SELECT c.name
FROM Customer c
WHERE NOT EXISTS(SELECT w.WID
FROM Woker w
WHERE NOT EXISTS(SELECT la
FROM look_after la
WHERE la.CID = c.CID
AND la.WID = w.WID));
我不知道代码意味着什么......有人能告诉我代码的作用吗? C是一名顾客,他将照顾一名工人。
答案 0 :(得分:8)
查询选择所有员工照顾的客户。
双not exists
是实现relational division的一种方式。
答案 1 :(得分:2)
以Andomar的优秀答案为例说明:
-- Some test data
DROP SCHEMA tmp CASCADE;
CREATE SCHEMA tmp ;
SET search_path=tmp;
CREATE TABLE persons
( person_id INTEGER NOT NULL PRIMARY KEY
, pname varchar
);
INSERT INTO persons( person_id, pname ) VALUES
(1 , 'Bob' ) ,(2 , 'Alice' ) ,(3 , 'Carol' )
;
CREATE TABLE movies
( movie_id INTEGER NOT NULL PRIMARY KEY
, mname varchar
);
INSERT INTO movies( movie_id, mname ) VALUES
(1, 'The Blues brothers' ), (2, 'Modern Times' ), (3, 'The Sound of Music' )
,(4, 'Amadeus' ), (5, 'Never say Never' )
;
-- people that have seen a particular movie
CREATE TABLE person_movie
( person_id INTEGER NOT NULL
, movie_id INTEGER NOT NULL
, PRIMARY KEY ( person_id, movie_id)
);
INSERT INTO person_movie( person_id, movie_id) VALUES
(1 ,5 ) ,(1 ,1 )
,(2 ,5 ) ,(2 ,4 ) ,(2 ,1 ) ,(2 ,3 ) ,(2 ,2 )
,(3 ,1 ) ,(3 ,3 )
;
-- Find the people that have seen ALL the movies
-- This is equivalent to:
-- Find persons for whom NO movie exists that (s)he has NOT seen
SELECT * FROM persons p
WHERE NOT EXISTS (
SELECT * FROM movies m
WHERE NOT EXISTS (
SELECT * FROM person_movie pm
WHERE pm.movie_id = m.movie_id
AND pm.person_id = p.person_id
)
);
-- similar: Find the movies that have been seen by ALL people
SELECT * FROM movies m
WHERE NOT EXISTS (
SELECT * FROM persons p
WHERE NOT EXISTS (
SELECT * FROM person_movie pm
WHERE pm.movie_id = m.movie_id
AND pm.person_id = p.person_id
)
);
结果:
person_id | pname
-----------+-------
2 | Alice
(1 row)
movie_id | mname
----------+--------------------
1 | The Blues brothers
(1 row)