MySQL命令行删除表中的一个字段


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 <表名> drop column <字段名>;

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!


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