Question

如何从“2012-02”开始选择mysql中的所有字段？

我尝试了什么：

mysql> SELECT title,id FROM dle_post WHERE approve='1' AND date = '2012-02-13 21:35:00';
+-------------------------------------------------+----+
| title                                           | id |
+-------------------------------------------------+----+
| test title                                      |  7 |
+-------------------------------------------------+----+


mysql> SELECT title,id FROM dle_post WHERE approve='1' AND date='2012-02%';
Empty set, 2 warnings (0.00 sec)

mysql> SHOW WARNINGS;
+---------+------+-----------------------------------------------------------------+
| Level   | Code | Message                                                         |
+---------+------+-----------------------------------------------------------------+
| Warning | 1292 | Incorrect datetime value: '2012-02%' for column 'date' at row 1 |
| Warning | 1292 | Incorrect datetime value: '2012-02%' for column 'date' at row 1 |
+---------+------+-----------------------------------------------------------------+

Answer 1

理想情况下，日期应视为日期，而不是字符串。此查询应返回2012年2月的所有行，并且应该优于LIKE %个查询：

SELECT title, id
FROM dle_post
WHERE approve = 1 
AND date >= '2012-02-01' 
AND date <  '2012-02-01' + INTERVAL 1 MONTH

Answer 2

更改为

SELECT title,id FROM dle_post WHERE approve='1' AND date LIKE '2012-02%';

Answer 3

使用like运算符检查这些值

SELECT title,id FROM dle_post WHERE approve='1' AND date LIKE '2012-02%';

Answer 4

你试过吗

date LIKE '2012-02%'

而不是

date='2012-02%'

您可能还需要使用DATE_FORMAT（日期，格式）将日期转换为字符串

请参阅：http://dev.mysql.com/doc/refman/5.5/en/date-and-time-functions.html#function_date-format

mysql：选择所有以2012-02开头的字段*

4 个答案: