Question

我有MySQL功能 2个参数，即 user_id 和 post_id

这是我的功能：

CREATE FUNCTION isliked(pid INT, uid INT)
RETURN TABLE
AS
RETURN (EXISTS (SELECT 1 FROM likedata ld WHERE post_id = pid AND user_id = uid
       )) as is_liked
END

我尝试使用以下查询调用它：

SELECT posts.id, posts.title, isliked(111,123)
FROM posts

它返回以下错误：

You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'RETURN TABLE
AS
RETURN (EXISTS (SELECT 1 FROM likedata ld WHERE post_id = pid AN' at line 2

这应该是返回结果http://sqlfiddle.com/#!9/91040/5 我是sql的新手，任何帮助都会很棒，提前谢谢

Answer 1

如果希望函数返回布尔值，请使用：

CREATE FUNCTION isliked(pid INT, uid INT)
RETURNS BIT
   RETURN ( EXISTS ( SELECT 1 FROM likedata ld WHERE post_id = pid AND user_id = uid ) )

带参数的MySQL函数出错

1 个答案: