作者将“两个人是同一个人”的要求表述为:
- 拥有相同的名字和
- 拥有相同数量的电话号码并且所有电话号码都相同。
所以这个问题比看起来要复杂一些(或者也许我只是想多了)。
示例数据和(一个丑陋的,我知道,但总的想法是存在的)一个示例查询,我在下面的测试数据上进行了测试,它似乎工作正常(我使用的是 Oracle 11g R2):
CREATE TABLE contact (
id NUMBER PRIMARY KEY,
name VARCHAR2(40))
;
CREATE TABLE phone_number (
id NUMBER PRIMARY KEY,
contact_id REFERENCES contact (id),
phone VARCHAR2(10)
);
INSERT INTO contact (id, name) VALUES (1, 'John');
INSERT INTO contact (id, name) VALUES (2, 'John');
INSERT INTO contact (id, name) VALUES (3, 'Peter');
INSERT INTO contact (id, name) VALUES (4, 'Peter');
INSERT INTO contact (id, name) VALUES (5, 'Mike');
INSERT INTO contact (id, name) VALUES (6, 'Mike');
INSERT INTO contact (id, name) VALUES (7, 'Mike');
INSERT INTO phone_number (id, contact_id, phone) VALUES (1, 1, '123'); -- John having number 123
INSERT INTO phone_number (id, contact_id, phone) VALUES (2, 1, '456'); -- John having number 456
INSERT INTO phone_number (id, contact_id, phone) VALUES (3, 2, '123'); -- John the second having number 123
INSERT INTO phone_number (id, contact_id, phone) VALUES (4, 2, '456'); -- John the second having number 456
INSERT INTO phone_number (id, contact_id, phone) VALUES (5, 3, '123'); -- Peter having number 123
INSERT INTO phone_number (id, contact_id, phone) VALUES (6, 3, '456'); -- Peter having number 123
INSERT INTO phone_number (id, contact_id, phone) VALUES (7, 3, '789'); -- Peter having number 123
INSERT INTO phone_number (id, contact_id, phone) VALUES (8, 4, '456'); -- Peter the second having number 456
INSERT INTO phone_number (id, contact_id, phone) VALUES (9, 5, '123'); -- Mike having number 456
INSERT INTO phone_number (id, contact_id, phone) VALUES (10, 5, '456'); -- Mike having number 456
INSERT INTO phone_number (id, contact_id, phone) VALUES (11, 6, '123'); -- Mike the second having number 456
INSERT INTO phone_number (id, contact_id, phone) VALUES (12, 6, '789'); -- Mike the second having number 456
-- Mike the third having no number
COMMIT;
-- does not meet the requirements described in the question - will return Peter when it should not
SELECT DISTINCT c.name
FROM contact c JOIN phone_number pn ON (pn.contact_id = c.id)
GROUP BY name, phone_number
HAVING COUNT(c.id) > 1
;
-- returns correct results for provided test data
-- take all people that have a namesake in contact table and
-- take all this person's phone numbers that this person's namesake also has
-- finally (outer query) check that the number of both persons' phone numbers is the same and
-- the number of the same phone numbers is equal to the number of (either) person's phone numbers
SELECT c1_id, name
FROM (
SELECT c1.id AS c1_id, c1.name, c2.id AS c2_id, COUNT(1) AS cnt
FROM contact c1
JOIN contact c2 ON (c2.id != c1.id AND c2.name = c1.name)
JOIN phone_number pn ON (pn.contact_id = c1.id)
WHERE
EXISTS (SELECT 1
FROM phone_number
WHERE contact_id = c2.id
AND phone = pn.phone)
GROUP BY c1.id, c1.name, c2.id
)
WHERE cnt = (SELECT COUNT(1) FROM phone_number WHERE contact_id = c1_id)
AND (SELECT COUNT(1) FROM phone_number WHERE contact_id = c1_id) = (SELECT COUNT(1) FROM phone_number WHERE contact_id = c2_id)
;
-- cleanup
DROP TABLE phone_number;
DROP TABLE contact;
查看 SQL Fiddle:http ://www.sqlfiddle.com/#!4/36cdf/1
已编辑
回答作者的评论:当然我没有考虑到这一点......这是一个修改后的解决方案:
-- new test data
INSERT INTO contact (id, name) VALUES (8, 'Jane');
INSERT INTO contact (id, name) VALUES (9, 'Jane');
SELECT c1_id, name
FROM (
SELECT c1.id AS c1_id, c1.name, c2.id AS c2_id, COUNT(1) AS cnt
FROM contact c1
JOIN contact c2 ON (c2.id != c1.id AND c2.name = c1.name)
LEFT JOIN phone_number pn ON (pn.contact_id = c1.id)
WHERE pn.contact_id IS NULL
OR EXISTS (SELECT 1
FROM phone_number
WHERE contact_id = c2.id
AND phone = pn.phone)
GROUP BY c1.id, c1.name, c2.id
)
WHERE (SELECT COUNT(1) FROM phone_number WHERE contact_id = c1_id) IN (0, cnt)
AND (SELECT COUNT(1) FROM phone_number WHERE contact_id = c1_id) = (SELECT COUNT(1) FROM phone_number WHERE contact_id = c2_id)
;
我们允许没有电话号码(LEFT JOIN)的情况,在外部查询中,我们现在比较人的电话号码的数量 - 它必须等于 0,或者是从内部查询返回的数字。