1. SQL Design Patterns Advanced SQL programming idioms

2. Genesis C++ world Advanced C++ Programming Styles and Idioms, by James O. Coplien Design Patterns: Elements of Reusable Object- Oriented Software by Erich Gamma et al SQL SQL for Smarties by Joe Celko SQL Cookbook by Anthony Molinaro The Art of SQL by Stephane Faroult, Peter Robson

3. What is a SQL Pattern? A common design vocabulary A documentation and learning aid An adjunct to existing design methods A target for refactoring Large range of granularity -- from very general design principles to language- specific idioms

4. List of Patterns Counting Conditional summation Integer generator String/Collection decomposition List Aggregate Enumerating pairs Enumerating sets Interval coalesce

5. Discrete interval sampling User-defined aggregate Pivot Symmetric difference Histogram Skyline query Relational division Outer union Complex constraint Nested intervals Transitive closure Hierarchical total

6. Symmetric Difference A = B ? Isn ’ t it Equality operator ?

7. Venn diagram A\B B\A A∩BA∩B (A \ B) ∪ (B \ A) (A ∪ B) \ (A ∩ B)

8. SQL Query ( select * from A minus select * from B ) union all ( select * from B minus select * from A )

9. Test create table A as select obj# id, name from sys.obj$ where rownum < 100000; create table B as select obj# id, name from sys.obj$ where rownum < 100010;

10. Execution Statistics

11. Anti Join Transformation convert_set_to_join = true: select * from A where (col1,col2,…) not in (select col1,col2,… from B) union all select * from B where (col1,col2,…) not in (select col1,col2,… from A)

12. Execution Statistics

13. Optimization continued … CREATE INDEX A_id_name ON A(id, name); CREATE INDEX B_id_name ON B(id, name); _hash_join_enabled = false _optimizer_sortmerge_join_enabled = false or /*+ use_nl(@"SEL$74086987" A) use_nl(@"SET$D8486D66" B)*/

14. Symmetric Difference via Aggregation select * from ( select id, name, sum(case when src=1 then 1 else 0 end) cnt1, sum(case when src=2 then 1 else 0 end) cnt2 from ( select id, name, 1 src from A union all select id, name, 2 src from B ) group by id, name )where cnt1 <> cnt2

15. Execution Statistics

16. Equality checking via Aggregation 1. Is there any difference? (Boolean). 2. What are the rows that one table contains, and the other doesn't? ||orahash267629 ||orahash512259 ||orahash334382||orahash592731 + + + = 1523431

17. Relational Division Name Steve Pete Kate x Language SQL Java = Name Language Steve SQL Pete Java Kate SQL Steve Java Pete SQL Kate Java JobApplicants JobRequirements ApplicantSkills

18. Dividend, Divisor and Quotient Name Language Steve SQL Pete Java Kate SQL Kate Java / ApplicantSkills Language SQL Java = JobRequirements ? Name Kate Remainder

19. Is it a common Pattern? Not a basic operator in RA or SQL Informally: “ Find job applicants who meet all job requirements ” compare with: “ Find job applicants who meet at least one job requirement ”

20. Set Union Query Given a set of sets, e.g {{1,3,5},{3,4,5},{5,6}} Find their union: SELECT DISTINCT element FROM Sets ID ELEMENT 1 1 1 3 1 5 2 3 2 4 2 5 3 5 3 6 Sets

21. Set Intersection Given a set of sets, e.g {{1,3,5},{3,4,5},{5,6}} Find their intersection? ID ELEMENT 1 1 1 3 1 5 2 3 2 4 2 5 3 5 3 6 Sets

22. It ’ s Relational Division Query! “ Find Element s which belong to all sets ” compare with: “ Find Element s who belong to at least one set ” ID ELEMENT 1 1 1 3 1 5 2 3 2 4 2 5 3 5 3 6 / ID 1 2 3 = ELEMENT 5

23. Implementation (1) π Name (ApplicantSkills) x JobRequirements Name Language Steve SQL Pete Java Kate SQL Steve Java Pete SQL Kate Java

24. Implementation (2) Applicants who are not qualified: π Name ( π Name (ApplicantSkills) x JobRequirements - ApplicantSkills )

25. Implementation (3) Final Query: π Name (ApplicantSkills) - π Name ( ApplicantSkills - π Name (ApplicantSkills) x JobRequirements )

26. Implementation in SQL (1) select distinct Name from ApplicantSkills minus select Name from ( select Name, Language from ( select Name from ApplicantSkills ), ( select Language from JobRequirements ) minus select Name, Language from ApplicantSkills )

27. Implementation in SQL (2) select distinct Name from ApplicantSkills i where not exists ( select * from JobRequirements ii where not exists ( select * from ApplicantSkills iii where iii.Language = ii.Language and iii.Name = i.Name )

28. Implementation in SQL (3) “ Name the applicants such that for all job requirements there exists a corresponding entry in the applicant skills ”  “ Name the applicants such that there is no job requirement such that there doesn ’ t exists a corresponding entry in the applicant skills ”  “ Name the applicants for which the set of all job skills is a subset of their skills ”

29. Implementation in SQL (4) select distinct Name from ApplicantSkills i where (select Language from JobRequirements ii where ii.Name = i.Name) in (select Language from ApplicantSkills)

30. Implementation in SQL (5) A ⊆ B  A \ B = ∅ select distinct Name from ApplicantSkills i where not exists ( select Language from ApplicantSkills minus select Language from JobRequirements ii where ii.Name = i.Name )

31. Implementation in SQL (6) select Name from ApplicantSkills s, JobRequirements r where s.Language = r.Language group by Name having count(*) = (select count(*) from JobRequirements)

32. Book