用一个字段很容易找到副本:
SELECT email, COUNT(email)
FROM users
GROUP BY email
HAVING COUNT(email) > 1所以如果我们有张桌子
ID NAME EMAIL
1 John asd@asd.com
2 Sam asd@asd.com
3 Tom asd@asd.com
4 Bob bob@asd.com
5 Tom asd@asd.com这个查询将给我们约翰,山姆,汤姆,汤姆,因为他们都有相同的email。
但是,我想要的是使用相同的email 和 name复制。
也就是说,我想要“汤姆”,“汤姆”。
我需要这样做的原因是:我犯了一个错误,允许插入重复的name和email值。现在我需要删除/更改副本,所以我需要首先找到它们。
发布于 2010-04-07 18:20:18
SELECT
name, email, COUNT(*)
FROM
users
GROUP BY
name, email
HAVING
COUNT(*) > 1只需对这两列进行分组。
注意:较早的ANSI标准是在组中有所有非聚合列,但随着“功能依赖”的思想发生了变化。
在关系数据库理论中,函数依赖是数据库关系中两组属性之间的约束。换句话说,函数依赖是描述关系中属性之间关系的约束。
支持是不一致的:
sql_mode=only_full_group_by:
发布于 2010-04-07 18:22:43
试试这个:
declare @YourTable table (id int, name varchar(10), email varchar(50))
INSERT @YourTable VALUES (1,'John','John-email')
INSERT @YourTable VALUES (2,'John','John-email')
INSERT @YourTable VALUES (3,'fred','John-email')
INSERT @YourTable VALUES (4,'fred','fred-email')
INSERT @YourTable VALUES (5,'sam','sam-email')
INSERT @YourTable VALUES (6,'sam','sam-email')
SELECT
name,email, COUNT(*) AS CountOf
FROM @YourTable
GROUP BY name,email
HAVING COUNT(*)>1产出:
name email CountOf
---------- ----------- -----------
John John-email 2
sam sam-email 2
(2 row(s) affected)如果要使用dups的if,请使用以下命令:
SELECT
y.id,y.name,y.email
FROM @YourTable y
INNER JOIN (SELECT
name,email, COUNT(*) AS CountOf
FROM @YourTable
GROUP BY name,email
HAVING COUNT(*)>1
) dt ON y.name=dt.name AND y.email=dt.email产出:
id name email
----------- ---------- ------------
1 John John-email
2 John John-email
5 sam sam-email
6 sam sam-email
(4 row(s) affected)若要删除重复项,请尝试:
DELETE d
FROM @YourTable d
INNER JOIN (SELECT
y.id,y.name,y.email,ROW_NUMBER() OVER(PARTITION BY y.name,y.email ORDER BY y.name,y.email,y.id) AS RowRank
FROM @YourTable y
INNER JOIN (SELECT
name,email, COUNT(*) AS CountOf
FROM @YourTable
GROUP BY name,email
HAVING COUNT(*)>1
) dt ON y.name=dt.name AND y.email=dt.email
) dt2 ON d.id=dt2.id
WHERE dt2.RowRank!=1
SELECT * FROM @YourTable产出:
id name email
----------- ---------- --------------
1 John John-email
3 fred John-email
4 fred fred-email
5 sam sam-email
(4 row(s) affected)发布于 2010-04-07 18:20:36
试试这个:
SELECT name, email
FROM users
GROUP BY name, email
HAVING ( COUNT(*) > 1 )https://stackoverflow.com/questions/2594829
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