php - 将变量传递到 mysql 的位置

Question

无法运行查询，在 @searchin 变量上抛出错误。可能很简单，但看不到。

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

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

Answer 1

这会起作用，但是您应该清理进入其中的数据

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