在MySQL中将多个ID传递给单个参数?
要将多个ID传递给单个参数,请使用FIND_IN_SET()。让我们首先创建一个表-
mysql> create table DemoTable1817 ( EmployeeName varchar(20), CountryName varchar(20) );
使用插入命令在表中插入一些记录-
mysql> insert into DemoTable1817 values('Chris','AUS'); mysql> insert into DemoTable1817 values('David','UK'); mysql> insert into DemoTable1817 values('Bob','US');
使用select语句显示表中的所有记录-
mysql> select * from DemoTable1817;
这将产生以下输出-
+--------------+-------------+ | EmployeeName | CountryName | +--------------+-------------+ | Chris | AUS| | David | UK | | Bob | US | +--------------+-------------+ 3 rows in set (0.00 sec)
这是将多个ID传递给MySQL中的单个参数的查询。
mysql> select * from DemoTable1817 where find_in_set(CountryName,'US,UK');
这将产生以下输出-
+--------------+-------------+ | EmployeeName | CountryName | +--------------+-------------+ | David | UK | | Bob | US | +--------------+-------------+ 2 rows in set (0.00 sec)