HW1 is due tonight HW2 groups are assigned. Outline today: - nested queries and witnesses - We start with a detailed example! - outer joins, nulls?
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1 L05: SQL 183
2 Announcements! HW1 is due tonight HW2 groups are assigned Outline today: - nested queries and witnesses - We start with a detailed example! - outer joins, nulls? 184
3 Small IMDB schema (SQLite) 300 Movie_genre Movie Movie_director Director mid id did id genre name mid fname year lname rank Cast aid mid role Actor id Director_genre did genre prob fname lname gender 185
4 Big IMDB schema (Postgres) Genre Movie Movie_directors Directors mid id did id genre name mid fname year lname Casts pid mid Actor role id fname lname gender 186
5 Theta joins What do these queries compute? R a 1 2 U a SELECT R.a, U.a FROM R, U? WHERE R.a < U.a SELECT R.a, U.a FROM R, U? WHERE R.a <= U.a A Theta-join allows for arbitrary comparison relationships (such as ). An equijoin is a theta join using the equality operator. 187
6 Theta joins What do these queries compute? R a 1 2 U a a b 4 SELECT FROM WHERE R.a, U.a as b R, U R.a < U.a SELECT FROM WHERE R.a, U.a as b R, U R.a >= U.a a b 2 2 A Theta-join allows for arbitrary comparison relationships (such as ). An equijoin is a theta join using the equality operator. 188
7 Witnesses: with joins (1/6) Product2 (pname, price, cid) 315 Company2 (cid, cname, city) Q: For each company, find the most expensive product + its price 189
8 Witnesses: with joins (2/6) Product2 (pname, price, cid) 315 Company2 (cid, cname, city) Q: For each company, find the most expensive product + its price Our Plan: 1. Compute max price in a subquery for a given company 190
9 Witnesses: with joins (2/6) Product2 (pname, price, cid) 315 Company2 (cid, cname, city) Q: For each company, find the most expensive product + its price Our Plan: 1. Compute max price in a subquery for a given company 1. SELECT max(p1.price) FROM Product2 P1 WHERE P1.cid = 1 191
10 Witnesses: with joins (3/6) Product2 (pname, price, cid) 315 Company2 (cid, cname, city) Q: For each company, find the most expensive product + its price Our Plan: 1. Compute max price in a subquery for a given company 2. Compute each product and its price, per company 1. SELECT max(p1.price) FROM Product2 P1 WHERE P1.cid = 1 192
11 Witnesses: with joins (3/6) Product2 (pname, price, cid) 315 Company2 (cid, cname, city) Q: For each company, find the most expensive product + its price Our Plan: 1. Compute max price in a subquery for a given company 2. Compute each product and its price, per company 2. SELECT C2.cname, P2.pname, P2.price FROM Company2 C2, Product2 P2 WHERE C2.cid = P2.cid 1. SELECT max(p1.price) FROM Product2 P1 WHERE P1.cid = 1 193
12 Witnesses: with joins (3/6) Product2 (pname, price, cid) 315 Company2 (cid, cname, city) Q: For each company, find the most expensive product + its price Our Plan: 1. Compute max price in a subquery for a given company 2. Compute each product and its price, per company 3. Compare the price with the max price 2. SELECT C2.cname, P2.pname, P2.price FROM Company2 C2, Product2 P2 WHERE C2.cid = P2.cid 1. SELECT max(p1.price) FROM Product2 P1 WHERE P1.cid = 1 194
13 Witnesses: with joins (4/6) Product2 (pname, price, cid) 315 Company2 (cid, cname, city) Q: For each company, find the most expensive product + its price Our Plan: 1. Compute max price in a subquery for a given company 2. Compute each product and its price, per company 3. Compare the price with the max price SELECT C2.cname, P2.pname, P2.price FROM Company2 C2, Product2 P2 WHERE C2.cid = P2.cid and P2.price = (SELECT max(p1.price) FROM Product2 P1 WHERE P1.cid = C2.cid) How many aliases do we actually need? 195
14 Witnesses: with joins (5/6) Product2 (pname, price, cid) 315 Company2 (cid, cname, city) Q: For each company, find the most expensive product + its price Our Plan: 1. Compute max price in a subquery for a given company 2. Compute each product and its price, per company and compare the price with the max price SELECT cname, pname, price FROM Company2, Product2 WHERE Company2.cid = Product2.cid and price = (SELECT max(price) FROM Product2 WHERE cid = Company2.cid) We need no single alias here. Next: can we eliminate the max operator in the inner query? 196
15 Witnesses: with joins (6/6) Product2 (pname, price, cid) 315 Company2 (cid, cname, city) Q: For each company, find the most expensive product + its price Our Plan: 1. Compute all prices in a subquery, for a given company 2. Compute each product and its price, per company and compare the price with the all prices SELECT cname, pname, price FROM Company2, Product2 WHERE Company2.cid = Product2.cid and price >= ALL (SELECT price FROM Product2 WHERE cid = Company2.cid) But: "ALL" does not work in SQLite L 197
16 Witnesses: with FROM (1/3) Product2 (pname, price, cid) 315 Company2 (cid, cname, city) Q: For each company, find the most expensive product + its price Another Plan: 1. Create a table that lists the max price for each company id 2. Join this table with the remaining tables 1. SELECT cid, max(price) as MP FROM Product2 GROUP BY cid Finding the maximum price for each company was easy. But we want the witnesses, i.e. the products with max price. 198
17 Witnesses: with FROM (2/3) Product2 (pname, price, cid) 315 Company2 (cid, cname, city) Q: For each company, find the most expensive product + its price Another Plan: 1. Create a table that lists the max price for each company id 2. Join this table with the remaining tables 2. SELECT C2.cname, P2.pname, X.MP FROM Company2 C2, Product2 P2, (SELECT cid, max(price) as MP FROM Product2 GROUP BY cid) as X WHERE C2.cid = P2.cid and C2.cid = X.cid and P2.price = X.MP Let's write the same query with a "WITH" clause 199
18 Witnesses: with FROM (3/3) Product2 (pname, price, cid) 315 Company2 (cid, cname, city) Q: For each company, find the most expensive product + its price Another Plan with WITH: 1. Create a table that lists the max price for each company id 2. Join this table with the remaining tables WITH X(cid, MP) as (SELECT cid, max(price) FROM Product2 GROUP BY cid) SELECT C2.cname, P2.pname, X.MP FROM Company2 C2, Product2 P2, X WHERE C2.cid = P2.cid and C2.cid = X.cid and P2.price = X.MP 200
19 Witnesses: with aggregates per group (1/8) First: How to get the product that is sold with maximum price? Purchase Product Price Quantity Product mp Bagel 3 20 Banana 4 Bagel 2 20 Banana 1 50 Banana 2 10 Banana SELECT product, max(price) as mp FROM WHERE GROUP BY HAVING??? 201
20 Witnesses: with aggregates per group (2/8) First: How to get the product that is sold with maximum price? 1) Find the maximum price Purchase Product Price Quantity (no name) Bagel Bagel 2 20 Banana 1 50 Banana 2 10 Banana SELECT max(price) FROM Purchase 202
21 Witnesses: with aggregates per group (3/8) First: How to get the product that is sold with maximum price? 2) Now you need to find product with price = maximum price Purchase Product Price Quantity Product mp Bagel 3 20 Banana 4 Bagel 2 20 Banana 1 50 Banana 2 10 Banana SELECT P2.product, P2.price as mp FROM Purchase P2 WHERE P2.price = ( SELECT max(price) FROM Purchase ) 203
22 Witnesses: with aggregates per group (4/8) First: How to get the product that is sold with maximum price? Another way to formulate this query Purchase Product Price Quantity Product mp Bagel 3 20 Banana 4 Bagel 2 20 Banana 1 50 Banana 2 10 Banana SELECT P2.product, P2.price as mp FROM Purchase P2 WHERE P2.price >= ALL ( SELECT price FROM Purchase ) 204
23 Witnesses: with aggregates per group (5/8) Second: How to get the product that is sold with max sales (=quanities sold)? Purchase Product Price Quantity Product sales Bagel 3 20 Banana 70 Bagel 2 20 Banana 1 50 Banana 2 10 Banana SELECT FROM WHERE GROUP BY HAVING??? 205
24 Witnesses: with aggregates per group (6/8) Second: How to get the product that is sold with max sales (=quanities sold)? 1: find the total sales (sum of quantity) for each product Purchase Product Price Quantity Product sales Bagel 3 20 Bagel 40 Bagel 2 20 Banana 70 Banana 1 50 Banana 2 10 Banana SELECT product, sum(quantity) as sales FROM Purchase GROUP BY product 206
25 Witnesses: with aggregates per group (7/8) Second: How to get the product that is sold with max sales? 2: we can use the same query as nested query Purchase Product Price Quantity (no name) Bagel Bagel Banana 1 50 Banana 2 10 Banana SELECT sum(quantity) FROM Purchase GROUP BY product 207
26 Witnesses: with aggregates per group (8/8) Second: How to get the product that is sold with max sales? 3: putting things together 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) >= ALL ( SELECT sum(quantity) FROM Purchase GROUP BY product ) Next: Can you write the query without the "ALL" quanitfier? 208
27 Witnesses: with aggregates per group (8/8) Second: How to get the product that is sold with max sales? Another way to formulate this query without "ALL" 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 ) 209
28 Understanding nested queries 210
29 More SQL Queries Sailors (sid, sname, rating, age) Reserves (sid, bid, day) Boats (bid, bname, color) 340 Query from: Ramakrishnan, Gehrke: Database management systems, 2nd ed (2000) 211
30 More nested Queries 1 Sailors (sid, sname, rating, age) Reserves (sid, bid, day) Boats (bid, bname, color) 340 Q: Find the names of sailors who have reserved a red boat. SELECT S.sname FROM Sailors S WHERE S.sid IN ( SELECT R.sid FROM Reserves R WHERE R.bid IN ( SELECT B.bid FROM Boats B WHERE B.color = 'red')) Query from: Ramakrishnan, Gehrke: Database management systems, 2nd ed (2000) 212
31 More nested Queries 2 Sailors (sid, sname, rating, age) Reserves (sid, bid, day) Boats (bid, bname, color) 340 Q: Find the names of sailors who have reserved a boat that is not red. SELECT S.sname FROM Sailors S WHERE S.sid IN ( SELECT R.sid FROM Reserves R WHERE R.bid IN not IN ( SELECT B.bid FROM Boats B WHERE B.color = 'red')) They must have reserved at least one boat in another color Query from: Ramakrishnan, Gehrke: Database management systems, 2nd ed (2000) 213
32 More nested Queries 3 Sailors (sid, sname, rating, age) Reserves (sid, bid, day) Boats (bid, bname, color) 340 Q: Find the names of sailors who have not reserved a red boat. SELECT S.sname FROM Sailors S WHERE S.sid notin ( SELECT R.sid FROM Reserves R WHERE R.bid not IN IN ( SELECT B.bid FROM Boats B WHERE B.color = 'red')) They can have reserved 0 or more boats in another color, but must not have reserved any red boat Query from: Ramakrishnan, Gehrke: Database management systems, 2nd ed (2000) 214
33 More nested Queries 4 Sailors (sid, sname, rating, age) Reserves (sid, bid, day) Boats (bid, bname, color) 340 = Find the names of sailors who have reserved only red boats Q: Find the names of sailors who have not reserved a boat that is not red. SELECT S.sname FROM Sailors S WHERE S.sid notin ( SELECT R.sid FROM Reserves R WHERE R.bid IN not IN ( SELECT B.bid FROM Boats B WHERE B.color = 'red')) Query from: Ramakrishnan, Gehrke: Database management systems, 2nd ed (2000) 215
34 More nested Queries 5 Sailors (sid, sname, rating, age) Reserves (sid, bid, day) Boats (bid, bname, color) 340 = Find the names of sailors who have reserved all red boats Q: Find the names of sailors so there is no red boat that is not reserved by him. SELECT S.sname FROM Sailors S WHERE not exists ( SELECT B.bid FROM Boats B WHERE B.color = 'red' AND not exists ( SELECT R.bid FROM Reserves R WHERE R.bid = B.bid AND R.sid = S.sid)) To understand semantics of nested queries, think of a nested loops evaluation: For each Sailors tuple, check the qualification by computing the subquery Query from: Ramakrishnan, Gehrke: Database management systems, 2nd ed (2000) 216
35 Once more: 1 Sailors (sid, sname, rating, age) Reserves (sid, bid, day) Boats (bid, bname, color) 340 Q: Find the names of sailors who have reserved a red boat. SELECT S.sname FROM Sailors S WHERE S.sid IN ( SELECT R.sid FROM Reserves R WHERE R.bid IN ( SELECT B.bid FROM Boats B WHERE B.color = 'red')) Query from: Ramakrishnan, Gehrke: Database management systems, 2nd ed (2000) 217
36 Once more: 2 Sailors (sid, sname, rating, age) Reserves (sid, bid, day) Boats (bid, bname, color) 340 Q: Find the names of sailors who have reserved a boat that is not red. SELECT S.sname FROM Sailors S WHERE S.sid IN ( SELECT R.sid FROM Reserves R WHERE R.bid not IN ( SELECT B.bid FROM Boats B WHERE B.color = 'red')) Query from: Ramakrishnan, Gehrke: Database management systems, 2nd ed (2000) 218
37 Once more: 3 Sailors (sid, sname, rating, age) Reserves (sid, bid, day) Boats (bid, bname, color) 340 Q: Find the names of sailors who have not reserved a red boat. SELECT S.sname FROM Sailors S WHERE S.sid notin ( SELECT R.sid FROM Reserves R WHERE R.bid IN ( SELECT B.bid FROM Boats B WHERE B.color = 'red')) Query from: Ramakrishnan, Gehrke: Database management systems, 2nd ed (2000) 219
38 Once more: 4 Sailors (sid, sname, rating, age) Reserves (sid, bid, day) Boats (bid, bname, color) 340 = Find the names of sailors who have reserved only red boats Q: Find the names of sailors who have not reserved a boat that is not red. SELECT S.sname FROM Sailors S WHERE S.sid notin ( SELECT R.sid FROM Reserves R WHERE R.bid not IN ( SELECT B.bid FROM Boats B WHERE B.color = 'red')) Query from: Ramakrishnan, Gehrke: Database management systems, 2nd ed (2000) 220
39 Once more: 5 Sailors (sid, sname, rating, age) Reserves (sid, bid, day) Boats (bid, bname, color) 340 = Find the names of sailors who have reserved all red boats Q: Find the names of sailors so there is no red boat that is not reserved by him. SELECT S.sname FROM Sailors S WHERE not exists ( SELECT B.bid FROM Boats B WHERE B.color = 'red' AND not exists ( SELECT R.bid FROM Reserves R WHERE R.bid = B.bid AND R.sid = S.sid)) Query from: Ramakrishnan, Gehrke: Database management systems, 2nd ed (2000) 221
40 QueryViz Input: Schema Your Input Specify or choose a pre-defined schema Employee and Department EMP(eid,name,sal,did) DEPT(did,dname,mgr) help Input Query Specify or choose an SQL Query Query 8 SELECT e1.name FROM EMP e1, EMP e2, DEPT d WHERE e1.did = d.did AND d.mgr = e2.eid AND e1.sal > e2.sal help Output: Visualization Submit QueryViz Result
41 Multiset operations (Intersect, Except) 223
42 Recall Multisets (Bags) Multiset X Tuple (1, a) (1, a) (1, b) (2, c) (2, c) (2, c) (1, d) (1, d) Equivalent Representations of a Multiset λ X = Count of tuple in X (Items not listed have implicit count 0) Multiset X Tuple λ(x) (1, a) 2 (1, b) 1 (2, c) 3 (1, d) 2 Note: In a set all counts are {0,1}. 224
43 Generalizing Set Operations to Multiset Operations Multiset X Multiset Y Multiset Z Tuple λ(x) Tuple λ(y) Tuple λ(z) (1, a) 2 (1, a) 5 (1, a) 2 = (1, b) 0 (1, b) 1 (1, b) 0 (2, c) 3 (2, c) 2 (2, c) 2 (1, d) 0 (1, d) 2 (1, d) 0 λ Z = min(λ X, λ Y ) For sets, this is intersection 225
44 Generalizing Set Operations to Multiset Operations Multiset X Multiset Y Multiset Z Tuple λ(x) Tuple λ(y) Tuple λ(z) (1, a) 2 (1, a) 5 (1, a) 7 = (1, b) 0 (1, b) 1 (1, b) 1 (2, c) 3 (2, c) 2 (2, c) 5 (1, d) 0 (1, d) 2 (1, d) 2 λ Z = λ X + λ Y For sets, this is union 226
45 Multiset Operations in SQL 227
46 Explicit Set Operators: INTERSECT r. A r. A = s. A r. A r. A = t. A} SELECT R.A FROM R, S WHERE R.A=S.A INTERSECT SELECT R.A FROM R, T WHERE R.A=T.A Q 1 Q 2 228
47 UNION r. A r. A = s. A r. A r. A = t. A} SELECT R.A FROM R, S WHERE R.A=S.A UNION SELECT R.A FROM R, T WHERE R.A=T.A Q 1 Q 2 Why aren t there duplicates? By default: SQL uses set semantics for INTERSECT and UNION! What if we want duplicates? 229
48 UNION ALL r. A r. A = s. A r. A r. A = t. A} SELECT R.A FROM R, S WHERE R.A=S.A UNION ALL SELECT R.A FROM R, T WHERE R.A=T.A Q 1 Q 2 ALL indicates Multiset operations 230
49 EXCEPT r. A r. A = s. A \{r. A r. A = t. A} SELECT R.A FROM R, S WHERE R.A=S.A EXCEPT SELECT R.A FROM R, T Q 1 Q 2 WHERE R.A=T.A What is the multiset version? 231
50 INTERSECT and EXCEPT* (SELECT R.a, R.b FROM R) INTERSECT (SELECT S.a, S.b FROM S) SELECT R.a, R.b FROM R WHERE EXISTS (SELECT * FROM S WHERE R.a=S.a and R.b=S.b) R(a,b) S(a,b) If R, S have no duplicates, then can write without sub-queries (HOW?) (SELECT R.A, R.B FROM R) EXCEPT (SELECT S.A, S.B FROM S) SELECT R.A, R.B FROM R WHERE NOT EXISTS (SELECT * FROM S WHERE R.A=S.A and R.B=S.B) *Not in all DBMSs. (SQLlite does not like the parentheses, Oracle uses "MINUS" instead of "EXCEPT")
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