使用从日期格式(如"10/12/2010")中提取的特定年份来更新表中的记录?
mysqlmysqli database
要使用特定年份来更新记录，请使用 YEAR() 方法，语法如下:
update yourTableName set yourColumnName1=yourValue1 where YEAR(str_to_date(yourColumnName2,'%d/%m/%Y'))=yourValue2;
首先我们创建一个表 −
mysql> create table DemoTable1924 ( UserName varchar(20), UserJoiningDate varchar(40) ); 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 DemoTable1924 values('Chris','10/12/2010');
Query OK, 1 row affected (0.00 sec)
mysql> insert into DemoTable1924 values('David','20/01/2011');
Query OK, 1 row affected (0.00 sec)
mysql> insert into DemoTable1924 values('Mike','20/01/2010');
Query OK, 1 row affected (0.00 sec)
mysql> insert into DemoTable1924 values('Carol','26/04/2013');
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 DemoTable1924;
这将产生以下输出 −
+----------+-----------------+ | UserName | UserJoiningDate | +----------+-----------------+ | Chris | 10/12/2010 | | David | 20/01/2011 | | Mike | 20/01/2010 | | Carol | 26/04/2013 | +----------+-----------------+ 4 rows in set (0.00 sec)
这是根据特定年份更新记录的查询 −
mysql> update DemoTable1924 set UserName='Robert' where YEAR(str_to_date(UserJoiningDate,'%d/%m/%Y'))=2010; Query OK, 2 rows affected (0.00 sec) Rows matched: 2 Changed: 2 Warnings: 0
让我们再次检查表记录 −
mysql> select * from DemoTable1924;
这将产生以下输出 −
+----------+-----------------+ | UserName | UserJoiningDate | +----------+-----------------+ | Robert | 10/12/2010 | | David | 20/01/2011 | | Robert | 20/01/2010 | | Carol | 26/04/2013 | +----------+-----------------+ 4 rows in set (0.00 sec)
