leetcode196 删除重复的电子邮箱 Delete Duplicate Emails

编写一个SQL查询来删除Person表中所有重复的电子邮件，在重复的邮件中只保留Id最小的邮件。

创建表和数据：

-- ----------------------------
-- Table structure for `person`
-- ----------------------------
DROP TABLE IF EXISTS `person`;
CREATE TABLE `person` (
 `Id` int(11) DEFAULT NULL,
 `Email` varchar(255) DEFAULT NULL
) ENGINE=InnoDB DEFAULT CHARSET=utf8mb4;
-- ----------------------------
-- Records of person
-- ----------------------------
INSERT INTO `person` VALUES (‘1‘,‘[email protected]‘);
INSERT INTO `person` VALUES (‘2‘,‘[email protected]‘);
INSERT INTO `person` VALUES (‘3‘,‘[email protected]‘);

解法：

1.按email分组，找到每组id最小的行。

(
SELECT MIN(P.id) AS `Id`,P.Email
FROM Person AS P
GROUP BY P.email
) AS P2

从原表中DELETE掉不在表2中的行。

DELETE P1
FROM Person AS P1,
(
SELECT MIN(P.id) AS `Id`,P.Email
FROM Person AS P
GROUP BY P.email
) AS P2
WHERE P1.Id != P2.Id AND P1.Email = P2.Email

注意：DELETE与FROM之间，只放置了P1。说明只删除P1中的行，不删除P2中的行。

FROM后，P1和P2叉积。当然也可以应用内连接。

DELETE P1
FROM Person AS P1
JOIN
(
SELECT MIN(P.id) AS `Id`,P.Email
FROM Person AS P
GROUP BY P.email
) AS P2
ON (P1.Id != P2.Id AND P1.Email = P2.Email)

2.既然DELETE 可以结合JOIN，直接表自连接，删除所有比当前行ID大的行。

DELETE P1
FROM Person AS P1 JOIN Person AS P2 ON (P1.email = P2.email AND P1.id > P2.id)

3.从集合的角度看。设，按email分组，找到每组id最小的行。 命名为集合A。那么，从全集U中保留集合A，删除U减去A的差集。再删除差集数据。应用LEFT JOIN。

DELETE U
FROM Person AS U
LEFT JOIN (
    SELECT MIN(id) AS `id`,email
    FROM Person
    GROUP BY email
) AS A ON (U.email = A.email AND U.id = A.id)
WHERE A.id IS NULL

原文地址：https://www.cnblogs.com/forever-fortunate/p/11723329.html