The MySQL command line deletes a field from a table

Take a look at the table structure before deleting:

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)

Execute the delete command, using the drop keyword.
The basic syntax is alter table < The name of the table > drop column < The field name > ;

The specific orders are as follows:

mysql > alter table test drop column birth1;
Query OK, 0 rows affected (0.13 sec)
Records: 0 Duplicates: 0 Warnings: 0

Look at the results after the deletion, is there no birth1 field?

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)

About MySQL SQL statement date function that calculates age according to birthday, this article introduces so much, hope to be helpful to you, thank you!