将变量传递到mysql所在的位置

Question

无法让查询运行，在@searchin变量上抛出错误。 可能很简单但看不到它。

set @search = "chip";
set @searchin = "CompanyName";

select * from con_search where @searchin like concat ('%',@search,'%')

Answer 1

这样可行，但是你应该对进入它的数据进行整理

set @search = 'chip';
set @searchin = 'CompanyName';
set @SQL = CONCAT("SELECT * FROM con_search WHERE `", @searchin, "` LIKE CONCAT('%'", @search, "'%');";
PREPARE stmt1 FROM @SQL;
EXECUTE stmt1;