Question

我目前在SEDE上创建了以下查询，以通过组合答案和分数来在每个标签中找到最重要的用户。可以在这里找到：FormControlDirective。但是，目前每个标签可以带回多个顶级用户，这是可以理解的，因为我还没有对此进行限制。

以下是查询：

SELECT TOP 50
       t.TagName,
       a.OwnerUserId AS [User Link],
       SUM(a.Score) / 10 AS Score,
       COUNT(a.Score) AS [Count],
       ((SUM(a.Score) / 10) + COUNT(a.Score)) / 2 AS Total
FROM Posts a, 
     Posts q
     INNER JOIN PostTags qt ON q.Id = qt.PostId
     INNER JOIN Tags t ON t.Id = qt.TagId
WHERE a.ParentId = q.Id
      AND a.PostTypeId = 2
      AND a.CommunityOwnedDate IS NULL
      AND a.OwnerUserId IS NOT NULL
GROUP BY a.OwnerUserId, t.TagName
ORDER BY ((SUM(a.Score) / 10) + COUNT(a.Score)) / 2 DESC

如何做到这一点，使其仅返回每个标签的最高用户？

Answer 1

我会将您的查询包装在CTE（公用表表达式）中，然后在第二个CTE上计算每个标签的最高得分，最后将两个CTE合并在一起以获取每个标签的头号用户。查询应如下所示：

with user_tag as ( -- score per user, per tag
  SELECT t.TagName,
       a.OwnerUserId AS [User Link],
       SUM(a.Score) / 10 AS Score,
       COUNT(a.Score) AS [Count],
       ((SUM(a.Score) / 10) + COUNT(a.Score)) / 2 AS Total
  FROM Posts a
  JOIN Posts q on a.ParentId = q.Id
  JOIN PostTags qt ON q.Id = qt.PostId
  JOIN Tags t ON t.Id = qt.TagId
  WHERE a.PostTypeId = 2
    AND a.CommunityOwnedDate IS NULL
    AND a.OwnerUserId IS NOT NULL
  GROUP BY a.OwnerUserId, t.TagName
),
max_score as ( -- max score per tag
  select TagName, max(Total) as max_score
  from user_tag
  group by TagName
)
select 
    u.*
  from user_tag u
  join max_score m on m.TagName = u.TagName
                  and m.max_score = u.Total

我不包含任何顺序，因为我不确定您想要行如何。如果首先有多个用户并获得了相同的分数，请考虑该查询将为同一标签显示多行。

SEDE查询以查找所有标签中的顶级用户

1 个答案: