1. SQL Review

2. Example Schema We will use these table definitions in our examples.
Sailors(sid: integer, sname: string, rating: integer, age: real)
Boats(bid: integer, bname: string, color: string)
Reserves(sid: integer, bid: integer, day: date)
Make reservations

3. Basic SQL Query SELECT [DISTINCT] target-list FROM relation-list
WHERE qualification
relation-list A list of relation names
target-list A list of attributes of relations in relation-list
qualification Comparisons (“Attr op const” or “Attr1 op Attr2,” where op is one of ˂, ˃, ≤, ≥, =, ≠ ) combined using AND, OR, and NOT.
DISTINCT is an optional keyword indicating that the answer should not contain duplicates.

4. Retrieve names of students and the courses they received an “A” grade
Querying Relations (1)
What does the following query compute?
Enrolled
SELECT S.name, E.cid
FROM Students S, Enrolled E
WHERE S.sid=E.sid AND E.grade=“A”
A student with “sid” has an entry in Enrolled
The Enrolled entry has a grade of “A”
Students
sid
name
login
age
gpa
53831
Zhang 19
3.2
53650
Smith 21
3.9
53666
Jones 20
3.5
Retrieve names of students and the courses they received an “A” grade

5. Querying Relations (2) Enrolled Students SELECT S.name, E.cid
sid
name
login
age
gpa
53831
Zhang 19
3.2
53650
Smith 21
3.9
53666
Jones 20
3.5
SELECT S.name, E.cid
FROM Students S, Enrolled E
WHERE S.sid=E.sid AND E.grade=“A”
S.name
E.cid
Smith
Topology112

6. Query Result Semantics of SQL ˂, ˃, ≤, ≥, =, ≠ R1 × R2 × R3 × · · ·
QUERY PROCESSOR
SELECT [DISTINCT] target-list
FROM relation-list
WHERE qualification
target-list
Define search space
relation-list
R1 × R2 × R3 × · · ·
qualification
This strategy is probably the least efficient way to compute a query! An optimizer will find more efficient strategies to compute the same answers.
Select rows
˂, ˃, ≤, ≥, =, ≠
Query Result
Select columns
Projection

7. Example of Conceptual Evaluation
SELECT S.sname
FROM Sailors S, Reserves R
WHERE S.sid=R.sid AND R.bid=103
Range variable
Reservation
Sailors

8. Example of Conceptual Evaluation
SELECT S.sname
FROM Sailors S, Reserves R
WHERE S.sid=R.sid AND R.bid=103
S×R
(sid)
sname
rating
age
bid
day 22
dustin 7
45.0
101
10/10/96 58
103
11/12/96 31
lubber 8
55.5
rusty 10
35.0
Remove
Irrelevant
columns
Disqualified
Answer

9. A Note on Range Variables
Really needed only if the same relation appears twice in the FROM clause. The previous query can be written in two ways:
Range variable
SELECT S.sname
FROM Sailors S, Reserves R
WHERE S.sid=R.sid AND R.bid=103
It is good style,
however, to use
range variables
always!
SELECT sname
FROM Sailors, Reserves
WHERE Sailors.sid=Reserves.sid
AND bid=103 OR

10. Aggregate Operators Significant extension of relational algebra
COUNT (*)
The number of rows in the relation
COUNT ([DISTINCT] A)
The number of (unique) values in the A column
SUM ([DISTINCT] A)
The sum of all (unique) values in the A column
AVG ([DISTINCT] A)
The average of all (unique) values in the A column
MAX (A)
The maximum value in the A column
MIN (A)
The minimum value in the A column
The number of rows in the relation

11. Aggregate Operators SELECT COUNT (*) FROM Sailors S SELECT S.sname
Count the number of sailors
Find the name of sailors with the highest rating
SELECT COUNT (*)
FROM Sailors S
Compute maximum rating
SELECT S.sname
FROM Sailors S
WHERE S.rating= (SELECT MAX(S2.rating)
FROM Sailors S2)
Find the average age of sailors with a rating of 10
Find the average of the distinct ages of sailors with a rating of 10
SELECT AVG (S.age)
FROM Sailors S
WHERE S.rating=10
Count the number of distinct ratings of sailors called “Bob”
SELECT AVG (DISTINCT S.age)
FROM Sailors S
WHERE S.rating=10
SELECT COUNT (DISTINCT S.rating)
FROM Sailors S
WHERE S.sname=‘Bob’

12. Find the average age of sailors with a rating of 10
GROUP BY and HAVING (1)
So far, we’ve applied aggregate operators to all (qualifying) tuples.
Relation
Aggregator
Qualifier 32
Aggregator
Find the average age of sailors with a rating of 10
SELECT AVG (S.age)
FROM Sailors S
WHERE S.rating=10
Qualifier

13. GROUP BY and HAVING (1)
So far, we’ve applied aggregate operators to all (qualifying) tuples.
Sometimes, we want to apply them to each of several groups of tuples.
Relation
Aggregator
Qualifier 32
Relation
Group 1
Aggregator 12
Group 2
Group 3 9 11

14. Queries With GROUP BY and HAVING
SELECT [DISTINCT] target-list
FROM relation-list
WHERE qualification
GROUP BY grouping-list
HAVING group-qualification
MIN(Attribute)
HAVING
GROUP BY
Output a table
Aggregator 12 9
Qualifier selecting groups
Group 1
Group 2
Group 3
SELECT
FROM
WHERE

15. Find the age of the youngest sailor with age ≥ 18, for each rating with at least 2 such sailors
SELECT S.rating, MIN (S.age)
FROM Sailors S
WHERE S.age >= 18
GROUP BY S.rating
HAVING COUNT (*) > 1
Input relation
Sailors
age
Answer
Disqualify
Only one group satisfies HAVING
4 rating groups
Only S.rating and S.age are mentioned in SELECT

16. Summary
SQL was an important factor in the early acceptance of the relational model; more natural than earlier, procedural query languages.
Relationally complete; in fact, significantly more expressive power than relational algebra.
Even queries that can be expressed in RA can often be expressed more naturally in SQL.
Many alternative ways to write a query; optimizer should look for most efficient evaluation plan.
In practice, users need to be aware of how queries are optimized and evaluated for best results.