L07: SQL: Advanced & Practice. CS3200 Database design (sp18 s2) 1/11/2018
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1 L07: SQL: Advanced & Practice CS3200 Database design (sp18 s2) 1/11/
2 Announcements! Please pick up your name card - always bring it - move closer to the front HW3 will be again SQL - individual submission, but discussion in teams Exam on laptop in class - practice exam in class next week Thursday Polls Outline today: - HW1 together - outer joins, nulls 255
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7 More on WITH 260
8 Recall: Witnesses: with aggregates per group (8/8) Second: How to get the product that is sold with max sales? Purchase Product Price Quantity Product sales Bagel 3 20 Banana 70 Bagel 2 20 Banana 1 50 Banana 2 10 Banana SELECT product, sum(quantity) as sales FROM Purchase GROUP BY product HAVING sum(quantity) = (SELECT max (Q) FROM (SELECT sum(quantity) Q FROM Purchase GROUP BY product) X ) 261
9 WITH clause 308 WITH X AS (SELECT product, SUM(quantity) sales FROM Purchase GROUP BY product) SELECT product, sum(quantity) as sales FROM Purchase GROUP BY product HAVING sum(quantity) = (SELECT max (Q) FROM (SELECT sum(quantity) Q FROM Purchase GROUP BY product) X ) 262
10 WITH clause 308 WITH X AS (SELECT product, SUM(quantity) sales FROM Purchase GROUP BY product) SELECT * FROM X WHERE SELECT product, sum(quantity) as sales FROM Purchase GROUP BY product HAVING sum(quantity) = (SELECT max (Q) FROM (SELECT sum(quantity) Q FROM Purchase GROUP BY product) X ) 263
11 WITH clause 308 WITH X AS (SELECT product, SUM(quantity) sales FROM Purchase GROUP BY product) SELECT * FROM X WHERE sales = (SELECT MAX (sales) FROM X) SELECT product, sum(quantity) as sales FROM Purchase GROUP BY product HAVING sum(quantity) = (SELECT max (Q) FROM (SELECT sum(quantity) Q FROM Purchase GROUP BY product) X ) 264
12 WITH clause 308 WITH X AS (SELECT product, SUM(quantity) sales FROM Purchase GROUP BY product) Y AS (SELECT MAX (sales) maxs FROM X) SELECT * FROM X WHERE sales = (SELECT maxs FROM Y)) SELECT product, sum(quantity) as sales FROM Purchase GROUP BY product HAVING sum(quantity) = (SELECT max (Q) FROM (SELECT sum(quantity) Q FROM Purchase GROUP BY product) X ) 265
13 Inner Joins vs. Outer Joins 266
14 Illustration An "inner join": English SELECT * FROM English, French WHERE eid = fid Same as: etext eid One 1 Two 2 Three 3 Four 4 Five 5 Six 6 SELECT * FROM English JOIN French ON eid = fid French fid ftext 1 Un 3 Trois 4 Quatre 5 Cinq 6 Siz 7 Sept 8 Huit etext eid fid ftext One 1 1 Un Three 3 3 Trois Four 4 4 Quatre Five 5 5 Cinq Six 6 6 Siz "JOIN" same as "INNER JOIN"
15 Illustration "FULL JOIN" same as "FULL OUTER JOIN" English etext eid One 1 Two 2 Three 3 Four 4 Five 5 Six 6 French fid ftext 1 Un 3 Trois 4 Quatre 5 Cinq 6 Siz 7 Sept 8 Huit 361 SELECT * FROM English FULL JOIN French ON English.eid = French.fid SELECT * FROM English JOIN French ON eid = fid etext eid fid ftext One 1 1 Un Two 2 NULL NULL Three 3 3 Trois Four 4 4 Quatre Five 5 5 Cinq Six 6 6 Siz NULL NULL 7 Sept NULL NULL 8 Huit SQLite does not support "FULL OUTER JOIN"s L (but "LEFT JOIN" ) 268
16 Illustration English etext eid One 1 Two 2 Three 3 Four 4 Five 5 Six 6 French 361 fid ftext 1 Un 3 Trois 4 Quatre 5 Cinq 6 Siz 7 Sept 8 Huit 2 1,3, 7,8 4-6 = (INNER) JOIN = FULL (OUTER) JOIN Source: Fig. 7-2, Hoffer et al., Modern Database Management, 10ed ed,
17 Detailed Illustration with Examples (follow the link) Check this web page for illustrating examples Source: 270
18 Let's practice SELECT <select_list> FROM A LEFT JOIN B ON A.key = B.key WHERE B.key IS NULL English etext eid One 1 Two 2 Three 3 Four 4 Five 5 Six 6 SELECT * FROM English WHERE eid NOT IN (SELECT fid FROM French) French fid ftext 1 Un 3 Trois 4 Quatre 5 Cinq 6 Siz 7 Sept 8 Huit 361 SELECT etext, eid FROM English LEFT JOIN French ON eid = fid WHERE fid IS NULL Source: 271
19 Missing sales Item(name, category) 334 Purchase2(iName, store, month) An "inner join": SELECT Item.name, Purchase2.store FROM Item, Purchase2 WHERE Item.name = Purchase2.iName Same as: SELECT Item.name, Purchase2.store FROM Item JOIN Purchase2 ON Item.name = Purchase2.iName Item Purchase2 Result Name Category iname Store Month Name Store Gizmo Gadget Gizmo Wiz 8 Gizmo Wiz Camera Photo Camera Ritz 8 Camera Ritz OneClick Photo Camera Wiz 9 Camera Wiz 272
20 Missing sales An "inner join": SELECT Item.name, Purchase2.store FROM Item, Purchase2 WHERE Item.name = Purchase2.iName Same as: SELECT Item.name, Purchase2.store FROM Item INNER JOIN Purchase2 ON Item.name = Purchase2.iName Item(name, category) 334 Purchase2(iName, store, month) "INNER JOIN" same as "JOIN" Item Purchase2 Result Name Category iname Store Month Name Store Gizmo Gadget Gizmo Wiz 8 Gizmo Wiz Camera Photo Camera Ritz 8 Camera Ritz OneClick Photo Camera Wiz 9 Camera Wiz Products that never sold will be lost L 273
21 Outer Joins Item(name, category) 334 Purchase2(iName, store, month) If we want to include the never-sold products, then we need an "outer join": SELECT Item.name, Purchase2.store FROM Item LEFT INNER OUTER JOIN Purchase2 JOIN Purchase2 ON ON Item.name = Purchase2.iName "LEFT OUTER JOIN" same as "LEFT JOIN" Item Purchase2 Result Name Category iname Store Month Name Store Gizmo Gadget Gizmo Wiz 8 Gizmo Wiz Camera Photo Camera Ritz 8 Camera Ritz OneClick Photo Camera Wiz 9 Camera Wiz Now we include those products J OneClick NULL 274
22 Outer Joins Item(name, category) 334 Purchase2(iName, store, month) Same question, but now only for those sold in month = 9: SELECT Item.name, Purchase2.store FROM Item LEFT OUTER JOIN Purchase2 ON Item.name = Purchase2.iName WHERE month = 9 Item Name Gizmo Camera OneClick Category Gadget Photo Photo Explanation: the filter ("month = 9") applies of the result of the outer join. Any tuple that has NULL as month, does not pass the filter Purchase2 iname Store Month Gizmo Wiz 8 Camera Ritz 8 Camera Wiz 9 The products disappeared *despite* outer join L Result Name Camera Store Wiz 275
23 Outer Joins Item(name, category) 334 Purchase2(iName, store, month) SELECT Item.name, Purchase2.store FROM Item LEFT OUTER JOIN Purchase2 ON (Item.name = Purchase2.iName WHERE AND month = 9) Item Name Gizmo Camera OneClick Category Gadget Photo Photo Explanation: now the filter ("month = 9") applies to the right side of the left join *before* joining. NULLs are appended only after filter, during join Same question, but now only for those sold in month = 9: Purchase2 iname Store Month Gizmo Wiz 8 Camera Ritz 8 Camera Wiz 9 parenthesis not required, and just for illustration Result Name Camera Gizmo OneClick Store Wiz NULL NULL Now they are back again J 276
24 Empty Group Problem Item(name, category) 334 Purchase2(iName, store, month) Compute, for each product, the total number of sales in Sept (= month 9) SELECT name, count(*) FROM Item, Purchase2 WHERE name = iname and month = 9 GROUP BY name What s wrong? 277
25 Empty Group Problem Item(name, category) 334 Purchase2(iName, store, month) Compute, for each product, the total number of sales in Sept (= month 9) We need to use an attribute from "Purchase2" to get the correct 0 count. Try "name" from "Item". SELECT name, count(store) FROM Item LEFT JOIN Purchase2 ON name = iname and month = 9 GROUP BY name Now we also get the products with 0 sales 278
26 Empty Group Problem Item(name, category) 334 Purchase2(iName, store, month) Compute, for each product, the total number of sales in Sept (= month 9) We need to use an attribute from "Purchase2" to get the correct 0 count. Try "name" from "Item". SELECT x.name, count(y.store) FROM Item x LEFT OUTER JOIN Purchase2 y ON x.name = y.iname and y.month = 9 GROUP BY x.name Now we also get the products with 0 sales 279
27 Outer Joins: summary Left (outer) join: - Include the left tuple even if there s no match Right (outer) join: - Include the right tuple even if there s no match Full (outer) join: - Include both left and right tuples even if there s no match (inner) join: - Include only the matches 280
28 Processing Multiple Tables Joins Join: a relational operation that causes two or more tables with a common domain to be combined into a single table or view Equi-join: a join in which the joining condition is based on equality between values in the common columns; common columns appear redundantly in the result table A Theta-join allows for arbitrary comparison relationships (e.g., ). An equijoin is a theta join using the equality operator. Natural join: an equi-join in which one of the duplicate columns is eliminated in the result table The common columns in joined tables are usually the primary key of the dominant table and the foreign key of the dependent table in 1:M relationships 281
29 Processing Multiple Tables Joins Left Outer join: a join in which rows from the left table that do not have matching values in common columns are nonetheless included in the result table (as opposed to inner join, in which rows must have matching values in order to appear in the result table) Union join ("Full outer join"): includes all columns from each table in the join, and an instance for each row of each table 282
30 More Practice 283
31 1. What does this query return? Actor id name gender 1 Alice f 2 Bob m Casts aid mid role 1 1 role role role 3 Movie id name 1 Kill Bill 2 Kill Bill SELECT movie.name FROM movie, casts, actor WHERE casts.aid = actor.id AND movie.id = casts.mid AND actor.id NOT IN (SELECT a2.id FROM actor a2 WHERE a2.gender = 'm')? 284
32 1. What does this query return? Actor id name gender 1 Alice f 2 Bob m Casts aid mid role 1 1 role role role 3 Movie id name 1 Kill Bill 2 Kill Bill SELECT movie.name FROM movie, casts, actor WHERE casts.aid = actor.id AND movie.id = casts.mid AND actor.id NOT IN (SELECT a2.id FROM actor a2 WHERE a2.gender = 'm') name Kill Bill 285
33 1. The "Universal Relation" Select * id name gender 1 Alice f 2 Bob m 2 Bob m aid mid role 1 1 role role role 3 id name 1 Kill Bill 1 Kill Bill 1 Kill Bill SELECT * FROM movie, casts, actor WHERE casts.aid = actor.id AND movie.id = casts.mid SELECT a2.id FROM actor a2 WHERE a2.gender = 'm' id 2 More information about the "universal relation": 286
34 2. How to get the number of casts for each movie? Actor Casts id name gender aid mid role 1 Alice f 1 1 role 1 2 Bob m 2 1 role 2 3 Charly m 2 1 role role 4 Movie id name 1 Kill Bill 2 Kill Bill SELECT m.name FROM WHERE GROUP BY? 287
35 2. How to get the number of casts for each movie? Actor Casts id name gender aid mid role 1 Alice f 1 1 role 1 2 Bob m 2 1 role 2 3 Charly m 2 1 role role 4 Movie id name 1 Kill Bill 2 Kill Bill SELECT m.name, count(c.aid) FROM movie m, casts c WHERE m.id = c.mid GROUP BY m.name name Kill Bill 4 (no name) 288
36 2. How to get the number of casts for each movie? Actor Casts id name gender aid mid role 1 Alice f 1 1 role 1 2 Bob m 2 1 role 2 3 Charly m 2 1 role role 4 Movie id name 1 Kill Bill 2 Kill Bill SELECT m.name, count(c.aid) FROM movie m, casts c WHERE m.id = c.mid GROUP BY m.id? 289
37 2. How to get the number of casts for each movie? Actor Casts id name gender aid mid role 1 Alice f 1 1 role 1 2 Bob m 2 1 role 2 3 Charly m 2 1 role role 4 Movie id name 1 Kill Bill 2 Kill Bill SELECT m.name, count(c.aid) FROM movie m, casts c WHERE m.id = c.mid GROUP BY m.id? Error on some databases! (E.g. Microsoft SQLserver) 290
38 2. How to get the number of casts for each movie? Actor Casts id name gender aid mid role 1 Alice f 1 1 role 1 2 Bob m 2 1 role 2 3 Charly m 2 1 role role 4 Movie id name 1 Kill Bill 2 Kill Bill SELECT m.name, count(c.aid) FROM movie m, casts c WHERE m.id = c.mid GROUP BY m.id name (no name) Kill Bill 3 Kill Bill 1? 291
39 2. How to get the number of casts for each movie? Actor Casts id name gender aid mid role 1 Alice f 1 1 role 1 2 Bob m 2 1 role 2 3 Charly m 2 1 role role 4 Movie id name 1 Kill Bill 2 Kill Bill SELECT m.name, count(c.aid) FROM movie m, casts c WHERE m.id = c.mid GROUP BY m.id, m.name name (no name) Kill Bill 3 Kill Bill 1 292
40 2. How to get the number of casts for each movie? Actor Casts id name gender aid mid role 1 Alice f 1 1 role 1 2 Bob m 2 1 role 2 3 Charly m 2 1 role role 4 Movie id name 1 Kill Bill 2 Kill Bill SELECT m.name, count(distinct c.aid) FROM movie m, casts c WHERE m.id = c.mid GROUP BY m.id, m.name name (no name) Kill Bill 2 Kill Bill 1 293
41 3. How to get the number of casts for '%Bill%'? Actor Casts id name gender aid mid role 1 Alice f 1 1 role 1 2 Bob m 2 1 role 2 3 Charly m 2 1 role role 4 Movie id name 1 Kill Bill 2 Kill Bill SELECT m.name, count(distinct c.aid) FROM movie m, casts c WHERE m.id = c.mid GROUP BY m.id, m.name? 294
42 3. How to get the number of casts for '%Bill%'? Actor Casts id name gender aid mid role 1 Alice f 1 1 role 1 2 Bob m 2 1 role 2 3 Charly m 2 1 role role 4 Movie id name 1 Kill Bill 2 Kill Bill SELECT m.name, count(distinct c.aid) FROM movie m, casts c WHERE m.id = c.mid AND m.name like '%Bill%' GROUP BY m.id, m.name 295
43 4. How to get the number of casts for each actor? Actor Casts id name gender aid mid role 1 Alice f 1 1 role 1 2 Bob m 2 1 role 2 3 Charly m 2 1 role role 4 Movie id name 1 Kill Bill 2 Kill Bill SELECT FROM WHERE GROUP BY? name (no name) Alice 1 Bob 2 Charly 1 296
44 4. How to get the number of casts for each actor? Actor Casts id name gender aid mid role 1 Alice f 1 1 role 1 2 Bob m 2 1 role 2 3 Charly m 2 1 role role 4 Movie id name 1 Kill Bill 2 Kill Bill SELECT a.name, count(*) FROM actor a, casts c WHERE a.id = c.aid GROUP BY a.id, a.name name (no name) Alice 1 Bob 2 Charly 1 297
45 4. How to get the number of casts for each actor? Actor Casts id name gender aid mid role 1 Alice f 1 1 role 1 2 Bob m 2 1 role 2 3 Charly m 2 1 role role 4 Movie id name 1 Kill Bill 2 Kill Bill SELECT a.name, count(distinct c.aid) FROM actor a, casts c WHERE a.id = c.aid GROUP BY a.id, a.name? 298
46 4. How to get the number of casts for each actor? Actor Casts id name gender aid mid role 1 Alice f 1 1 role 1 2 Bob m 2 1 role 2 3 Charly m 2 1 role role 4 Movie id name 1 Kill Bill 2 Kill Bill SELECT a.name, count(distinct c.aid) FROM actor a, casts c WHERE a.id = c.aid GROUP BY a.id, a.name name Alice 1 Bob 1 (no name) Will always show 1 (since there is Charly 1 only one distinct aid per group grouped by aid *by def.*) 299
47 4. How to get the number of movies for each actor? Actor Casts Movie id name gender aid mid role id name 1 Alice f 1 1 role 1 1 Kill Bill 2 Bob m 2 1 role 2 2 Kill Bill 3 Charly m 2 1 role role 4 SELECT FROM WHERE GROUP BY? name (no name) Alice 1 Bob 1 Charly 1 300
48 4. How to get the number of movies for each actor? Actor Casts Movie id name gender aid mid role id name 1 Alice f 1 1 role 1 1 Kill Bill 2 Bob m 2 1 role 2 2 Kill Bill 3 Charly m 2 1 role role 4 SELECT a.name, count(distinct c.mid) FROM actor a, casts c WHERE a.id = c.aid GROUP BY a.id, a.name name (no name) Alice 1 Bob 1 Charly 1 301
49 4. How to get the number of casts for each male actor? Actor Casts Movie id name gender aid mid role id name 1 Alice f 1 1 role 1 1 Kill Bill 2 Bob m 2 1 role 2 2 Kill Bill 3 Charly m 2 1 role role 4 SELECT FROM WHERE AND GROUP BY? name (no name) Bob 2 Charly 1 302
50 4. How to get the number of casts for each male actor? Actor Casts Movie id name gender aid mid role id name 1 Alice f 1 1 role 1 1 Kill Bill 2 Bob m 2 1 role 2 2 Kill Bill 3 Charly m 2 1 role role 4 SELECT a.name, count(distinct a.id) FROM actor a, casts c WHERE a.id = c.aid AND a.gender = 'm' GROUP BY a.id, a.name name (no name) Bob 2 Charly 1 303
51 5. How to get male actors with number of casts> 1 Actor Casts id name gender aid mid role 1 Alice f 1 1 role 1 2 Bob m 2 1 role 2 3 Charly m 2 1 role role 4 Movie id name 1 Kill Bill 2 Kill Bill SELECT FROM WHERE AND GROUP BY? name Bob 2 (no name) 304
52 5. How to get male actors with number of casts> 1 Actor Casts id name gender aid mid role 1 Alice f 1 1 role 1 2 Bob m 2 1 role 2 3 Charly m 2 1 role role 4 Movie id name 1 Kill Bill 2 Kill Bill SELECT a.name, count(*) FROM actor a, casts c WHERE a.id = c.aid AND a.gender = 'm' GROUP BY a.id, a.name HAVING count(*) > 1 name Bob 2 (no name) 305
53 SQL injection 306
54 Simple SQL Query Product PName Price Category Manufacturer Gizmo $19.99 Gadgets GizmoWorks Powergizmo $29.99 Gadgets GizmoWorks SingleTouch $ Photography Canon MultiTouch $ Household Hitachi SELECT PName FROM Product WHERE category='gadgets' PName Gizmo Powergizmo 307
55 Parameterized SQL Query Varies between DBMSs. The following is the semantics for SQL server. You do not need to know that for exams Product PName Price Category Manufacturer Gizmo $19.99 Gadgets GizmoWorks Powergizmo $29.99 Gadgets GizmoWorks SingleTouch $ Photography Canon MultiTouch $ Household Hitachi VARCHAR(50) = 'Gadgets' SELECT PName FROM Product WHERE category=@num PName Gizmo Powergizmo 308
56 What happened here? Source:
57 It's called SQL injection: Version 1 Source:
58 It's called SQL injection: Version 2 Source:
59 SQL injection Source: 312
60 TJX Credit Card Numbers Theft In 2006, a $17.5 billion Fortune 500 firm Unauthorized intrusion resulting in lost of credit/debit card information - From May 2006 to January 2007? - From July 2005 to December 2006? TJX s overall losses: $1.35 billion $4.5 billion 313
Please pick up your name card
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