1 / 26

CPS216: Advanced Database Systems Notes 02:Query Processing (Overview)

CPS216: Advanced Database Systems Notes 02:Query Processing (Overview). Shivnath Babu. Query Processing. Declarative SQL Query  Query Plan.

florencestanley
Download Presentation

CPS216: Advanced Database Systems Notes 02:Query Processing (Overview)

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. CPS216: Advanced Database SystemsNotes 02:Query Processing (Overview) Shivnath Babu

  2. Query Processing Declarative SQL Query  Query Plan NOTE: You will not be tested on how well you know SQL. Understanding the SQL introduced in class will be sufficient (a primer follows). SQL is described in Chapter 6, GMUW. Focus: Relational System (i.e., data is organized as tables, or relations)

  3. Course Outline • 50% of the class is about core DBMS concepts • Query execution, query optimization, transactions, recovery, etc. • Textbook material • 50% of the class is on “what is happening today in data management” • New developments on textbook material • Web search – Google, Yahoo! • Solid State Drives (Flash Drives) • Simplifying database mgmt., self-tuning databases • Data streams • Data integration (structured data + unstructured data) • Data mining

  4. Course Logistics • Reference: Database Systems: The Complete Book, by H. Garcia-Molina, J. D. Ullman, and J. Widom • Web site: http://www.cs.duke.edu/courses/fall08/cps216 • Grading: • Project 30% • Homework Assignments 20% • Midterm 20% • Final 30%

  5. SQL Primer We will focus on SPJ, or Select-Project-Join Queries Select <attribute list> From <relation list> Where <condition list> Example Filter Query over R(A,B,C): Select B From R Where R.A = “c”  R.C > 10

  6. SQL Primer (contd.) We will focus on SPJ, or Select-Project-Join-Queries Select <attribute list> From <relation list> Where <condition list> Example Join Query over R(A,B,C) and S(C,D,E): Select B, D From R, S Where R.A = “c”  S.E = 2  R.C = S.C

  7. Answer B D 2 x R A B C S C D E a 1 10 10 x 2 b 1 20 20 y 2 c 2 10 30 z 2 d 2 35 40 x 1 e 3 45 50 y 3 Select B,D From R,S Where R.A = “c”  S.E = 2  R.C=S.C

  8. How do we execute this query? Select B,D From R,S Where R.A = “c”  S.E = 2  R.C=S.C - Do Cartesian product - Select tuples - Do projection One idea

  9. Bingo! Got one... R X S R.A R.B R.C S.C S.D S.E a 1 10 10 x 2 a 1 10 20 y 2 . . c 2 10 10 x 2 . . Select B,D From R,S Where R.A = “c” S.E = 2  R.C=S.C

  10. Relational Algebra - can be used to describe plans Ex: Plan I B,D sR.A=“c” S.E=2  R.C=S.C X R S

  11. Relational Algebra Primer (Chapter 5, GMUW) Select:sR.A=“c” R.C=10 Project: B,D Cartesian Product: R XS Natural Join: R S

  12. Relational Algebra - can be used to describe plans Ex: Plan I B,D sR.A=“c” S.E=2  R.C=S.C X R S OR: B,D [sR.A=“c” S.E=2  R.C = S.C (RXS)]

  13. Another idea: B,D sR.A = “c”sS.E = 2 R(A,B,C) S(C,D,E) Plan II natural join Select B,D From R,S Where R.A = “c”  S.E = 2  R.C=S.C

  14. R S A B C s (R) s(S) C D E a 1 10 A B C C D E 10 x 2 b 1 20 c 2 10 10 x 2 20 y 2 c 2 10 20 y 2 30 z 2 d 2 35 30 z 2 40 x 1 e 3 45 50 y 3 Select B,D From R,S Where R.A = “c”  S.E = 2  R.C=S.C

  15. Plan III Use R.A and S.C Indexes (1) Use R.A index to select R tuples with R.A = “c” (2) For each R.C value found, use S.C index to find matching tuples (3) Eliminate S tuples S.E  2 (4) Join matching R,S tuples, project B,D attributes, and place in result

  16. =“c” <c,2,10> <10,x,2> check=2? output: <2,x> next tuple: <c,7,15> R S A B C C D E a 1 10 10 x 2 b 1 20 20 y 2 c 2 10 30 z 2 d 2 35 40 x 1 e 3 45 50 y 3 c 7 15 A C I1 I2

  17. Overview of Query Processing SQL query parse parse tree Query rewriting Query Optimization statistics logical query plan Physical plan generation physical query plan execute Query Execution result

  18. Example Query Select B,D From R,S Where R.A = “c”  R.C=S.C

  19. Example: Parse Tree <Query> <SFW> SELECT <SelList> FROM <FromList> WHERE <Cond> <Attribute> <SelList> <RelName> <FromList> <Cond> AND <Cond> B <Attribute> R <RelName> <Attr> <Op> <Const> D S R.A = “c” Select B,D From R,S Where R.A = “c”  R.C=S.C <Attr> <Op> <Attr> R.C = S.C

  20. Along with Parsing … • Semantic checks • Do the projected attributes exist in the relations in the From clause? • Ambiguous attributes? • Type checking, ex: R.A > 17.5 • Expand views

  21. SQL query Initial logical plan parse Rewrite rules parse tree Query rewriting Logical plan statistics logical query plan Physical plan generation “Best” logical plan physical query plan execute result

  22. Initial Logical Plan B,D Select B,D From R,S Where R.A = “c”  R.C=S.C R.A = “c” Λ R.C = S.C X R S Relational Algebra: B,D [sR.A=“c” R.C = S.C (RXS)]

  23. Apply Rewrite Rule (1) B,D B,D R.C = S.C R.A = “c” Λ R.C = S.C R.A = “c” X X R S R S B,D [sR.C=S.C [R.A=“c”(R X S)]]

  24. Apply Rewrite Rule (2) B,D B,D R.C = S.C R.C = S.C R.A = “c” X R.A = “c” S X R S R B,D [sR.C=S.C [R.A=“c”(R)] X S]

  25. Apply Rewrite Rule (3) B,D B,D R.C = S.C Natural join R.A = “c” X S R.A = “c” S R R B,D [[R.A=“c”(R)] S]

  26. SQL query Initial logical plan parse Rewrite rules parse tree Query rewriting Logical plan statistics logical query plan Physical plan generation “Best” logical plan physical query plan execute result

More Related