使用 date() 函数按日期匹配来更新 MySQL 表格列?
以下是匹配 date() 函数和更新一列的语法 -
update yourTableName set yourColumnName=yourValue where date(yourColumnName)=curdate();
让我们首先创建一张表 -
mysql> create table DemoTable1816 ( Name varchar(20), JoiningDate datetime ); Query OK, 0 rows affected (0.00 sec)
使用 insert 命令在表中插入一些记录 -
mysql> insert into DemoTable1816 values('Chris','2019-11-29 12:34:50');
Query OK, 1 row affected (0.00 sec)
mysql> insert into DemoTable1816 values('David','2019-11-30 11:00:00');
Query OK, 1 row affected (0.00 sec)
mysql> insert into DemoTable1816 values('Mike','2018-11-30 10:20:30');
Query OK, 1 row affected (0.00 sec)使用 select 语句显示表中的所有记录 -
mysql> select * from DemoTable1816;
这将生成以下输出 -
+-------+---------------------+ | Name | JoiningDate | +-------+---------------------+ | Chris | 2019-11-29 12:34:50 | | David | 2019-11-30 11:00:00 | | Mike | 2018-11-30 10:20:30 | +-------+---------------------+ 3 rows in set (0.00 sec)
以下是通过匹配日期更新一列的查询 -
mysql> update DemoTable1816 set Name='Robert' where date(JoiningDate)=curdate(); Query OK, 1 row affected (0.00 sec) Rows matched: 1 Changed: 1 Warnings: 0
让我们再次查看表记录 -
mysql> select * from DemoTable1816;
这将生成以下输出 -
+--------+---------------------+ | Name | JoiningDate | +--------+---------------------+ | Chris | 2019-11-29 12:34:50 | | Robert | 2019-11-30 11:00:00 | | Mike | 2018-11-30 10:20:30 | +--------+---------------------+ 3 rows in set (0.00 sec)
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