1. 1 3. Relational Algebra and SQL Example: Let the following relations describe point sets: A(x, y), B(x, y), J(x, y) 2D points in the plane H(x, y, z), I(x, y, z) 3D points in space F(z) line on the z-axis K(y,z) 2D points in (y, z) plane

2. 2 3.1 Relation algebra operators A  B A  B A \ B A and B

3. 3 A  F J K

4. 4  x,y H  y,z H  2x+z = 0 I

5. 5 Example: Find the SSN and tax for each person. π SSN,Tax σ wages+interest+capital_gain = income Taxrecord × Taxtable Example: Find the area of Lincoln reached by a radio station. ( π X,Y ( σ Name=“Lincoln” Town ) )  ( π X,Y Broadcast )

6. 6 3.2 SQL SELECT a 1,..., a n FROM R 1, R 2, …, R m WHERE Con 1, …,Con k This means: π a 1,..., a n ( σ Con 1 ( … (σ Con k ( R 1 × R 2 × … × R m ))…))

7. 7 Example: Find the SSN and tax for each person. SELECT SSN, Tax FROM Taxrecord, Taxtable WHERE wages + interest + capital_gain = income

8. 8 AS – keyword used to rename relations Two SQL expressions can be combined by: INTERSECT UNION MINUS – set difference

9. 9 Example: Find the names of the streets that intersect. SELECT S.NAME, T.NAME FROM Streets AS S, Streets AS T WHERE S.X = T.X and S.Y = T.Y

10. 10 Example: Assume we have the relations: Broadcast ( Radio, X, Y ) Town ( Name, X, Y ) Find the parts of Lincoln, NE that can be reached by at least one Radio station. (SELECT X, Y FROM Town WHERE Name = “Lincoln”) INTERSECT (SELECT X, Y FROM Broadcast)

11. 11 Another way of connecting SQL expressions is using the IN keyword. SELECT …….. FROM …….. WHERE a IN ( SELECT b FROM ….. WHERE ….. )

12. 12 SQL with aggregation – SELECT aggregate_function FROM ……. WHERE …… aggregate_function – Max (c 1 a 1 + ……..+ c n a n ) where a i are attributes Min (c 1 a 1 + ……..+ c n a n ) and c i are constants Sum(a) where a is an attribute that is Avg(a) constant in each constraint tuple Count(a)

13. 13 Example: Package(Serial_No, From, Destination, Weight) Postage (Weight, Fee) Find the total postage of all packages sent from Omaha. SELECT Sum(Fee) FROM Package, Postage WHERE Package.Weight = Postage.Weight AND Package.From = “ Omaha “

14. 14 GROUP BY – SELECT a 1, …, a n, aggregate_function FROM ….. WHERE …… GROUP BY a 1,..., a k Evaluates basic SQL query Groups the tuples according to different values of a 1,..,a k Applies the aggregate function to each group separately {a 1, …, a k }  {a 1, …, a n }

15. 15 Example: Find the total postage sent out from each city. SELECT Package.From, Sum(Postage.Fee) FROM Package, Postage WHERE Package.Weight = Postage.Weight GROUP BY Package.From