合并 2 个具有类似 where 子句但不同聚合函数目标的查询

Question

我有 2 个单独的查询可以正常工作。鉴于它们相似，我想将它们合并为一个高性能查询。看起来很简单，因为 where 子句是相似的。但是 sum、count 和 min 函数都适用于不同的行并妨碍了。

上下文：

• 用户可以对位置进行评分（或评分）并获得积分
• 当用户 B 首次提交分数时，用户 A 可以推荐用户 B 并获得推荐积分
• 积分在特定日期后过期
• 目标是建立一个用户排行榜以及他们在特定位置（地区/国家）的得分和推荐总分
• 位置参数用“Massachusetts”、“United States”和 scoreDateTime 到期日期的硬值填充，不幸的是在两个选择子查询中重复。

问题：

如何重新组织以下查询以组合约束？必须有一种方法可以从特定日期后特定位置的分数列表开始。唯一的麻烦是获取用户 B 的第一个得分日期，并且仅在过期日期之后才向用户 A 提供推荐积分。

select scoring.userId, scoring.points + referring.points as leaderPoints
from (
    select   userId, sum(ratingPoints) as points
    from     scores s, locations l
    where    s.locationId = l.locationId and
             l.locationArea = 'Massachusetts' and
             l.locationCountry = 'United States' and
             s.scoreDateTime > '2016-04-16 18:50:53.154' and
             s.userId != 0
    group by s.userId
) as scoring

join (
    select u1.userId, count(*) * 20 as points
    from users u0
    join users u1 on u0.userId = u1.userId
    join users u2 on u2.referredByEmail = u1.emailAddress
    join scores s on u2.userId = s.userId
    join locations l on s.locationId = l.locationId
    where    l.locationArea = 'Massachusetts' and
             l.locationCountry = 'United States' and
             scoreDateTime = (
                 select min(scoreDateTime)
                 from   scores
                 where  userId = u2.userId
             ) and
             scoreDateTime >= '2016-04-16 18:50:53.154'
    group by u1.userId
) as referring on scoring.userId = referring.userId
order by leaderPoints desc
limit 10;

Answer 1

这是未经测试的代码，但它应该可以解决问题。 Cross Apply 是为了可读性...它会损害性能，但这似乎不是一个特别需要处理的查询，所以我会保留它。

请尝试一下，如果您有任何问题，请告诉我。

SELECT  U.UserID, 
        ISNULL(SUM(CASE WHEN S.UserID IS NULL THEN 0 ELSE S.ratingPoints END), 0) AS [Rating Points], 
        ISNULL(SUM(CASE WHEN SS.userID IS NULL THEN 0 ELSE 20 END), 0) AS [Referral Points]
FROM Users U
LEFT OUTER JOIN scores S
    ON S.userID = U.userID
    AND S.scoreDateTime >= '2016-04-16 18:50:53.154'
LEFT OUTER JOIN locations L
    ON S.locationID = L.locationID
    AND L.locationArea = 'Massachusetts'
    AND L.LocationCountry = 'United States'
LEFT OUTER JOIN Users U2
    ON U2.referredByEmail = U.emailAddress
LEFT OUTER JOIN scores SS
    ON SS.userID = U2.userID
LEFT OUTER JOIN locations LL
    ON SS.locationID = LL.locationID
    AND LL.locationArea = 'Massachusetts'
    AND LL.locationCountry = 'United States'
    AND SS.scoreDateTime >= '2016-04-16 18:50:53.154'
    AND SS.scoreDateTime = 
        (
            SELECT MIN(scoreDateTime)
            FROM scores
            where userID = U2.userID
        )
GROUP BY U.userID

编辑：

修改答案以删除交叉应用

Answer 2

感谢 Stan Shaw，但我无法让您的查询在 MySQL 上运行以测试结果。但是，我确实注意到了原始查询未涵盖的特殊情况。用户可以从他们自己没有提交分数的区域获得参考点。只要新用户在该区域得分，他们就会在该区域获得推荐积分。

这是我使用的最后一个查询。我无法以看似高效的方式整合重复的 where 子句。

select userId, sum(points) as leaderPoints
from (

    select   s.userId, sum(s.ratingPoints) as points
    from     scores s, locations l
    where    s.locationId = l.locationId and  
             l.locationArea = 'Georgia' and  
             l.locationCountry = 'United States' and  
             s.scoreDateTime >= '2016-04-05 03:00:00.000' and  
             s.userId != 1
    group by userId

    union

    select   u1.userId, 20 as points
    from     users u0, users u1, users u2, scores s, locations l
    where    u0.userId = u1.userId and
             u2.referredByEmail = u1.emailAddress and
             u2.userId = s.userId and
             s.locationId = l.locationId and
             l.locationArea = 'Georgia' and  
             l.locationCountry = 'United States' and  
             scoreDateTime >= '2016-04-05 03:00:00.000' and
             scoreDateTime = (
                 select min(scoreDateTime) 
                 from   scores  
                 where  userId = u2.userId  
             )

) as pointsEarned
group by userId
order by leaderPoints desc
limit 10
order by leaderPoints desc  
limit 100;