先看看删除之前的表结构：

mysql> select * from test;

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

| t_id | t_name | t_password                       | t_birth    | birth      | birth1     | birth2     |

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

|    1 | name1  | 12345678901234567890123456789012 | NULL       | 1990-01-01 | 0000-00-00 | 2013-01-01 |

|    2 | name2  | 12345678901234567890123456789012 | 2013-01-01 | NULL       | 0000-00-00 | 2013-01-01 |

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

2 rows in set (0.00 sec)

执行删除命令，使用drop关键字。

基本的语法为：alter table drop column ;

具体的命令如下：

mysql> alter table test drop column birth1;

Query OK, 0 rows affected (0.13 sec)

Records: 0  Duplicates: 0  Warnings: 0

看看删除后的结果，是不是已经没有birth1字段了？

mysql> select * from test;

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

| t_id | t_name | t_password                       | t_birth    | birth      | birth2     |

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

|    1 | name1  | 12345678901234567890123456789012 | NULL       | 1990-01-01 | 2013-01-01 |

|    2 | name2  | 12345678901234567890123456789012 | 2013-01-01 | NULL       | 2013-01-01 |

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

2 rows in set (0.00 sec)

关于MySQL中根据生日计算年龄的SQL语句日期函数，本文就介绍这么多，希望对大家有所帮助，谢谢！