1. 1 Database Systems Lecture #7 Yan Pan School of Software, SYSU 2011

2. 2 Agenda Subqueries, etc.

3. 3 Relation operators Basic operators:  Selection:   Projection:   Cartesian Product:  Other set-theoretic ops:  Union:   Intersection:  Difference: - Additional operators:  Joins (natural, equijoin, theta join, semijoin)  Renaming:   Grouping…

4. 4 New-style join types Cross joins (simplest):  FROM A CROSS JOIN B Inner joins (regular joins):  FROM A [INNER] JOIN B ON … Natural join:  FROM A NATURAL JOIN B;  Joins on common fields and merges Outer joins (later)  No dangling rows

5. 5 SQL e.g. with tuple vars Reps(ssn, name, etc.) Clients(ssn, name, rssn) Q: Who are George’s clients, in SQL? Conceptually:   Clients.name (  Reps.name=“George” and Reps.ssn=rssn (Reps x Clients))

6. 6 New topic: Subqueries Powerful feature of SQL: one clause can contain other SQL queries  Anywhere where a value or relation is allowed Several ways:  Selection  single constant (scalar) in SELECT  Selection  single constant (scalar) in WHERE  Selection  relation in WHERE  Selection  relation in FROM

7. 7 Standard multi-table example Purchase(prodname, buyerssn, etc.) Person(name, ssn, etc.) Q: What did Christo buy? As usual, need to AND on equality identifying ssn’s row and buyerssn’s row SELECT Purchase.prodname FROM Purchase, Person WHERE buyerssn = ssn AND name = 'Christo'

8. 8 Subquery motivation Purchase(prodname, buyerssn, etc.) Person(name, ssn, etc.) Q: What did Christo buy? Natural intuition:  Go find Christo ’s ssn  Then find purchases SELECT ssn FROM Person WHERE name = 'Christo' SELECT Purchase.prodname FROM Purchase WHERE buyerssn = Christo’s-ssn

9. 9 Subqueries Subquery: copy in Christo ’s selection for his ssn: The subquery returns one value, so the = is valid If it returns more (or fewer), we get a run-time error SELECT Purchase.prodname FROM Purchase WHERE buyerssn = (SELECT ssn FROM Person WHERE name = 'Christo') SELECT Purchase.prodname FROM Purchase WHERE buyerssn = (SELECT ssn FROM Person WHERE name = 'Christo')

10. 10 Operators on subqueries Several new operators applied to (unary) selections: 1. IN R 2. EXISTS R 3. UNIQUE R 4. s > ALL R 5. s > ANY R 6. x IN R > is just an example op Each expression can be negated with NOT

11. Guifeng Zheng, DBMS, SS/SYSU 11 Subqueries with IN Product(prodname,maker), Person(name,ssn), Purchase(buyerssn,product) Q: Find companies Martha bought products from Strategy: 1. Find Martha’s ssn 2. Find products listed with that ssn as buyer 3. Find company names of those products SELECT DISTINCT Product.maker FROM Product WHERE prodname IN (SELECT product FROM Purchase WHERE buyerssn = (SELECT ssn FROM Person WHERE name = 'Martha')) SELECT DISTINCT Product.maker FROM Product WHERE prodname IN (SELECT product FROM Purchase WHERE buyerssn = (SELECT ssn FROM Person WHERE name = 'Martha'))

12. 12 Subqueries returning relations Equivalent to: Or: SELECT DISTINCT Product.maker FROM Product, Purchase, People WHERE prodname = product AND buyerssn = ssn AND name = 'Martha' SELECT DISTINCT Product.maker FROM Product, Purchase, People WHERE prodname = product AND buyerssn = ssn AND name = 'Martha' SELECT DISTINCT Product.maker FROM Product JOIN Purchase ON prodname=product JOIN People ON buyerssn=ssn WHERE name = 'Martha' SELECT DISTINCT Product.maker FROM Product JOIN Purchase ON prodname=product JOIN People ON buyerssn=ssn WHERE name = 'Martha'

13. 13 FROM subqueries Motivation for another way:  suppose we’re given Martha’s purchases  Then could just cross with Products to get product makers  Substitute (named) subquery for Martha’s purchases SELECT maker FROM Product, (SELECT product FROM Purchase WHERE buyerssn = (SELECT ssn FROM Person WHERE name = 'Martha')) Marthas WHERE Product.name = Marthas.product SELECT maker FROM Product, (SELECT product FROM Purchase WHERE buyerssn = (SELECT ssn FROM Person WHERE name = 'Martha')) Marthas WHERE Product.name = Marthas.product

14. 14 Complex RA Expressions Scenario: 1. Purchase(pid, seller-ssn, buyer-ssn, etc.) 2. Person(ssn, name, etc.) 3. Product(pid, name, etc.) Q: Who (give names) bought gizmos from Dick? Where to start? Purchase uses pid, ssn, so must get them…

15. 15 Complex RA Expressions Person Purchase Person Product  name='Dick'  name='Gizmo'  pid  ssn seller-ssn=ssnpid=pidbuyer-ssn=Person.ssn  name

16. 16 Translation to SQL We’re converting the tree on the last slide into SQL The result of the query should be the names indicated above One step at a time, we’ll make the query more complete, until we’ve translated the English-language description to an actual SQL query We’ll also simplify the query when possible (the names of the people who bought gadgets from Dick)

17. 17 Translation to SQL Blue type = actual SQL Black italics = description of subquery Note: the subquery above consists of purchase records, except with the info describing the buyers attached  In the results, the column header for name will be 'buyer' SELECT DISTINCT name buyer FROM (the info, along with buyer names, for purchases of gadgets sold by Dick) SELECT DISTINCT name buyer FROM (the info, along with buyer names, for purchases of gadgets sold by Dick)

18. 18 Translation to SQL Note: the subquery in this version is being given the name P2 We’re pairing our rows from Person with rows from P2 SELECT DISTINCT name buyer FROM (SELECT * FROM Person, (the purchases of gadgets from Dick) P2 WHERE Person.ssn = P2.buyer-ssn) SELECT DISTINCT name buyer FROM (SELECT * FROM Person, (the purchases of gadgets from Dick) P2 WHERE Person.ssn = P2.buyer-ssn)

19. 19 Translation to SQL We simplified by combining the two SELECTs SELECT DISTINCT name buyer FROM Person, (the purchases of gadgets from Dick) P2 WHERE Person.ssn = P2.buyer-ssn SELECT DISTINCT name buyer FROM Person, (the purchases of gadgets from Dick) P2 WHERE Person.ssn = P2.buyer-ssn

20. Guifeng Zheng, DBMS, SS/SYSU 20 Translation to SQL P2 is still the name of the subquery It’s just been filled in with a query that contains two subqueries Outer parentheses are bolded for clarity SELECT DISTINCT name buyer FROM Person, (SELECT * FROM Purchases WHERE seller-ssn = (Dick’s ssn) AND pid = (the id of gadget)) P2 WHERE Person.ssn = P2.buyer-ssn SELECT DISTINCT name buyer FROM Person, (SELECT * FROM Purchases WHERE seller-ssn = (Dick’s ssn) AND pid = (the id of gadget)) P2 WHERE Person.ssn = P2.buyer-ssn

21. 21 Translation to SQL Now the subquery to find Dick’s ssn is filled in SELECT DISTINCT name buyer FROM Person, (SELECT * FROM Purchases WHERE seller-ssn = (SELECT ssn FROM Person WHERE name='Dick') AND pid = (the id of gadget)) P2 WHERE Person.ssn = P2.buyer-ssn SELECT DISTINCT name buyer FROM Person, (SELECT * FROM Purchases WHERE seller-ssn = (SELECT ssn FROM Person WHERE name='Dick') AND pid = (the id of gadget)) P2 WHERE Person.ssn = P2.buyer-ssn

22. 22 Translation to SQL And now the subquery to find Gadget’s product id is filled in, too Note: the SQL simplified by using subqueries  Not used in relational algebra SELECT DISTINCT name buyer FROM Person, (SELECT * FROM Purchases WHERE seller-ssn = (SELECT ssn FROM Person WHERE name='Dick') AND pid = (SELECT pid FROM Product WHERE name='Gadget')) P2 WHERE Person.ssn = P2.buyer-ssn SELECT DISTINCT name buyer FROM Person, (SELECT * FROM Purchases WHERE seller-ssn = (SELECT ssn FROM Person WHERE name='Dick') AND pid = (SELECT pid FROM Product WHERE name='Gadget')) P2 WHERE Person.ssn = P2.buyer-ssn

23. 23 George’s-neighbors with subqueries People(ssn, name, street, city, state, state) Q: Who lives on George’s street? A: First, find George:   name='George' (People) And get George’s street/city/state:   street (  name='George' (People)) Look up people on that street…

24. 24 Next: ALL op Employees(name, job, divid, salary) Find which employees are paid more than all the programmers SELECT name FROM Employees WHERE salary > ALL (SELECT salary FROM Employees WHERE job='programmer') SELECT name FROM Employees WHERE salary > ALL (SELECT salary FROM Employees WHERE job='programmer')

25. Guifeng Zheng, DBMS, SS/SYSU 25 ANY/SOME op Employees(name, job, divid, salary) Find which employees are paid more than at least one vice president SELECT name FROM Employees WHERE salary > ANY (SELECT salary FROM Employees WHERE job='VP') SELECT name FROM Employees WHERE salary > ANY (SELECT salary FROM Employees WHERE job='VP')

26. Guifeng Zheng, DBMS, SS/SYSU 26 ANY/SOME op Employees(name, job, divid, salary) Find which employees are paid more than at least one vice president SELECT name FROM Employees WHERE salary > SOME (SELECT salary FROM Employees WHERE job='VP') SELECT name FROM Employees WHERE salary > SOME (SELECT salary FROM Employees WHERE job='VP')

27. 27 Existential/Universal Conditions Employees(name, job, divid, salary) Division(name, id, head) Find all divisions with an employee whose salary is > 100000 Existential: easy! SELECT DISTINCT Division.name FROM Employees, Division WHERE salary > 100000 AND divid=id SELECT DISTINCT Division.name FROM Employees, Division WHERE salary > 100000 AND divid=id

28. 28 Existential/Universal Conditions Employees(name, job, divid, salary) Division(name, id, head) Find all divisions in which everyone makes > 100000 Existential: easy!

29. 29 Existential/universal with IN 2. Select the divisions we didn’t find: 1. Find the other divisions: in which someone makes <= 100000: SELECT name FROM Division WHERE id IN (SELECT divid FROM Employees WHERE salary <= 100000 SELECT name FROM Division WHERE id IN (SELECT divid FROM Employees WHERE salary <= 100000 SELECT name FROM Division WHERE id NOT IN (SELECT divid FROM Employees WHERE salary <= 100000 SELECT name FROM Division WHERE id NOT IN (SELECT divid FROM Employees WHERE salary <= 100000

30. 30 Next: correlated subqueries Acc(name,bal,type…) Q: Who has the largest balance? Can we do this with subqueries?

31. 31 Acc(name,bal,type,…) Q: Find holder of largest account SELECT name FROM Acc WHERE bal >= ALL (SELECT bal FROM Acc) SELECT name FROM Acc WHERE bal >= ALL (SELECT bal FROM Acc) Correlated Queries

32. 32 Correlated Queries So far, subquery executed once;  result used for higher query More complicated: correlated queries “[T]he subquery… [is] evaluated many times, once for each assignment of a value to some term in the subquery that comes from a tuple variable outside the subquery” (Ullman, p286). Q: What does this mean? A: That subqueries refer to vars from outer queries

33. 33 Acc(name,bal,type,…) Q2: Find holder of largest account of each type SELECT name, type FROM Acc WHERE bal >= ALL (SELECT bal FROM Acc WHERE type=type) SELECT name, type FROM Acc WHERE bal >= ALL (SELECT bal FROM Acc WHERE type=type) Correlated Queries correlation

34. 34 Acc(name,bal,type,…) Q2: Find holder of largest account of each type Note: 1. scope of variables SELECT name, type FROM Acc as a1 WHERE bal >= ALL (SELECT bal FROM Acc WHERE type=a1.type) SELECT name, type FROM Acc as a1 WHERE bal >= ALL (SELECT bal FROM Acc WHERE type=a1.type) Correlated Queries correlation

35. 35 New topic: R.A./SQL Set Operators Relations are sets  have set-theoretic ops  Venn diagrams Union: R1  R2 Example:  ActiveEmployees  RetiredEmployees Difference: R1 – R2 Example:  AllEmployees – RetiredEmployees = ActiveEmployees Intersection: R1  R2 Example:  RetiredEmployees  UnionizedEmployees

36. 36 Set operations - example NameAddressGenderBirthdate Fisher123 MapleF9/9/99 Hamill456 OakM8/8/88 NameAddressGenderBirthdate Fisher123 MapleF9/9/99 Ford345 PalmM7/7/77 R: S: NameAddressGenderBirthdate Fisher123 MapleF9/9/99 Hamill456 OakM8/8/88 Ford345 PalmM7/7/77 R  S:

37. 37 Set operations - example NameAddressGenderBirthdate Fisher123 MapleF9/9/99 Hamill456 OakM8/8/88 NameAddressGenderBirthdate Fisher123 MapleF9/9/99 Ford345 PalmM7/7/77 R: S: R  S: NameAddressGenderBirthdate Fisher123 MapleF9/9/99

38. 38 Set operations - example NameAddressGenderBirthdate Fisher123 MapleF9/9/99 Hamill456 OakM8/8/88 NameAddressGenderBirthdate Fisher123 MapleF9/9/99 Ford345 PalmM7/7/77 R: S: R - S: NameAddressGenderBirthdate Hamill456 OakM8/8/88

39. 39 Set ops in SQL UNION, INTERSECT, EXCEPT Oracle SQL uses MINUS rather than EXCEPT These ops applied to queries: (SELECT name FROM Person WHERE City = 'New York') INTERSECT (SELECT custname FROM Purchase WHERE store='Kim''s') (SELECT name FROM Person WHERE City = 'New York') INTERSECT (SELECT custname FROM Purchase WHERE store='Kim''s')

40. 40 Boat examples Reserve(ssn,bmodel,color) Q: Find ssns of sailors who reserved red boats or green boats SELECT DISTINCT ssn FROM reserve WHERE color = 'red' OR color = 'green' SELECT DISTINCT ssn FROM reserve WHERE color = 'red' OR color = 'green'

41. 41 Boat examples Reserve(ssn,bmodel,color) Q: Find ssns of sailors who reserved red boats and green boats SELECT DISTINCT ssn FROM reserve WHERE color = 'red' AND color = 'green' SELECT DISTINCT ssn FROM reserve WHERE color = 'red' AND color = 'green'

42. Guifeng Zheng, DBMS, SS/SYSU 42 Boat examples Reserve(ssn,bmodel,color) Q: Find ssns of sailors who reserved red boats and green boats SELECT DISTINCT r1.ssn FROM reserve r1, reserve r2 WHERE r1.ssn = r2.ssn AND r1.color = 'red' AND r2.color = 'green' SELECT DISTINCT r1.ssn FROM reserve r1, reserve r2 WHERE r1.ssn = r2.ssn AND r1.color = 'red' AND r2.color = 'green'

43. 43 Boat examples Reserve(ssn,bmodel,color) Q: Find ssns of sailors who reserved red boats and green boats (SELECT DISTINCT ssn FROM reserve WHERE color = 'red') INTERSECT(SELECT DISTINCT ssn FROM reserve WHERE color = 'green') (SELECT DISTINCT ssn FROM reserve WHERE color = 'red') INTERSECT(SELECT DISTINCT ssn FROM reserve WHERE color = 'green')

44. 44 Boat examples Reserve(ssn,bmodel,color) Q: Find ssns of sailors who reserved red boats or green boats (SELECT DISTINCT ssn FROM reserve WHERE color = 'red') UNION (SELECT DISTINCT ssn FROM reserve WHERE color = 'green') (SELECT DISTINCT ssn FROM reserve WHERE color = 'red') UNION (SELECT DISTINCT ssn FROM reserve WHERE color = 'green')

45. 45 Boat examples Reserve(ssn,bmodel,color) Q: Find ssns of sailors who reserved red boats but not green boats (SELECT DISTINCT ssn FROM reserve WHERE color = 'red') EXCEPT (SELECT DISTINCT ssn FROM reserve WHERE color = 'green') (SELECT DISTINCT ssn FROM reserve WHERE color = 'red') EXCEPT (SELECT DISTINCT ssn FROM reserve WHERE color = 'green')

46. 46 (SELECT name, address FROM Cust1) UNION (SELECT name FROM Cust2) (SELECT name, address FROM Cust1) UNION (SELECT name FROM Cust2) Union-Compatibility Situation: Cust1(name,address,…), Cust2(name,…) Want: report of all customer names and addresses (if known) Can’t do: Both tables must have same sequence of types  Applies to all set ops

47. 47 Union-Compatibility Situation: Cust1(name,address,…), Cust2(name,…) Want: report of all customer names and addresses (if known) But can do: Resulting field names taken from first table (SELECT name, address FROM Cust1) UNION (SELECT name, '(N/A)' FROM Cust2) (SELECT name, address FROM Cust1) UNION (SELECT name, '(N/A)' FROM Cust2) Result(name, address)

48. 48 New topic: Nulls in SQL If we don’t have a value, can put a NULL Null can mean several things:  Value does not exists  Value exists but is unknown  Value not applicable But null is not the same as 0

49. 49 Null Values x = NULL  4*(3-x)/7 = NULL x = NULL  x + 3 – x = NULL x = NULL  3 + (x-x) = NULL x = NULL  x = 'Joe' is UNKNOWN In general: no row using null fields appear in the selection test will pass the test  With one exception Pace Boole, SQL has three boolean values:  FALSE=0  TRUE=1  UNKNOWN=0.5

50. 50 Null values in boolean expressions C1 AND C2= min(C1, C2) C1 OR C2= max(C1, C2) NOT C1= 1 – C1 height > 6 = UNKNOWN  UNKNOWN OR weight > 190 = UNKOWN  (age < 25) AND UNKNOWN = UNKNOWN E.g. age=20 height=NULL weight=180 SELECT * FROM Person WHERE (age < 25) AND (height > 6 OR weight > 190) SELECT * FROM Person WHERE (age < 25) AND (height > 6 OR weight > 190)

51. 51 Comparing null and non-nulls The schema specifies whether null is allowed for each attribute  NOT NULL to forbid  Nulls are allowed by default Unexpected behavior: Some Persons are not included! The “trichotomy law” does not hold! SELECT * FROM Person WHERE age = 25 SELECT * FROM Person WHERE age = 25

52. 52 Testing for null values Can test for NULL explicitly:  x IS NULL  x IS NOT NULL But:  x = NULL is never true Now it includes all Persons SELECT * FROM Person WHERE age = 25 OR age IS NULL SELECT * FROM Person WHERE age = 25 OR age IS NULL

53. 53 Null/logic review TRUE AND UNKNOWN = ? TRUE OR UNKNOWN = ? UNKNOWN OR UNKNOWN = ? X = NULL = ?

54. 54 Next: Outer join Like inner join except that dangling tuples are included, padded with nulls Left outerjoin: dangling tuples from left are included  Nulls appear “on the right” Right outerjoin: dangling tuples from right are included  Nulls appear “on the left”

55. 55 Cross join - example NameAddressGenderBirthdate Hanks123 Palm RdM01/01/60 Taylor456 Maple AvF02/02/40 Lucas789 Oak StM03/03/55 NameAddressNetworth Spielberg246 Palm Rd10M Taylor456 Maple Av20M Lucas789 Oak St30M MovieStar MovieExec

56. 56 NameAddressG.BirthdateNameAddressNet Hanks123 Palm RdM01/01/60 Taylor456 Maple AvF02/02/40Taylor456 Maple Av20M Lucas789 Oak StM03/03/55Lucas789 Oak St30M Spielberg246 Palm Rd10M

57. Guifeng Zheng, DBMS, SS/SYSU 57 Outer Join - Example SELECT * FROM MovieStar LEFT OUTER JOIN MovieExec ON MovieStart.name=MovieExec.name SELECT * FROM MovieStar RIGHT OUTER JOIN MovieExec ON MovieStart.name=MovieExec.name NameAddressG.BirthdateNameAddressNet Hanks123 Palm RdM01/01/60Null Taylor456 Maple AvF02/02/40Taylor456 Maple Av20M Lucas789 Oak StM03/03/55Lucas789 Oak St30M Null Spielberg246 Palm Rd10M NameAddressG.BirthdateNameAddressNet Hanks123 Palm RdM01/01/60Null Taylor456 Maple AvF02/02/40Taylor456 Maple Av20M Lucas789 Oak StM03/03/55Lucas789 Oak St30M Null Spielberg246 Palm Rd10M

58. 58 Outer Join - Example NameAddressGenderBirthdate Hanks123 Palm RdM01/01/60 Taylor456 Maple AvF02/02/40 Lucas789 Oak StM03/03/55 NameAddressNetworth Spielberg246 Palm Rd10M Taylor456 Maple Av20M Lucas789 Oak St30M MovieStarMovieExec SELECT * FROM MovieStar FULL OUTER JOIN MovieExec ON MovieStart.name=MovieExec.name NameAddressG.BirthdateNameAddressNet Hanks123 Palm RdM01/01/60Null Taylor456 Maple AvF02/02/40Taylor456 Maple Av20M Lucas789 Oak StM03/03/55Lucas789 Oak St30M Null Spielberg246 Palm Rd10M

59. 59 New-style outer joins Outer joins may be left, right, or full  FROM A LEFT [OUTER] JOIN B;  FROM A RIGHT [OUTER] JOIN B;  FROM A FULL [OUTER] JOIN B; OUTER is optional  If OUTER is included, then FULL is the default Q: How to remember left v. right? A: It indicates the side whose rows are always included

60. 60 Next: Grouping & Aggregation In SQL:  aggregation operators in SELECT,  Grouping in GROUP BY clause Recall aggregation operators:  sum, avg, min, max, count strings, numbers, dates  Each applies to scalars  Count also applies to row: count(*)  Can DISTINCT inside aggregation op: count(DISTINCT x) Grouping: group rows that agree on single value  Each group becomes one row in result

61. 61 Aggregation functions Numerical: SUM, AVG, MIN, MAX Char: MIN, MAX  In lexocographic/alphabetic order Any attribute: COUNT  Number of values SUM(B) = 10 AVG(A) = 1.5 MIN(A) = 1 MAX(A) = 3 COUNT(A) = 4 AB 12 34 12 12

62. 62 Straight aggregation In R.A.  sum(x)  total (R) In SQL: Just put the aggregation op in SELECT NB: aggreg. ops applied to each non-null val  count(x) counts the number of nun-null vals in field x  Use count(*) to count the number of rows SELECT SUM(x) total FROM R SELECT SUM(x) total FROM R

63. 63 Straight aggregation example COUNT applies to duplicates, unless otherwise stated: Better: Can we say: same as Count(*), except excludes nulls SELECT Count(category) FROM Product WHERE year > 1995 SELECT Count(category) FROM Product WHERE year > 1995 SELECT COUNT(DISTINCT category) FROM Product WHERE year > 1995 SELECT COUNT(DISTINCT category) FROM Product WHERE year > 1995 SELECT category, COUNT(category) FROM Product WHERE year > 1995 SELECT category, COUNT(category) FROM Product WHERE year > 1995

64. 64 Straight aggregation example Purchase(product, date, price, quantity) Q: Find total sales for the entire database: Q: Find total sales of bagels: SELECT SUM(price * quantity) FROM Purchase SELECT SUM(price * quantity) FROM Purchase SELECT SUM(price * quantity) FROM Purchase WHERE product = 'bagel' SELECT SUM(price * quantity) FROM Purchase WHERE product = 'bagel'

65. 65 Largest balance again Acc(name,bal,type) Q: Who has the largest balance? Q: Who has the largest balance of each type? Can we do these with aggregation functions?

66. 66 Straight grouping Group rows together by field values Produces one row for each group  I.e., by each (combin. of) grouped val(s)  Don’t select non-grouped fields Reduces to DISTINCT selections: SELECT product FROM Purchase GROUP BY product SELECT product FROM Purchase GROUP BY product SELECT DISTINCT product FROM Purchase SELECT DISTINCT product FROM Purchase

67. 67 Grouping & aggregation Sometimes want to group and compute aggregations by group  Aggregation op applied to rows in group,  not to all rows in table Purchase(product, date, price, quantity) Find total sales for products that sold for > 0.50: SELECT product, SUM(price*quantity) total FROM Purchase WHERE price >.50 GROUP BY product SELECT product, SUM(price*quantity) total FROM Purchase WHERE price >.50 GROUP BY product

68. 68 Illustrated G&A example Purchase

69. 69 First compute the FROM-WHERE Then GROUP BY product: Illustrated G&A example

70. 70 Finally, aggregate and select: Illustrated G&A example SELECT product, SUM(price*quantity) total FROM Purchase WHERW price >.50 GROUP BY product SELECT product, SUM(price*quantity) total FROM Purchase WHERW price >.50 GROUP BY product

71. 71 Illustrated G&A example GROUP BY may be reduced to (a possibly more complicated) subquery: SELECT product, SUM(price*quantity) total FROM Purchase WHERE price >.50 GROUP BY product SELECT product, SUM(price*quantity) total FROM Purchase WHERE price >.50 GROUP BY product SELECT DISTINCT x.product, (SELECT SUM(y.price*y.quantity) FROM Purchase y WHERE x.product = y.product AND y.price >.50) total FROM Purchase x WHERE x.price >.50 SELECT DISTINCT x.product, (SELECT SUM(y.price*y.quantity) FROM Purchase y WHERE x.product = y.product AND y.price >.50) total FROM Purchase x WHERE x.price >.50

72. 72 For every product, what is the total sales and max quantity sold? Multiple aggregations SELECT product, SUM(price * quantity) SumSales, MAX(quantity) MaxQuantity FROM Purchase WHERE price >.50 GROUP BY product SELECT product, SUM(price * quantity) SumSales, MAX(quantity) MaxQuantity FROM Purchase WHERE price >.50 GROUP BY product

73. 73 Another grouping/aggregation e.g. Movie(title, year, length, studioName) Q: How many total minutes of film have been produced by each studio? Strategy: Divide movies into groups per studio, then add lengths per group

74. 74 Another grouping/aggregation e.g. TitleYearLengthStudio Star Wars1977120Fox Jedi1980105Fox Aviator2004800Miramax Pulp Fiction1995110Miramax Lost in Translation 200395Universal SELECT studio, sum(length) totalLength FROM Movies GROUP BY studio SELECT studio, sum(length) totalLength FROM Movies GROUP BY studio

75. 75 Another grouping/aggregation e.g. TitleYearLengthStudio Star Wars1977120Fox Jedi1980105Fox Aviator2004800Miramax Pulp Fiction1995110Miramax Lost in Translation 200395Universal SELECT studio, sum(length) length FROM Movies GROUP BY studio SELECT studio, sum(length) length FROM Movies GROUP BY studio

76. 76 Another grouping/aggregation e.g. TitleYearLengthStudio Star Wars1977120Fox Jedi1980105Fox Aviator2004800Miramax Pulp Fiction1995110Miramax Lost in Translation 200395Universal StudioLength Fox225 Miramax910 Universal95 SELECT studio, sum(length) totalLength FROM Movies GROUP BY studio SELECT studio, sum(length) totalLength FROM Movies GROUP BY studio

77. 77 Grouping/aggregation example StarsIn(SName,Title,Year) Q: Find the year of each star’s first movie Q: Find the span of each star’s career  Look up first and last movies SELECT sname, min(year) firstyear FROM StarsIn GROUP BY sname SELECT sname, min(year) firstyear FROM StarsIn GROUP BY sname

78. 78 Account types again Acc(name,bal,type) Q: Who has the largest balance of each type? Can we do this with grouping/aggregation?

79. 79 G & A for constructed relations Movie(title,year,producerSsn,length) MovieExec(name,ssn,netWorth) Can do the same thing for larger, non-atomic relations Q: How many mins. of film did each producer make?  What happens to non-producer movie-execs? SELECT name, sum(length) total FROM Movie, MovieExec WHERE producerSsn = ssn GROUP BY name SELECT name, sum(length) total FROM Movie, MovieExec WHERE producerSsn = ssn GROUP BY name

80. 80 HAVING clauses Sometimes want to limit which rows may be grouped Q: How many mins. of film did each rich producer make?  Rich = netWorth > 10000000 Q: Is HAVING necessary here? A: No, could just add rich req. to WHERE SELECT name, sum(length) total FROM Movie, MovieExec WHERE producerSsn = ssn GROUP BY name HAVING netWorth > 10000000 SELECT name, sum(length) total FROM Movie, MovieExec WHERE producerSsn = ssn GROUP BY name HAVING netWorth > 10000000

81. 81 HAVING clauses Sometimes want to limit which rows may be grouped Q: How many mins. of film did each rich producer make?  Old = made movies before 1930 Q: Is HAVING necessary here? SELECT name, sum(length) total FROM Movie, MovieExec WHERE producerSsn = ssn GROUP BY name HAVING min(year) < 1930 SELECT name, sum(length) total FROM Movie, MovieExec WHERE producerSsn = ssn GROUP BY name HAVING min(year) < 1930