MYSQL时间函数用法宝典

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)