如何在MySQL带有给定日期

Question

使用比较毫秒

mysql Ver 14.12 Distrib 5.0.45, for redhat-linux-gnu (i686) using readline 5.0 

我有一个表像这样定义：

+-----------------------+--------------+------+-----+---------+-------+ 
| Field     | Type   | Null | Key | Default | Extra | 
+-----------------------+--------------+------+-----+---------+-------+ 
| id     | bigint(20) | NO | PRI |   |  | 
| user_id    | bigint(20) | NO | MUL |   |  | 
| directory_id   | bigint(20) | NO | MUL |   |  | 
| attribute_name  | varchar(255) | NO |  |   |  | 
| attribute_value  | varchar(255) | YES |  | NULL |  | 
| attribute_lower_value | varchar(255) | YES |  | NULL |  | 
+-----------------------+--------------+------+-----+---------+-------+ 

选择了我最后的验证时间ATTRIBUTE_VALUE返回

+-------------------+-----------------+ 
| attribute_name | attribute_value | 
+-------------------+-----------------+ 
| lastAuthenticated | 1330380013284 | 
+-------------------+-----------------+ 

如果我使用http://www.epochconverter.com/验证该值，它说

Assuming that this timestamp is in milliseconds: 
GMT: Mon, 27 Feb 2012 22:00:13 GMT 

但如果我尝试的代码

mysql> select from_unixtime('1330380013284'); 
+--------------------------------+ 
| from_unixtime('1330380013284') | 
+--------------------------------+ 
| NULL       | 
+--------------------------------+ 

任何人谁立即知道我在这里失踪以下行？

Answer 1

您需要通过1000分以及删除引号，否则你的日期是遥远

SELECT FROM_UNIXTIME(theField/1000); 

SELECT FROM_UNIXTIME(1330380013284/1000); 

参见文档 - 注意MySQL的方法返回秒，不毫秒就像你正在使用：

MySQL docs for UNIX_TIMESTAMP

If called with no argument, returns a Unix timestamp 
(SECONDS since '1970-01-01 00:00:00' UTC) as an unsigned integer 

Answer 2

删除'，你必须通过一个数字。