问题：执行如下语句时出现大量Warning 1292

mysql> UPDATE show_info INNER JOIN show_like ON show_like.show_id = show_info.show_id SET show_info.status = 0 WHERE show_like.show_id = 1 AND show_like.cid = ‘taobao_mecity‘;

Query OK, 0 rows affected, 99 warnings (0.00 sec)

Rows matched: 0  Changed: 0  Warnings: 0

mysql> show warnings;

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

| Level   | Code | Message                                                               |

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

| Warning | 1292 | Truncated incorrect DOUBLE value: ‘25a3c516a4c15eda917963e48a254‘  |

| Warning | 1292 | Truncated incorrect DOUBLE value: ‘fd3de128303ed731e2e3dda447908‘  |

| Warning | 1292 | Truncated incorrect DOUBLE value: ‘b1f7f42f3e184812b5dff70fb3a35‘  |

| Warning | 1292 | Truncated incorrect DOUBLE value: ‘a4e7d5114b0f79987720de910fcf3‘  |

| Warning | 1292 | Truncated incorrect DOUBLE value: ‘0b19c482a49346eaedde3309e43a9‘  |

| Warning | 1292 | Truncated incorrect DOUBLE value: ‘284f24a76fb532ad60612580b30b2‘  |

| Warning | 1292 | Truncated incorrect DOUBLE value: ‘ffbee81cd49dc9925fa269acd9eff‘  |

解决：

desc show_like 发现show_id 是字符串.将sql中的show_id = 1 改为show_id = ‘1‘即可。