1. Structured Query Language
SQL
Structured Query Language

2. Query Language SQL is a query language
Used to examine data in the database
SQL queries do not change the contents of the database (no side-effects!)
The result of an SQL query is printed to the screen, not stored in the database

3. Basic SQL query structure
SELECT Attributes
FROM relations
WHERE condition
For example:
SELECT sid,sname
FROM students
WHERE sid=1122

4. Query Components A query can contain the following clauses
select
from
where
group by
having
order by
Only select and from are obligatory
Order of clauses is always as above

5. Very Basic SQL Query
SELECT [Distinct] Attributes
FROM relation
Attributes: The attributes or values which will appear in the query result (For example: id, name).
DISTINCT: Optional keyword to delete duplicates
Relation: Relation to perform the query on.
Example:
Select studentID, studentName
From students

6. Select studentID, studentName From students Result:
StudentDept.
StudentName
StudentAge
1123
Math
Moshe 25
2245
Computers
Mickey 26
55611
Menahem 29
Select studentID, studentName
From students
Result:
StudentID
StudentName
1123
Moshe
2245
Mickey
55611
Menahem

7. Basic SQL Query
SELECT [Distinct] Attributes
FROM relation
WHERE condition
condition: A Boolean condition (For example: Eid>21, or Ename=‘Yuval’ ). Only tuples which return ‘true’ for this condition will appear in the result

8. Select studentID, studentName From students Where StudentDept=‘Math’
StudentAge
1123
Math
Moshe 25
2245
Computers
Mickey 26
55611
Menahem 29
Select studentID, studentName
From students
Where StudentDept=‘Math’
Result:
StudentID
StudentName
1123
Moshe
55611
Menahem

9. SQL and relational algebra
SELECT Distinct A1,…,An
FROM R1,…,Rm
WHERE C;
A1,…,An (C(R1 x…x Rm))

10. Basic SQL Query Important! The evaluation order, conceptually, is:
SELECT [Distinct] attributes
FROM relations
WHERE condition;
Important! The evaluation order, conceptually, is:
Compute the cross product of the tables in relations.
Delete all rows that do not satisfy condition.
Delete all columns that do not appear in attributes.
If Distinct is specified eliminate duplicate rows.

11. Example Tables Used Boats bid bname color 101 103 Nancy Gloria red
green
Sailors
sid
sname
rating
age 22 31 58
Dustin
Lubber
Rusty 7 8 10
45.0
55.5
35.0
Reserves
sid
bid
day 22 58
101
103
10/10/96
11/12/96

12. All sailors who have reserved a boat
What does this compute?
Select sname
from sailors, reserves
Where sailors.sid=reserves.sid
All sailors who have reserved a boat
Sailors
sid
sname
rating
age 22 31 58
Dustin
Lubber
Rusty 7 8 10
45.0
55.5
35.0
Reserves
sid
bid
day 22 58
101
103
10/10/96
11/12/96
Computation on upcoming slides…

13. Stage 1: Sailors x Reserves
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

14. Stage 2: “where sailors.sid=reserves.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

15. Stage 2: “where sailors.sid=reserves.sid”
sname
rating
age
bid
day 22
Dustin 7
45.0
101
10/10/96 58
Rusty 10
35.0
103
11/12/96

16. Stage 3: “select sname”
Sailors
Reserves
sid
sname
rating
age
bid
day 22
Dustin 7
45.0
101
10/10/96 58
Rusty 10
35.0
103
11/12/96

17. Stage 3: “select sname”
sname
Dustin
Rusty
Final answer

18. Example Query Q: What does this compute? SELECT sname, age
FROM Sailors
WHERE rating>7;
Q: What does this compute?
Name and age of sailors with rating > 7

19. Example Query Q: What does this compute? SELECT DISTINCT sname
FROM Sailors, Reserves
WHERE Sailors.sid = Reserves.sid and
bid = 103;
Q: What does this compute?
Names of sailors who reserved boat number 103

20. WHERE Sailors.sid = Reserves.sid and
bid = 103;
Sailors
Reserves
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

21. Select sname Sailors Reserves 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

22. A Few SELECT Options Select all columns: Rename selected columns:
FROM Sailors;
Rename selected columns:
SELECT sname AS Sailors_Name
Applying functions (e.g., Mathematical manipulations)
SELECT (age-5)*2

23. The WHERE Clause Numerical and string comparison:
!=,<>,=, <, >, >=, <=, between(val1 AND val2)
Logical components: AND, OR
Null verification: IS NULL, IS NOT NULL
Checking against a list with IN, NOT IN.

24. Examples SELECT sname FROM Sailors
WHERE age>=40 AND rating IS NOT NULL ;
SELECT sid, sname
FROM sailors
WHERE sid IN (1223, 2334, 3344) or
sname between(‘George’ and ‘Paul’);

25. The LIKE Operator A pattern matching operator (regular expression)
Basic format: colname LIKE pattern
Example:
_ is a single character
% is 0 or more characters
SELECT sid
FROM Sailors
WHERE sname LIKE ‘R_%y’;

26. Relation naming Naming relations is good style
SELECT S.sname
FROM Sailors S, Reserves R
WHERE S.sid = R.sid and
R.bid = 103;
Naming relations is good style
It is necessary if the same relation appears twice in the FROM clause

27. Explanation in the next slides
Example Query
SELECT S.sname
FROM Sailors S, Reserves R
WHERE S.sid = R.sid and
R.bid != 103;
Q: Does this return the names of sailors who did not reserve boat 103?
A: No! it returns the names of sailors who reserved a boat other than boat 103
Explanation in the next slides

28. SELECT S.sname
FROM Sailors S, Reserves R
WHERE S.sid = R.sid and
R.bid != 103;
Sailors
sid
sname
rating
age 22 31
Dustin
Lubber 7 8
45.0
55.5
Reserves
sid
bid
day 22 31
101
103
104
10/10/07
11/12/07
12/2/07

29. Sailors
Reserves
sid
sname
rating
age
bid
day 22
Dustin 7
45.0
101
10/10/07
103
11/12/07 31
104
12/2/07
Lubber 8
55.5

30. Sailors
Reserves
sid
sname
rating
age
bid
day 22
Dustin 7
45.0
101
10/10/07
103
11/12/07 31
104
12/2/07
Lubber 8
55.5

31. But Dustin did order boat 103!
Sailors
Reserves
sid
sname
rating
age
bid
day 22
Dustin 7
45.0
101
10/10/07 31
Lubber 8
55.5
104
12/2/07
sname
Dustin
Lubber
But Dustin did order boat 103!

32. When there is a sailor who reserved more than a single boat
SQL query
SELECT S.sid
FROM Sailors S, Reserves R
WHERE S.sid = R.sid;
When would adding DISTINCT give a different result?
When there is a sailor who reserved more than a single boat

33. Are any of these the same?
SELECT S.sid
FROM Sailors S, Reserves R
WHERE S.sid = R.sid;
SELECT DISTINCT R.sid
FROM Sailors S, Reserves R
WHERE S.sid = R.sid;
SELECT R.sid
FROM Reserves R
No. The first will return a Sailor sid as many times as he reserved a boat. The second will return sailors who reserved boats, without repetition. The third will return sids in Reserves. The first and the third will be the same if every sid in reserves appears in sailors (i.e., if there is a foreign key constraint)
Sailors
sid
sname
rating
age
Reserves
sid
bid
day

34. How would you find sailors who have reserved more than one boat?
Example Query
How would you find sailors who have reserved more than one boat?
SELECT S.sname
FROM Sailors S, Reserves R1,
Reserves R2
WHERE S.sid = R1.sid and
R1.sid=R2.sid and R1.bid!=R2.bid;

35. SQL query Q: What does this return? SELECT S.sname
FROM Sailors S, Reserves R, Boats B
WHERE S.sid = R.sid and
R.bid = B.bid and
B.color = 'red'
Q: What does this return?
Names of sailors who reserved red boats

36. SQL query Q: How would you find the colors of boats reserved by Bob?
SELECT distinct B.color
FROM Sailors S, Reserves R, Boats B
WHERE S.sname = ‘Bob’ and
S.sid = R.sid and R.bid = B.bid

37. Order Of the Result
The ORDER BY clause can be used to sort results by one or more columns
The default sorting, when ORDER BY is used, is in ascending order
Can specify ASC or DESC
SELECT sname, rating, age
FROM Sailors S
WHERE age > 50
ORDER BY rating ASC, age DESC

38. What does this return? What would happen if we replaced or by and ?
SELECT DISTINCT S.sname
FROM Sailors S, Reserves R, Boats B
WHERE S.sid = R.sid and
R.bid = B.bid and
(B.color = 'red' or
B.color='green')
What would happen if we replaced or by and ?
Returns sailors who reserved red or green boats. Replacing or with and would result in sailors who reserved a boat that is both red and green. Since no boat is both red and green (each boat has a single color), the result will be empty.
We would get no results!
Then how can we find sailors who have reserved both a green and a red boat?

39. Sailors who’ve reserved red and green boats
SELECT S.sname
FROM Sailors S, Reserves R1, Reserves R2
Boats B1, Boats B2
WHERE S.sid = R1.sid and R1.bid = B1.bid and B1.color = ‘red’ and
S.sid = R2.sid and R2.bid = B2.bid and B2.color = ‘green’;

40. Other Relational Algebra Operators
So far, we have seen selection, projection and Cartesian product
How do we do operators UNION and MINUS?

41. Three SET Operators [Query] UNION [Query] [Query] EXCEPT [Query]
[Query] INTERSECT [QUERY]
Note: The operators remove duplicates by default!

42. Sailors who’ve reserved red or green boat
SELECT S.sname
FROM Sailors S, Boats B, Reserves R
WHERE S.sid = R.sid and R.bid = B.bid and B.color = ‘red’
UNION
WHERE S.sid = R.sid and R.bid = B.bid and B.color = ‘green’;
Would INTERSECT give us sailors who reserved both red and green boats?
Be careful when intersecting between queries that do not return a key!
Almost, but not quite because sname is not unique…

43. Multiset (Bag) Operators
Union without removing duplicates:
UNION ALL
SELECT DISTINCT sname
FROM Sailors S
UNION ALL

44. Nested Queries

45. Nested Queries
A query is nested if one of its clauses contains a query
Queries can be nested in the following clauses:
Select
From
Where
Having

46. Nested Queries (cont)
A sub-query of a nested query is correlated if it refers to relations appearing in the outer portion of the query
We start by discussing subqueries in the WHERE clause
Common operators used to correlate are: IN, ANY/ALL, EXISTS

47. Remember!
The WHERE clause is evaluated for each tuple in the Cartesian Product formed by the FROM clause, and a Boolean answer is returned
The subquery is used to define the Boolean answer!

48. Nested queries in WHERE
Subqueries with multiple results:
SELECT S.sname
FROM Sailors S
WHERE S.sid IN (SELECT R.sid
FROM Reserves R
WHERE R.bid = 103);
We would return names of sailors who did not reserve boat 103
What would happen if we wrote NOT IN?

49. In/Not In Format The format of these subqueries is always:
Attribute or value
In / Not in
Subquery that returns a single column
Returns true if the attribute value is in / not in the result of the subquery

50. Names of sailors who did not reserve a red boat
What does this produce?
SELECT S.sname
FROM Sailors S
WHERE S.sid NOT IN
(SELECT R.sid
FROM Reserves R
WHERE R.bid IN
(SELECT B.bid
FROM Boats B
WHERE B.color='red'))
Names of sailors who did not reserve a red boat

51. Any/All Format The format of these subqueries is always:
Attribute or value
Arithmetic comparison operator
Any / All
Subquery that returns a single column
Returns true if the attribute value satisfies the arithmetic operator with respect to any/all of the query results

52. Set-Comparison Queries
SELECT *
FROM Sailors S1
WHERE S1.age > ANY (SELECT S2.age
FROM Sailors S2);
We can also use op ALL (op is >, <, =, >=, <=, or <>).

53. Exists/Not Exists Format
The format of these subqueries is always:
Exists / Not Exists
Subquery that returns any number of columns
Returns true if the subquery returns a non-empty (resp. empty) result

54. Correlated Nested Queries
SELECT S.sid
FROM Sailors S
WHERE EXISTS (SELECT *
FROM Reserves R
WHERE R.bid = 103 and
S.sid = R.sid);
S not in subquery, refers to outer loop
Sid of sailors who reserved boat 103
Q: What if we wrote NOT EXISTS?
A: We would get sid of sailors who did not reserve boat 103

55. Exists and Not Exists Differs from In and Not In Exists:
For every tuple in the outer loop, the inner loop is tested. If the inner loop produces a result, the outer tuple is added to the result.

56. How would you find the names of sailors who have reserved a red boat but not a green boat?
SELECT SS.sname from sailors SS where SS.sid in ( SELECT R1.sid
FROM Reserves R1, Boats B1
WHERE R1.bid=B1.bid and B1.color=‘red’
EXCEPT
SELECT R2.sid
FROM Reserves R2, Boats B2
WHERE R2.bid=B2.bid and B2.color=‘green’ );

57. Rewrite using not in
SELECT SS.sname from sailors SS where SS.sid in ( SELECT R1.sid
FROM Reserves R1, Boats B1
WHERE R1.bid=B1.bid and B1.color=‘red’ and
R1.sid not in (
SELECT R2.sid
FROM Reserves R2, Boats B2
WHERE R2.bid=B2.bid and B2.color=‘green’
));

58. How would you find the sailors who have reserved all boats?

59. Remember: Algebraic Operator of Division
Consider: A(X,Y) and B(Y).
Then AB =
In general, we require that the set of fields in B be contained in those of A.

60. Reminder: Suppliers from A who supply All Parts from B (1)
sno
pno S1 S2 S3 S4 P1 P2 P3 P4
sno P2
pno B1  =
S1, S2, S3, S4 A

61. Reminder: Suppliers from A who supply All Parts from B (2)
sno
pno S1 S2 S3 S4 P1 P2 P3 P4
sno P2 P4
pno B2  =
S1, S4 A

62. Reminder: Suppliers from A who supply All Parts from B (3)
sno
pno S1 S2 S3 S4 P1 P2 P3 P4
sno P1 P2 P4
pno B3  = S1 A

63. Sailors who Reserved all Boats
So what is the result of this?
Reserves(sid,bid) Boats(bid)
Sailors who Reserved all Boats
Sailor S whose "set of boats reserved" contains the "set of all boats"

64. What is the strategy for finding sailors who have reserved all boats?
The sailors for which there does not exist a boat which they have not reserved

65. Sailors who reserved all boats (Division 1)
Sailors for which there does not exist a boat that they did not reserve
SELECT sid
FROM Sailors S
WHERE NOT EXISTS
(SELECT B.bid
FROM Boats B
WHERE B.bid NOT IN
(SELECT R.bid
FROM Reserves R
WHERE R.sid = S.sid));

66. Sailors who reserved all boats (Division 2)
Sailors for which there does not exist a boat that they did not reserve
SELECT S.sid
FROM Sailors S
WHERE NOT EXISTS(
SELECT B.bid
FROM Boats B
SELECT R.bid
FROM Reserves R
WHERE R.bid=B.bid and
R.sid=S.sid))

67. Sailors who reserved all boats (Division 3)
Sailors for which there does not exist a boat for which there is no reservation in Reserves
SELECT S.sid
FROM Sailors S
WHERE NOT EXISTS((SELECT B.bid
FROM Boats B)
EXCEPT
(SELECT R.bid
FROM Reserves R
WHERE R.sid = S.sid));

68. Aggregation

69. Aggregate Operators The aggregate operators available in SQL are:
COUNT(*)
COUNT([DISTINCT] A)
SUM([DISTINCT] A)
AVG([DISTINCT] A)
MAX(A)
MIN(A)

70. Some Examples  SELECT COUNT(*) FROM Sailors S SELECT COUNT(sid)
SELECT AVG(S.age)
FROM Sailors S
WHERE S.rating=10
SELECT COUNT(distinct color)
FROM Boats

71. Find Average Age for each Rating
So far, aggregation has been applied to all tuples that passed the WHERE clause test.
How can we apply aggregation to groups of tuples?

72. Find Average Age for each Rating
SELECT AVG(age)
FROM Sailors
GROUP BY rating;

73. Basic SQL Query SELECT [Distinct] attributes FROM relation-list
WHERE condition
GROUP BY grouping-attributes
HAVING group-condition;
attributes: must appear in grouping-attributes or aggregation operators
group-condition: Constrains groups. Can only constrain attributes appearing in grouping-attributes or in aggregation operators

74. Evaluation- important!
SELECT [Distinct] attributes
FROM relation-list
WHERE condition
GROUP BY grouping-attributes
HAVING group-condition;
Compute cross product of relation-list
Tuples failing condition are thrown away
Tuples are partitioned into groups by values of grouping-attributes
The group-condition is applied to eliminate groups
One answer in generated for each group!

75. 40 55.5 34 SELECT AVG(age) FROM Sailors GROUP BY rating; Sailors sid
Sname
rating
age 22 31 58 63 78 84
Dustin
Lubber
Rusty
Fluffy
Morley
Popeye 7 8 10
45.0
55.5
35.0
44.0
31.0
33.0
Sailors
sid
sname
rating
age 22 63 78 31 58 84
Dustin
Fluffy
Morley
Lubber
Rusty
Popeye 7 8 10
45.0
44.0
31.0
55.5
35.0
33.0 40
55.5 34

76. SELECT AVG(age)
FROM Sailors
Where age<50
GROUP BY rating
Having count(*)>2;
Step 1
Sid
Sname
rating
age 22 31 58 63 78 84
Dustin
Lubber
Rusty
Fluffy
Morley
Popeye 7 8 10
45.0
55.5
35.0
44.0
31.0
33.0
Sid
Sname
rating
age 22 58 63 78 84
Dustin
Rusty
Fluffy
Morley
Popeye 7 10
45.0
35.0
44.0
31.0
33.0

77. SELECT AVG(age)
FROM Sailors
Where age<50
GROUP BY rating
Having count(*)>2;
Step 2
Sid
sname
rating
age 22 63 78 58 84
Dustin
Fluffy
Morley
Rusty
Popeye 7 10
45.0
44.0
31.0
35.0
33.0

78. SELECT AVG(age)
FROM Sailors
Where age<50
GROUP BY rating
Having count(*)>2;
Step 3
Sid
sname
rating
age 22 63 78 58 84
Dustin
Fluffy
Morley
Rusty
Popeye 7 10
45.0
44.0
31.0
35.0
33.0
Sid
sname
rating
age 22 63 78
Dustin
Fluffy
Morley 7
45.0
44.0
31.0

79. 40 SELECT AVG(age) FROM Sailors Where age<50 GROUP BY rating
Having count(*)>2;
Step 4
Sid
sname
rating
age 22 63 78
Dustin
Fluffy
Morley 7
45.0
44.0
31.0
Final Answer: 40

80. Find name and age of oldest Sailor..?
Wrong! Why?
SELECT S.sname, MAX(S.age)
FROM Sailors S
SELECT S.sname, MAX(S.age)
FROM Sailors S
GROUP BY S.sname
Wrong! Why?
First one is syntactically incorrect. Sname cannot appear in the select clause (outside of an aggregation operator) since it does not appear in the Group By
Second one will return the maximum age for each sname (since tuples are grouped by sname)

81. Find name and age of oldest Sailor
SELECT S.sname, S.age
FROM Sailors S
WHERE S.age =
(SELECT MAX(S2.age)
FROM Sailors S2)
Right!!
How else can this be done?
HINT: Use ALL
SELECT S.sname, S.age
FROM Sailors S
WHERE S.age >= ALL (SELECT S2.age
FROM Sailors S2)

82. What does this return?
SELECT B.bid, COUNT(*)
FROM Boats B, Reserves R
WHERE R.bid=B.bid and B.color=‘red’
GROUP BY B.bid
This would not be legal, since color does not appear in the group by clause
What would happen if we put the condition about the color in the HAVING clause?

83. What would happen if we put the condition about the color in the HAVING clause?
SELECT B.bid, COUNT(*)
FROM Boats B, Reserves R
WHERE R.bid=B.bid
GROUP BY B.bid, B.color
HAVING B.color=‘red’
This is legal, and is equivalent to the earlier version, since grouping by bid or by bid and color, creates the same groups. (bid determines color)

84. Names of Boats that were not Reserved on more than 5 days
What does this return?
SELECT bname
FROM Boats B, Reserves R
WHERE R.bid=B.bid
GROUP BY bid, bname
HAVING count(DISTINCT day) <= 5
Names of Boats that were not Reserved on more than 5 days
Can we move the condition in the HAVING to the WHERE?
No! Aggregate functions are not allowed in WHERE

85. The Color for which there are the most boats..?
SELECT color
FROM Boats B
GROUP BY color
HAVING max(count(bid))
What is wrong with this?
How would you fix it?
Technically, we cannot apply an aggregate function to the result of an aggregate function.
This also would not make sense, since max returns a number, and HAVING is supposed to return a Boolean value
Fixed version in next slide

86. The Color for which there are the most boats
SELECT color
FROM Boats B
GROUP BY color
HAVING count(bid) >= ALL
(SELECT count(bid)
FROM Boats
GROUP BY Color)

87. Aggregation Instead of Exists
Aggregation can take the place of exists.
What does this return?
SELECT color
FROM Boats B1
WHERE NOT EXISTS(
SELECT *
FROM Boats B2
WHERE B1.bid <> B2.bid
AND B1.color=B2.color)
The color of boats which have a unique color (no other boats with the same color)

88. Aggregation Instead of Exists
SELECT color
FROM Boats B1
GROUP BY color
HAVING count(bid) = 1
Somewhat simpler…

89. Subqueries in the FROM and in the SELECT clauses

90. A Complex Query We would like to create a table containing 3 columns:
Sailor id
Sailor age
Age of the oldest Sailor (same value in all rows)
How can this be done?

91. What We Want: Sailors sid sname rating age 22 31 58 Dustin Lubber
Rusty 7 8 10
45.0
55.5
35.0
Result
sid
age
Max-age 22 31 58
45.0
55.5
35.0

92. Attempt 1 SELECT S.sid, S.age, MAX(S.age) FROM Sailors S;
Syntactically incorrect. Sid and age do not appear in the group by
Why is this wrong?

93. Attempt 2 SELECT S.sid, S.age, MAX(S.age) FROM Sailors S
GROUP BY S.id, S.age;
Returns the same values in the last 2 columns
Why is this wrong?

94. Solution 1: Subquery in FROM
SELECT S.sid, S.age, M.mx
FROM Sailors S,(SELECT MAX(S2.age) as mx
FROM Sailors S2) M;
We can put a query in the FROM clause instead of a table
The query in the FROM clause must be renamed with a range variable (M in this case).

95. Solution 2: Subquery in SELECT
SELECT S.sid, S.age, (SELECT MAX(S2.age)
FROM Sailors S2)
FROM Sailors S;
A query in the SELECT clause must return at most one value for each row returned by the outer query.

96. Another Example of a Sub-query in SELECT
SELECT S.sid, S.age, (SELECT MAX(S2.age)
FROM Sailors S2
WHERE S2.age<S.age)
FROM Sailors S;
For each sailor, his age, and the age of the oldest sailor younger than him
What does this query return?
Note the use of S (defined in the outer query) within the inner query.

97. Result: Sailors sid sname rating age 22 31 58 Dustin Lubber Rusty 7 810
45.0
55.5
35.0
Result
sid
age
(Select…) 22 31 58
45.0
55.5
35.0
null

98. Another Example of a Sub-query in FROM??
SELECT S.sid, S.age, M.mx
FROM Sailors S, (SELECT MAX(S2.age) as mx
FROM Sailors S2
WHERE S2.age<S.age);
Cannot correlate within the FROM clause. Query evaluation is not iterating over tuples at this point, so it does not make sense
Why is this wrong?

99. Translating RA to SQL

100. RA is strictly less expressive than SQL
Every query in relational algebra can be equivalently written in SQL
There are SQL queries that cannot be expressed using relational algebra
Examples?
We now present a procedure for translating RA to SQL
Note: This is not the most efficient translation

101. Assumptions
To make the presentation simpler, assume we are translating a relational algebra expression E into SQL where:
E does not use the same relation twice
No two relations have any attributes with the same names
Easy to overcome these assumptions using RA renaming and SQL aliasing.

102. Translation By Induction on Structure of E
Induction on the number of relational algebra operators appearing in E.
Base case: 0 operators.
E is simply a relation R
SELECT DISTINCT *
FROM R

103. Translation By Induction on Structure of E
Induction Step: Assume that for all E with less than k operators, there is an SQL expression equivalent to E. We show for k
Must consider several cases, depending on the “last operator” using in E:
, , , , -

104. Sub-query in the FROM Clause!
Last Operator is 
E = C(E1), where C is a Boolean condition
Let S be an SQL expression equivalent to E1 (there is one by the induction hypothesis)
E is equivalent to
SELECT DISTINCT *
FROM S
WHERE C
Sub-query in the FROM Clause!

105. Sub-query in the FROM Clause!
Last Operator is 
E = A1,..,Ak(E1), where A1,…,Ak are attributes
Let S be an SQL expression equivalent to E1 (there is one by the induction hypothesis)
E is equivalent to
SELECT DISTINCT A1,…,Ak
FROM S
Sub-query in the FROM Clause!

106. Last Operator is  E = E1  E2
Let S1,S2 be SQL expressions equivalent to E1 and E2
E is equivalent to S1
UNION S2

107. Sub-queries in the FROM Clause!
Last Operator is 
E = E1  E2
Let S1,S2 be SQL expressions equivalent to E1 and E2
E is equivalent to
SELECT *
FROM S1, S2
Sub-queries in the FROM Clause!

108. Last Operator is - E = E1 - E2
Let S1,S2 be SQL expressions equivalent to E1 and E2
E is equivalent to S1
EXCEPT S2

109. Example Translate sname,color(rating<10(Sailors  Boats))
Select distinct sname, color
From (select *
from (select * from
(select * from Sailors) T1, (select * from Boats) T2) T3
where rating < 20) T4

110. Null Values

111. What does NULL Mean?
There are different interpretations to a value of NULL
Value Unknown: I know that there is a value that belongs here, but I don’t know what it is.
Example: Birthday attribute
Value Inapplicable: There is no value that makes sense here
Example: Spouse attribute for unmarried person

112. What does NULL Mean? (cont)
Value Withheld: We are not entitled to know this value
Example: phone number attribute

113. Null Values in Expressions
Two important rules:
When we operate on NULL and any other value, (including another NULL), using an arithmetic operator like * or +, the result is always NULL
When we compare NULL and any other value (including another NULL) , using a comparison operator like = or >, the result is UNKNOWN.
The correct way to determine if an attribute x has value NULL is using x IS NULL or x IS NOT NULL, which will return true or false

114. 3 Valued Logic: True, False, UnknownB
A and B
A or B
True
False
Unknown A
Not A
True
False
Unknown
Only tuples for which the WHERE clause has value TRUE are used to create tuples in the result

115. What does this return? SELECT * FROM Sailors WHERE sname = sname
Returns all rows where sname is not null

116. What does this return?
SELECT * FROM Sailors WHERE rating > 5 or rating <= 5
Returns all rows where rating is not null

117. What do these return? SELECT sname, rating * 0 FROM Sailors
SELECT sname, rating - rating
FROM Sailors
1) Returns name, 0 when rating is not null, and name, null when rating is null
2) Same

118. Nulls in Aggregation Functions
count(*): counts all rows (even rows that are all null values)
count(A): counts non-null A-s. returns 0 if all As are null
sum(A), avg(A), min(A), max(A)
ignore null values of A
if A only contains null value, the result is null

119. Distinct and Group By
Rows are considered identical, for group by and distinct, if they have all the same non-null values and both have null values in the same columns
Distinct removes duplicates of such rows
Such rows form a single group when using GROUP BY

120. Example R B C 1 null 2 3 4 5 SELECT count(*), count(c), min(c), sum(c)
FROM (SELECT c
FROM R
WHERE c IS NULL or c <> NULL
GROUP BY c)
Returns 1, 0, null, null

121. Join Operators in the FROM Clause
“Syntactic Sugar”

122. Shorthand for Conditional Join
SELECT S1.sname, S2.sname
FROM Sailors S1, Sailors S2
WHERE S1.sid != S2.sid and
S1.sname = ‘Rusty’
SELECT S1.sname, S2.sname
FROM Sailors S1 JOIN Sailors S2 on
S1.sid != S2.sid
WHERE S1.sname = ‘Rusty’

123. Shorthand for Equi-Join
SELECT S.sname,
FROM Sailors S, Reserves R
WHERE S.sid = R.sid and
S.age > 20
SELECT S.sname,
FROM Sailors S JOIN Reserves R USING (sid)
WHERE S.age > 20

124. Shorthand for Natural Join
SELECT S.sname,
FROM Sailors S, Reserves R
WHERE S.sid = R.sid and
S.age > 20
SELECT S.sname,
FROM Sailors S NATURAL JOIN Reserves R
WHERE S.age > 20
Requires equality on all common fields

125. Outer Join

126. Left Outer Join The left outer join of R and S contains:
all the tuples in the join of R and S
all the tuples in R that did not join with tuples from S, padded with null values
SELECT Sailors.sid, Reserves.bid
FROM Sailors NATURAL LEFT OUTER JOIN Reserves

127. Result Sailors sid sname rating age 22 31 58 Dustin Lubber Rusty 7 810
45.0
55.5
35.0
Reserves
sid
bid
day 22 58
101
103
10/10/96
11/12/96
Result
sid
bid 31 22 58
null
101
103

128. Right Outer Join, Full Outer Join
The right outer join of R and S contains:
all the tuples in the join of R and S
all the tuples in S that did not join with tuples from R, padded with null values
The full outer join of R and S contains:
all the tuples in the left outer join of R and S
all the tuples in the right outer join of R and S

129. Express the Left Outer Join in SQL, without the Left Outer Join Operator
Suppose we have R(A,B) and S(B,C).
Can you write a query that returns the left outer join of R and S, that does not use the left outer join operator?
Select *
From R, S
Where R.b = S.b
UNION
Select R.a, R.b, null
From R
Where R.b not in (select b from S)

130. ALL and ANY: Special Cases

131. Query 1: What does this return?
SELECT * FROM Sailors WHERE age > ANY (SELECT age WHERE sname=‘Joe’)
Returns sailors that are older than any sailor named joe. (If there is no Joe, no sailors will be returned)

132. Query 2: What does this return?
SELECT * FROM Sailors WHERE age > ALL (SELECT age WHERE sname=‘Joe’)
Returns sailors that are older than all sailors named joe. (If there is no Joe, all sailors will be returned)

133. Query Containment
We say that a query Q1 contains query Q2, if for all databases D, the result of applying Q1 to D contains the result of applying Q2 to D.
Does Query 2 contain Query 1?
Does Query 1 contain Query 2?
(the answer to both questions is no – but why?)
No, a sailor might be older than one joe, but younger than another
No, if there are no joes, then query 1 returns no answers

134. Query Q3: What does this return?
SELECT *
FROM Sailors
WHERE age = ANY (SELECT age
WHERE sname=‘Joe’)
Sailors who are the same age as any joe

135. Query Q4: What does this return?
SELECT *
FROM Sailors
WHERE age IN (SELECT age
WHERE sname=‘Joe’)
Equivalent to Q3

136. Query Q5: What does this return?
SELECT *
FROM Sailors
WHERE age <> ANY (SELECT age
WHERE sname=‘Joe’)
Sailors whose age differs with that of any joe

137. Query Q6: What does this return?
SELECT *
FROM Sailors
WHERE age NOT IN (SELECT age
WHERE sname=‘Joe’)
Sailros whose age differs from all joes
Not equivalent to Q5 – why?

138. Views

139. What is a View? A view is a virtual table A view is defined by a query
The result of the query is the contents of the virtual table
always update with respect to the database
does not exist, is computed every time referenced
Changing a table (insert/update/delete) automatically changes the view

140. Defining a View CREATE VIEW <view-name> AS <view-def>;
Where view-def is an SQL query
Example:
CREATE VIEW GreatSailors AS
SELECT sid, sname
FROM Sailors
WHERE rating>=9

141. Defining a View Another example: CREATE VIEW SailorsDates AS
SELECT sid, date
FROM Sailors S, Reservations R
WHERE S.sid = R.sid

142. Querying a View
Once you have defined a view, you can use it in a query (in the same way that you use a relation)
SELECT sid
FROM GreatSailors
WHERE sname = ‘Joe’

143. Querying a View
You can use a view and a regular relation together in a query
SELECT bname
FROM GreatSailors G, Reservations R, Boats B
WHERE G.sid = R.sid and R.bid = B.bid

144. Understanding Queries using Views
When writing a query with a view it is as if the expression defining the view is a sub-query is the FROM clause
SELECT bname
FROM GreatSailors G, Reservations R, Boats B
WHERE G.sid = R.sid and R.bid = B.bid
SELECT bname
FROM (SELECT sid
FROM Sailors
WHERE rating >=9) G, Reservations R, Boats B
WHERE G.sid = R.sid and R.bid = B.bid

145. What are views good for? (1)
Simplifying complex queries: Here is an example allows the user to "pretend" that there is a single table in the database
CREATE VIEW SRB as
SELECT S.sid, sname, rating, age, R.bid, day,
bname, color
FROM Sailors S, Boats B, Reserves R
WHERE S.sid = R.sid and R.bid = B.bid

146. What are views good for? (1)
Now: Find snames of Sailors who reserved red boats on 1/11/09 using SRB
SELECT sname
FROM SRB
WHERE color = ‘red’ and day = ‘1/11/09’

147. What are views good for? (2)
Security issues – preventing unauthorized access. Example: hiding the rating value
CREATE VIEW SailorInfo
SELECT sname, sid, age
FROM Sailors
grant SELECT on SailorInfo to shimon;

148. Modifying Views
Sometimes it is possible to insert into, delete from, or update, a view !!!
Actually, the user request is translated into a modification of the base tables (the tables used in the view definition)
Modifications are possible only when the view is updatable

149. Updatable Views
There are complex rules determining when a view is updatable
Basically, updates are possible when the view is defined by selecting (SELECT, not SELECT DISTINCT) from a single relation R such that:
The WHERE clause does not involve R in a subquery
The FROM clause contains only the one relation R
The list in the SELECT clause must include enough attributes such that for every tuple inserted through the view, we can fill the other attributes with NULL or a default value

150. Interestingly, we won’t see this tuple when we query GreatSailors
Inserting Example
CREATE VIEW GreatSailors AS SELECT sid, sname FROM Sailors WHERE rating>=9
Interestingly, we won’t see this tuple when we query GreatSailors
INSERT INTO GreatSailors
VALUES(113, ‘Sam’)
INSERT INTO Sailors(sid, sname)
VALUES(113, ‘Sam’)

151. IMPORTANT NOTE There is no relation GreatSailors
Insertion actually affects the table over which GreatSailors is defined, i.e., Sailors
Similarly, deletion and updates will affect the underlying tables…

152. Deleting Example
CREATE VIEW GreatSailors AS SELECT sid, sname FROM Sailors WHERE rating>=9
We add the where condition from the view definition to make sure that only tuples appearing in the view are deleted
DELETE FROM GreatSailors
WHERE sname = ‘John’
DELETE FROM Sailors
WHERE sname = ‘John’ and rating>=9

153. Updating Example
CREATE VIEW GreatSailors AS SELECT sid, sname FROM Sailors WHERE rating>=9
Update GreatSailors
SET sname = ‘Abraham’
WHERE sname = ‘John’
Update Sailors
SET sname = ‘Abraham’
WHERE sname = ‘John’and rating>=9

154. Postgres Support Postgres does not support updatable views
Can achieve the same effect using triggers…

155. Recursion

156. Flights Flight(airline, from, to) Airline Frm To El Al Tel Aviv
New York
Continental
Los Angeles
Air Canada
Toronto
Montreal

157. Can you find?
How can you find all places that you can get to by a direct flight from Tel Aviv?
SELECT to
FROM Flights
WHERE frm = ‘Tel Aviv’

158. Can you find?
How can you find all places that you can get to by a flight with one stop-over from Tel Aviv?
SELECT F2.to
FROM Flights F1, Flights F2
WHERE F1.frm = ‘Tel Aviv’ and F1.to = F2.frm

159. Can you find?
How can you find all places that you can get to by a flight with zero or one stop-over from Tel Aviv?
SELECT to
FROM Flights
WHERE frm = ‘Tel Aviv’
UNION
SELECT F2.to
FROM Flights F1, Flights F2
WHERE F1.frm = ‘Tel Aviv’ and F1.to = F2.frm

160. Why can’t you find?
How can you find all places that you can get to by a flight any number of stop-overs from Tel Aviv?
Problem: How many times should Flights appear in the FROM clause?
Not possible with features seen so far

161. Recursion in SQL
The SQL-99 standard allows us to define temporary relations which can be recursive
WITH R AS <definition of R> <query involving R>
Or more generally:
WITH
[RECURSIVE] R1 AS <definition of R1> , …, [RECURSIVE] Rn AS <definition of Rn>
<query involving R1, .., Rn>

162. Example
WITH RECURSIVE Reaches(frm,to) AS (SELECT frm, to FROM Flights) UNION (SELECT R1.frm, R2.to FROM Reaches R1, Reaches R2 WHERE R1.to = R2.frm) SELECT to FROM Reaches WHERE frm=‘Tel Aviv’

163. Fix-Point Semantics
The value of Reaches is derived by repeatedly evaluating its definition until no changes are made
Before starting evaluation, Reaches is empty
Then, its definition is repeatedly evaluated, and the result defines Reaches
This continues until no more changes appear

164. Flights Reaches: Step 1 WITH RECURSIVE Reaches(frm,to) AS
(SELECT frm, to FROM Flights)
UNION
(SELECT R1.frm, R2.to
FROM Reaches R1, Reaches R2
WHERE R1.to = R2.frm)
SELECT to FROM Reaches WHERE frm=‘Tel Aviv’
Flights
Reaches: Step 1
Airline
Frm To
El Al
Tel Aviv
New York
Continental
Los Angeles
Air Canada
Toronto
Montreal
Frm To
Tel Aviv
New York
Los Angeles
Toronto
Montreal

165. Flights Reaches: Step 1 Reaches: Step 2 Airline Frm To El Al Tel Aviv
New York
Continental
Los Angeles
Air Canada
Toronto
Montreal
Frm To
Tel Aviv
New York
Los Angeles
Toronto
Montreal
Frm To
Tel Aviv
New York
Los Angeles
Toronto
Montreal
Reaches: Step 2

166. Flights Reaches: Step 1 Reaches: Step 3 Airline Frm To El Al Tel Aviv
New York
Continental
Los Angeles
Air Canada
Toronto
Montreal
Frm To
Tel Aviv
New York
Los Angeles
Toronto
Montreal
Frm To
Tel Aviv
New York
Los Angeles
Toronto
Montreal
Reaches: Step 3

167. Query will return values in Red
Flights
Reaches: Step 1
Airline
Frm To
El Al
Tel Aviv
New York
Continental
Los Angeles
Air Canada
Toronto
Montreal
Frm To
Tel Aviv
New York
Los Angeles
Toronto
Montreal
Frm To
Tel Aviv
New York
Los Angeles
Toronto
Montreal
Reaches:
Final Value
Query will return values in Red

168. Mutually Recursive Relations
We can define several recursive queries, which can use one another in their definitions.
A dependency graph has a node for each relation defined, and an edge from one node to another if the first uses the second in its definition
In particular, in the previous example, there would be an edge from Reaches to itself
R and S are mutually recursive, if there is a cycle in the graph involving nodes R and S

169. P and Q are mutually recursive
Example
WITH RECURSIVE P(x) AS (SELECT * FROM R) EXCEPT (SELECT * FROM Q) RECURSIVE Q(x) AS (SELECT * FROM P) SELECT * FROM P R P Q
P and Q are mutually recursive

170. Problematic Recursion
Complicated recursions are allowed
However, sometimes the result may not be well defined
These cases are not allowed by SQL
Before defining exactly what is not allowed, we consider an example

171. Is there a Fix Point?
Recall that the result of a defined relation is derived by simply evaluating it again and again until it no longer changes.
However, what happens if this process never terminates?

172. What is in P and Q if R has the single tuple (0)?
Example
WITH RECURSIVE P(x) AS (SELECT * FROM R) EXCEPT (SELECT * FROM Q) RECURSIVE Q(x) AS (SELECT * FROM P) SELECT * FROM P
What is in P and Q if R has the single tuple (0)?
At the end of the first round of recursive evaluation, P and Q both have the tuple (0). At the end of the second round, they are both empty, at the end of the third round they both contain the tuple (0), …

173. Monotonicity Requirement
R can be defined using a mutually recursive relation S only if R is monotone in S, i.e.,
Adding an arbitrary tuple to S might add tuples to R, or might leave R unchanged, but can never cause a tuple to be deleted from R

174. Monotonicity Requirement
In the previous example, R uses the mutually recursive relation Q, but R is not monotonic in Q (adding tuples to Q can cause tuples to be removed from R)
Therefore, this type of recursion is not allowed in SQL