection list>
FROM <relation names>
WHERE <conditions>
GROUP BY <grouping columns>
HAVING <grouping conditions>
ORDER BY <order columns>;
optional
Optional clauses if added must be in the order above
Order of execution
FROM  Check which relations are used
WHERE  Filter records based on conditions
GROUP BY  Form groups
HAVING  Filter groups based on conditions
ORDER BY  Sort the data
SELECT  Form the projection list (output columns)

11. Example: Order of Execution
Student
sNumber
sName
address
pNumber 1
Dave
320FL 2
Greg 3
Matt 4
Jan
500MA 5
SELECT pNumber,address, count(sName) AS CNT, sum(sNumber) AS SUM
FROM Student
WHERE sNumber > 1
GROUP BY pNumber, address
HAVING sum(sNumber) > 3; 1 2 3 4
pNumber
address
CNT
SUM 2
500MA 1 4

12. Questions optional SELECT <projection list>
FROM <relation names>
WHERE <conditions>
GROUP BY <grouping columns>
HAVING <grouping conditions>
ORDER BY <order columns>;
optional

13. Example Queries
Report the branch-cities having more than 5 branches, and order them by the number branches
Report the output relation O(CustomerName, Num_Loans, Num_Accounts) only for customers who have loans or accounts

14. Example Queries
Report the branch-cities having more than 5 branches, and order them by the number branches
SELECT branch_city, count(branch_name)
FROM Branch
GROUP BY branch_city
Having count(branch_name) > 5
Order By count(branch_name);

15. Example Queries
Report the output relation O(CustomerName, Num_Loans, Num_Accounts) only for customers who have loans or accounts
SELECT customer_name, count(loan_number), count(account_number)
FROM Depositor D, Borrower B
WHERE D.customer_name = B.customer_name
GROUP BY customer_name; X
Join will cause values to replicate

16. Fixing Previous Example
The join keeps only customers
Who appear in both sides
Report the output relation O(CustomerName, Num_Loans, Num_Accounts) only for customers who have loans and accounts
SELECT customer_name, count(Distinct loan_number), count(Distinct account_number)
FROM Depositor D, Borrower B
WHERE D.customer_name = B.customer_name
GROUP BY customer_name;
 Applies the aggregation over the distinct values
 Can be used with others, e.g., Min, Max, Avg, …

17. Fixing Previous Example
In this case, we need Outer Join
Report the output relation O(CustomerName, Num_Loans, Num_Accounts) only for customers who have loans or accounts
SELECT customer_name, count(Distinct loan_number), count(Distinct account_number)
FROM Depositor D Full Outer Join Borrower B On D.customer_name = B.customer_name
GROUP BY customer_name;
 Notice the syntax is slightly different

18. More in SELECT Statement
Special handling for NULL values
Queries inside Insert/Update/Delete
Nested subqueries

19. Null Values Null means ‘unknown’ value
Any expression containing Null returns Null
5 + null  null
‘ABC’ || null  null
Null in predicates returns UNKNOWN
Predicates usually return TRUE or FALSE

20. Having Null in the data is problematic and needs special care…
Example
Student
sNumber
sName
address
pNumber 1
Dave
320FL 2
Greg
null 3
Matt 4
Jan
500MA
sNumber 1 2 3
SELECT sNumber
FROM Student
WHERE address = ‘320FL’;
May or may not appear
Having Null in the data is problematic and needs special care…

21. Use of “IS NULL” or “IS NOT NULL”
Check if a value is null or not
SELECT sNumber
FROM Student
WHERE address is null;
Select student numbers where the address is null
SELECT sNumber
FROM Student
WHERE address is not null
AND address = ‘320FL’;
Remember:
SELECT sNumber
FROM Student
WHERE address = null; X
The returned value here is unknown

22. Use of “NVL” Function NVL( exp1, exp2)
If exp1 is null return exp2, otherwise return expr1
Can be used in projection list or in predicates
SELECT sNumber
FROM Student
WHERE nvl(address, ‘n/a’) <> ‘n/a’
AND address ‘320FL’;
SELECT sNumber, nvl(address, ‘N/A’)
FROM Student;
sNumber
address 1
320FL 2
N/A 3 4
500MA

23. Null with Grouping & Aggregation
Null is ignored with all aggregates, e.g., SUM, AVG, MIN, MAX except COUNT()
Grouping
Null is considered as a separate group

24. Example Student SELECT address, sum(pNumber) as sum, count(*) as cnt
sNumber
sName
address
pNumber 1
Dave
320FL 2
Greg
null 3
Matt 4
Jan
500MA
SELECT address, sum(pNumber) as sum, count(*) as cnt
FROM Student
GROUP BY address;
address
sum
cnt
320FL 1
null 2
500MA

25. More in SELECT Statement
Special handling for NULL values
Queries inside Insert/Update/Delete
Nested subqueries

26. Reminder About: Insert, Update, Delete
This is performed using Data Manipulation Language of SQL (DML)
Insertion
Insert into Students values (‘1111’, …);
Deletion
Delete from Students;
Delete from Students Where sid = ‘1111’;
Update
Update Students Set GPA = GPA + 0.4;
Update Students Set GPA = GPA Where sid = ‘1111’;

27. Use of Select Inside Insert
All records returned from the Select will be inserted
Notice that there is no keyword “values” in this case
INSERT INTO suppliers (supplier_id, supplier_name)
SELECT account_no, name
FROM externals
Where code = 1;
Number of columns and data types should match

28. Use of Select Inside Delete
Student
Registration
sNumber
sName
address 1
Dave
320FL 2
Greg
null 3
Matt 4
Jan
500MA
sNumber
courseID
grade 4
DB1 A 2 F 3
DB2
Delete from Student all those who have grade ‘F’
Delete
From Student
Where sNumber in (Select sNumber from Registration Where grade = ‘F’);
Students number 2 & 4 will be deleted…

29. Use of Select Inside Delete
Student
Registration
sNumber
sName
address 1
Dave
320FL 2
Greg
null 3
Matt 4
Jan
500MA
sNumber
courseID
grade 4
DB1 A 2 F 3
DB2
Delete from Student all those who do not have registration
Delete
From Student
Where sNumber not in (Select sNumber from Registration);
Student number 1 will be deleted…

30. Use of Select Inside Update
Student
Registration
sNumber
sName
address 1
Dave
320FL 2
Greg
null 3
Matt 4
Jan
500MA
sNumber
courseID
grade 4
DB1 A 2 F 3
DB2
Update the grades of student ‘Matt’ to B
Update Registration
Set grade = ‘B’
Where sNumber in (Select sNumber from Student Where sName =‘Matt’);

31. Remember… In Delete and Update
From Student
Where sNumber not in (Select sNumber from Registration);
Update Registration
Set grade = ‘B’
Where sNumber in (Select sNumber from Student Where sName =‘Matt’);
In Delete and Update
- The ‘From’ clause always has one table
- The subquery can be only added in the ‘Where’ clause

32. More in SELECT Statement
Special handling for NULL values
Queries inside Insert/Update/Delete
Nested subqueries

33. Nested Subquery
SQL provides a mechanism for the nesting of subqueries.
A subquery is a SELECT statement expression that is nested within another query
Subquery can appear in FROM or WHERE clauses

34. Nested Subquery in FROM Clause
SELECT *
FROM Student, (inner SELECT) AS q
WHERE …
Table built on the fly
Use the inner SELECT like any other table
It is just built on the fly
Inner SELECT can be a full statement with all clauses
ORDER BY clause does not make sense in the inner select

35. Example Subquery 1 is computed on the fly
It is treated as a normal table after that

36. Back to this Example Find another way ??
Report the output relation O(CustomerName, Num_Loans, Num_Accounts) only for customers who have loans and accounts
SELECT customer_name, count(Distinct loan_number), count(Distinct account_number)
FROM Depositor D, Borrower B
WHERE D.customer_name = B.customer_name
GROUP BY customer_name;

37. Back to this Example
Report the output relation O(CustomerName, Num_Loans, Num_Accounts) only for customers who have loans and accounts
SELECT D.customer_name, Num_Loan, Num_Acc
FROM ( Select customer_name, count(*) As Num_Acc
From Depositor Group By customer_name) As D,
( Select customer_name, count(*) As Num_Loan
From Borrower Group By customer_name) As B
WHERE D.customer_name = B.customer_name;

38. Back to this Example Find another way ??
Report the output relation O(CustomerName, Num_Loans, Num_Accounts) only for customers who have loans Or accounts
SELECT customer_name, count(Distinct loan_number), count(Distinct account_number)
FROM Depositor D, Borrower B
WHERE D.customer_name = B.customer_name
GROUP BY customer_name;

39. Back to this Example
Report the output relation O(CustomerName, Num_Loans, Num_Accounts) only for customers who have loans Or accounts
SELECT D.customer_name, Num_Loan, Num_Acc
FROM ( Select customer_name, count(*) As Num_Acc
From Depositor Group By customer_name) As D
Full Outer Join
( Select customer_name, count(*) As Num_Loan
From Borrower Group By customer_name) As B
On D.customer_name = B.customer_name;

40. Nested Subquery in WHERE Clause
2- Then, execute this statement once pNumber from the first step is known
(outer SELECT)
SELECT *
FROM Student
WHERE pNumber =
(SELECT pNumber
FROM Professor
WHERE pName = ‘Mike’);
1- Execute this statement first to get the pNumber (inner SELECT)
Since the predicates has = :
The inner statement must return one record with one column
In this case, DBMS will automatically convert the relation to a single scalar value
Otherwise an error is generated

41. Example: Subqueries Retuning Scalar Value
Student
Professor
sNumber
sName
address
pNum 1
Dave
320FL 2
Greg 3
Matt
pNumber
pName
address 1 MM
141FL 2 ER
201FL
Select students of professor ‘MM’
SELECT sNumber, sName
FROM Student
WHERE pNum =
(SELECT pNumber
FROM Professor
WHERE pName=‘MM’);
sNumber
sName 1
Dave 2
Greg
CS3431

42. SubQuery Returning a Relation (General Case)
Outer Select (S)
SELECT sNumber, sName
FROM Student
WHERE pNum OP
(SELECT pNumber
FROM Professor
WHERE pName=‘MM’);
Inner Select (R)
Predicates may include any of (OP above) :
s in R  True if tuple s appears in R
s not in R  True if tuple s does not appear in R
s = R  R must return a single value (otherwise invalid op)
Exists R  True if R is not empty
Not Exists R  True if R is empty

43. Example 1: Subqueries Returning Relations
Student
Professor
sNumber
sName
address
pNum 1
Dave
320FL 2
Greg 3
Matt 4
Sam
30IN
pNumber
pName
address 1 MM
141FL 2 ER
201FL 3 XY
30WA
Select students of professors with address like ‘%FL’
SELECT sNumber, sName
FROM Student
WHERE pNum IN
(SELECT pNumber
FROM Professor
WHERE address Like ‘%FL’);
sNumber
sName 1
Dave 2
Greg 3
Matt
CS3431

44. Example 2: Subqueries Returning Relations
Student
Professor
sNumber
sName
address
pNum 1
Dave
320FL 2
Greg 3
Matt 4
Sam
30IN
pNumber
pName
address 1 MM
141FL 2 ER
201FL 3 XY
30WA
SELECT sNumber, sName
FROM Student
WHERE Exists
(SELECT pNumber
FROM Professor
WHERE address Like ‘%FL’);
sNumber
sName 1
Dave 2
Greg 3
Matt 4
Sam
Always true because it is not empty
CS3431

45. Example 3: Subqueries Registration  R Step 1
Q: Find the student ID taking the largest number of courses
Registration  R
sID
courseID … 1
CS101 2
CS300
CS202 4
CS500
CS203
CS303
SELECT sID, count(courseID) as CNT
FROM R
Group By sID;
Step 1
sID
CNT 1 3 2 4
CS3431

46. Example 3: Subqueries Registration  R Step 2
Q: Find the student ID taking the largest number of courses
Registration  R
sID
CNT 1 3
sID
courseID … 1
CS101 2
CS300
CS202 4
CS500
CS203
CS303
Step 2
SELECT sID, count(courseID) as CNT
FROM R
Group By sID
Having count(courseID) =
( SELECT Max(CNT) As Max
From (SELECT count(courseID) As CNT
Group By sID) q )
Max 3
sID
CNT 1 3 2 4
CNT 3 1 2
CS3431

47. Comparison Using ALL and ANY
Outer Select (S)
SELECT sNumber, sName
FROM Student
WHERE pNum OP
(SELECT pNumber
FROM Professor
WHERE pName=‘MM’);
Inner Select (R)
We took: Exists, IN, NOT IN
s > ALL R  True if s > all values in R
s > ANY R  True if s > any value in R
‘>’ can be any of the other comparison operators, e.g., <, <=, >=, =, <>
R must be relation with single column

48. Example Student Professor SELECT sNumber, sName FROM Student
address
pNum 1
Dave
320FL 2
Greg 3
Matt 4
Sam
30IN
pNumber
pName
address 1 MM
141FL 2 ER
201FL 3 XY
30WA
SELECT sNumber, sName
FROM Student
WHERE pNum >= ALL
(SELECT pNumber
FROM Professor
WHERE address Like ‘%FL’);
sNumber
sName 3
Matt 4
Sam
This inner select returns 1 , 2