你這個(gè)問題是否可以理解為“驗(yàn)證表里的某一不確定的字段的值，沒有重復(fù)”？

如果是的話，那么你就可以先查出表的總記錄數(shù)，然后在查出該字段去重之后的總記錄數(shù)，

對(duì)比兩次結(jié)果，如果數(shù)量相同則沒有重復(fù)；如果數(shù)量不相等，則代表有重復(fù)的內(nèi)容。

下面是一個(gè)示例：

mysql> select * from user;

+----+------+-----------+

| id | name | address |

+----+------+-----------+

| 1 | aaa | beijing |

| 2 | bbb | shanghai |

| 3 | ccc | hangzhou |

| 5 | aaa | chongqing |

+----+------+-----------+

4 rows in set (0.00 sec)

mysql> select (select count(*) from user) = (select count(*) from (select distinct name from user) t);

+-----------------------------------------------------------------------------------------+

| (select count(*) from user) = (select count(*) from (select distinct name from user) t) |

+-----------------------------------------------------------------------------------------+

| 0 |

+-----------------------------------------------------------------------------------------+

1 row in set (0.01 sec)

mysql> delete from user where id=5;

Query OK, 1 row affected (0.05 sec)

mysql> select * from user;

+----+------+----------+

| id | name | address |

+----+------+----------+

| 1 | aaa | beijing |

| 2 | bbb | shanghai |

| 3 | ccc | hangzhou |

+----+------+----------+

3 rows in set (0.00 sec)

mysql> select (select count(*) from user) = (select count(*) from (select distinct name from user) t);

+-----------------------------------------------------------------------------------------+

| (select count(*) from user) = (select count(*) from (select distinct name from user) t) |

+-----------------------------------------------------------------------------------------+

| 1 |

+-----------------------------------------------------------------------------------------+

1 row in set (0.00 sec)

總結(jié)

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