leetcode196 刪除重複的電子郵箱 Delete Duplicate Emails

編寫一個SQL查詢來刪除Person表中全部重複的電子郵件，在重複的郵件中只保留Id最小的郵件。

 

建立表和數據：

 

-- ----------------------------
-- 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','john@example.com');
INSERT INTO `person` VALUES ('2','bob@example.com');
INSERT INTO `person` VALUES ('3','john@example.com');

 

解法：

1.按email分組，找到每組id最小的行。

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

從原表中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與FROM之間，只放置了P1。說明只刪除P1中的行，不刪除P2中的行。

FROM後，P1和P2叉積。固然也能夠應用內鏈接。

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 能夠結合JOIN，直接表自鏈接，刪除全部比當前行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.從集合的角度看。設，按email分組，找到每組id最小的行。 命名爲集合A。那麼，從全集U中保留集合A，刪除U減去A的差集。再刪除差集數據。應用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