通过使用date()函数匹配日期来更新MySQL表列?
以下是将date与date()函数进行匹配并更新列的语法-
update yourTableName set yourColumnName=yourValue where date(yourColumnName)=curdate();
让我们首先创建一个表-
create table DemoTable1816 ( Name varchar(20), JoiningDate datetime );
使用插入命令在表中插入一些记录-
insert into DemoTable1816 values('Chris','2019-11-29 12:34:50'); insert into DemoTable1816 values('David','2019-11-30 11:00:00'); insert into DemoTable1816 values('Mike','2018-11-30 10:20:30');
使用select语句显示表中的所有记录-
select * from DemoTable1816;
这将产生以下输出-
+-------+---------------------+ | 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)
这是通过匹配日期来更新列的查询-
update DemoTable1816 set Name='Robert' where date(JoiningDate)=curdate(); Rows matched: 1 Changed: 1 Warnings: 0
让我们再次检查表记录-
select * from DemoTable1816;
这将产生以下输出-
+--------+---------------------+ | 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)