html5-img
1 / 15

3. Relational Algebra and SQL

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

yeo-walsh
Download Presentation

3. Relational Algebra and SQL

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


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

  3. A  F J K

  4.  y,z H  x,y H  2x+z = 0 I

  5. Example: Find the SSN and tax for each person. πSSN,Taxσwages+interest+capital_gain = income Taxrecord × Taxtable Example: Find the area of Lincoln reached by a radio station. ( πX,Y ( σName=“Lincoln” Town ) )  ( πX,Y Broadcast )

  6. 3.2 SQL SELECT a1, ..., an FROM R1, R2, …, Rm WHERE Con1, …,Conk This means: πa1, ..., an( σ Con1( … (σ Conk ( R1× R2 × … ×Rm))…))

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

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

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

  12. SQL with aggregation – SELECT aggregate_function FROM ……. WHERE …… aggregate_function– Max (c1a1 + ……..+ cnan) where ai are attributes Min (c1a1 + ……..+ cnan) and ci are constants Sum(a) where a is an attribute that is Avg(a) constant in each constraint tuple Count(a)

  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. GROUP BY– SELECT a1, …, an, aggregate_function FROM ….. WHERE …… GROUP BY a1, ..., ak • Evaluates basic SQL query • Groups the tuples according to different values of a1,..,ak • Applies the aggregate function to each group separately • {a1, …, ak}  {a1, …, an}

  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

More Related