全部博文(842)
分类: Mysql/postgreSQL
2012-06-09 10:35:15
MYSQL时间函数用法宝典
测试表:
root@test 16:50>desc t1;
+-------+----------+------+-----+---------+-------+
| Field | Type | Null | Key | Default | Extra |
+-------+----------+------+-----+---------+-------+
| id | int(11) | YES | | NULL | |
| t1 | datetime | YES | | NULL | |
+-------+----------+------+-----+---------+-------+
2 rows in set (0.00 sec)
root@test 16:50>select * from t1;
+------+---------------------+
| id | t1 |
+------+---------------------+
| 1 | 2012-05-13 12:27:12 |
| 2 | 2012-05-13 12:27:12 |
| 3 | 2012-05-13 12:27:12 |
+------+---------------------+
(1),
root@test 16:50>SELECT DATE_ADD(t1,INTERVAL -1 YEAR) FROM t1;
+-------------------------------+
| DATE_ADD(t1,INTERVAL -1 YEAR) |
+-------------------------------+
| 2011-05-13 12:27:12 |
| 2011-05-13 12:27:12 |
| 2011-05-13 12:27:12 |
+-------------------------------+
root@test 17:21>SELECT DATE_ADD('2008-01-02', INTERVAL 31 DAY);
+-----------------------------------------+
| DATE_ADD('2008-01-02', INTERVAL 31 DAY) |
+-----------------------------------------+
| 2008-02-02 |
+-----------------------------------------+
(2),root@test 17:00>SELECT DATE_FORMAT(t1, '%H:%i:%s') from t1;
+-----------------------------+
| DATE_FORMAT(t1, '%H:%i:%s') |
+-----------------------------+
| 12:27:12 |
| 12:27:12 |
| 12:27:12 |
+-----------------------------+
(3)datediff(expr1,expr2)用于计算两个时间的相差的天数
root@test 17:01>select datediff(t1,now()) from t1;
+--------------------+
| datediff(t1,now()) |
+--------------------+
| -23 |
| -23 |
| -23 |
+--------------------+
4,返回当前日期curdate(),curtime()返回当前时间,now()返回当前的日期和时间。
5,返回日期中的年份year(),返回日期中月份month(),返回日期中的天day(),返回日期中时间time()。
root@test 17:08>select day(t1) from t1;
+---------+
| day(t1) |
+---------+
| 13 |
| 13 |
| 13 |
+---------+
3 rows in set (0.00 sec)
root@test 17:13>select time(t1) from t1;
+----------+
| time(t1) |
+----------+
| 12:27:12 |
| 12:27:12 |
| 12:27:12 |
+----------+
3 rows in set (0.00 sec)
root@test 17:13>select month(t1) from t1;
+-----------+
| month(t1) |
+-----------+
| 5 |
| 5 |
| 5 |
+-----------+
3 rows in set (0.00 sec)
root@test 17:13>select year(t1) from t1;
+----------+
| year(t1) |
+----------+
| 2012 |
| 2012 |
| 2012 |
+----------+
3 rows in set (0.00 sec)
6,week()计算当前日期为本年度的多少周
root@test 17:16>select week('2012-06-05');
+--------------------+
| week('2012-06-05') |
+--------------------+
| 23 |
+--------------------+
1 row in set (0.00 sec)