Question

我在上述查询的accepted_join_id中发现输入的user_id给出错误，因为语法错误，意外的'$ user_id'（T_VARIABLE）

$query=DB::select('select activity_id,accepted_join_id from table_user_create_activity WHERE FIND_IN_SET('$user_id',?) and activity_id=?',[$user_id,$accepted_join_id]);

Answer 1

如果您想将php中的$ user_id值注入查询中，请尝试以下操作：

$query=DB::select('select activity_id,accepted_join_id from table_user_create_activity WHERE FIND_IN_SET('.$user_id.',?) and activity_id=?',[$user_id,$accepted_join_id]);

或者如果$ user_id在查询中，则应为：

$query=DB::select('select activity_id,accepted_join_id from table_user_create_activity WHERE FIND_IN_SET("$user_id",?) and activity_id=?',[$user_id,$accepted_join_id]);

FIND_IN_SET查询无法在laravel 5.7中运行

1 个答案: