我的 MySQL 函数有 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
我试着用下面的查询调用它:
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
应该是这样的返回结果http://sqlfiddle.com/#!9/91040/5 我是 sql 新手,任何帮助都会很棒,提前致谢
最佳答案
如果您希望函数返回 bool 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 ) )
关于mysql - 带参数的MySQL函数出错,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/50559637/