将SQL中的IN和NOT IN组合为单个结果

Question

对于模糊的标题，我深表歉意。 我试图写一个返回匹配值（从IN产生的）的别名列的查询以及与不匹配值的别名柱（使用NOT IN）。 我希望结果集具有：userId | 火柴| 不匹配。 目前，我有以下查询按预期它返回匹配。 我在获取结果集中的不匹配时遇到了麻烦-即从NOT IN语句

SET @userId = 9;
SELECT ug.user_id, COUNT(DISTINCT goal_id) as matches
FROM user_goal ug
WHERE ug.user_id!=@userId
AND goal_id IN (SELECT iug.goal_id FROM user_goal iug WHERE user_id=@userId)
GROUP BY user_id ORDER BY matches DESC LIMIT 4

因此，NOT IN看起来像这样：

goal_id NOT IN(SELECT uggg.goal_id FROM user_goal uggg WHERE user_id=@userId) AS nonmatches

我只是不确定如何在查询中合并NOT IN语句，所以我得到了所有结果

Answer 1

将其按另一个参数分组，修复子查询中有目标时将使用的一个参数

Answer 2

我认为您需要一个UNION：

SET @userId = 9;

SELECT ug.user_id, COUNT(DISTINCT goal_id) as matches
FROM user_goal ug
WHERE ug.user_id!=@userId
AND goal_id IN (SELECT iug.goal_id FROM user_goal iug WHERE user_id=@userId)
GROUP BY user_id ORDER BY matches DESC LIMIT 4

union all

SELECT ug.user_id, COUNT(DISTINCT goal_id) as matches
FROM user_goal ug
WHERE ug.user_id!=@userId
AND goal_id NOT IN(SELECT uggg.goal_id FROM user_goal uggg WHERE user_id=@userId)
GROUP BY user_id ORDER BY matches DESC LIMIT 4

Answer 3

试试这个（如果mysql有CTE，可以简化并进一步优化）：

select u.user_id, the_matches.matches, the_nonmatches.nonmatches
from user u 
left join 
(    
    SELECT ug.user_id, COUNT(DISTINCT goal_id) as matches
    FROM user_goal ug
    WHERE ug.user_id!=@userId
    AND goal_id IN (SELECT iug.goal_id FROM user_goal iug WHERE user_id=@userId)
    GROUP BY user_id 
    ORDER BY matches DESC LIMIT 4
) as the_matches on the_matches.user_id = u.user_id
left join
(
    SELECT ug.user_id, COUNT(DISTINCT goal_id) as nonmatches
    FROM user_goal ug
    WHERE ug.user_id!=@userId
    AND goal_id NOT IN(SELECT uggg.goal_id FROM user_goal uggg WHERE user_id=@userId) 
    GROUP BY user_id 
) as the_nonmatches on the_nonmatches.user_id = u.user_id