有沒有一種方法可以根據WHERE表達式來計算加權總和？

Question

我想知道我的堆棧溢出的聲譽排除所有獲得或失去從提問和回答regex問題（—感覺「髒錢」給我！）

堆棧交易所Data Explorer，讓我來計算的口碑這個。但是，我只能使用專有數據庫和C++包裝類進行編程，所以我對自己的SQL技能並不太自信。儘管如此，我給它一個鏡頭，並最終想出了a query that provided my answer：

-- (approximate) reputation gained/lost on a specified tag 
-- only counts post upvotes and downvotes 

DECLARE @UserId  int = ##UserId## 
DECLARE @QuestionsUp int = 0; 
DECLARE @QuestionsDown int = 0; 
DECLARE @AnswersUp  int = 0; 
DECLARE @AnswersDown int = 0; 
DECLARE @Tag   nvarchar(25) = 'regex'; 

SELECT 
    @QuestionsUp = COUNT(*) 
FROM Tags 
    INNER JOIN PostTags ON PostTags.TagId = Tags.id 
    INNER JOIN Posts ON Posts.ParentId = PostTags.PostId 
    INNER JOIN Votes ON Votes.PostId = Posts.Id and VoteTypeId = 2 
WHERE 
    Posts.OwnerUserId = @UserId and 
    Posts.PostTypeId = 1 and 
    Tags.TagName = @Tag 

SELECT 
    @QuestionsDown = COUNT(*) 
FROM Tags 
    INNER JOIN PostTags ON PostTags.TagId = Tags.id 
    INNER JOIN Posts ON Posts.ParentId = PostTags.PostId 
    INNER JOIN Votes ON Votes.PostId = Posts.Id and VoteTypeId = 3 
WHERE 
    Posts.OwnerUserId = @UserId and 
    Posts.PostTypeId = 1 and 
    Tags.TagName = @Tag 

SELECT 
    @AnswersUp = COUNT(*) 
FROM Tags 
    INNER JOIN PostTags ON PostTags.TagId = Tags.id 
    INNER JOIN Posts ON Posts.ParentId = PostTags.PostId 
    INNER JOIN Votes ON Votes.PostId = Posts.Id and VoteTypeId = 2 
WHERE 
    Posts.OwnerUserId = @UserId and 
    Posts.PostTypeId = 2 and 
    Tags.TagName = @Tag 

SELECT 
    @AnswersDown = COUNT(*) 
FROM Tags 
    INNER JOIN PostTags ON PostTags.TagId = Tags.id 
    INNER JOIN Posts ON Posts.ParentId = PostTags.PostId 
    INNER JOIN Votes ON Votes.PostId = Posts.Id and VoteTypeId = 3 
WHERE 
    Posts.OwnerUserId = @UserId and 
    Posts.PostTypeId = 2 and 
    Tags.TagName = @Tag 

SELECT @QuestionsUp * 5 + 
     @AnswersUp * 10 + 
     (@QuestionsDown + @AnswersDown) * -2 

這不可能是最好的人能做到，雖然。四個單獨的查詢只是將5個問題的問題分爲加分問題，10個問答問題，以及-2個問題解答問題和答案？有沒有辦法壓縮這個查詢來執行一次運行？

（請隨時免費的，如果你有關於語法，格式，好的做法，這樣的任何輔助的建議發表評論。另外，請不要評論，如果有其他的方法來獲得或失去特有的標籤—我的天堂美譽— 「噸因素）

Answer 1

你總是可以嘗試像

SELECT 
    SUM(
     CASE 
      WHEN VoteTypeId = 2 AND Posts.PostTypeId = 1 
       THEN 1 
      ELSE 0 
     END 
    ) QuestionsUp, 
    SUM(
     CASE 
      WHEN VoteTypeId = 3 AND Posts.PostTypeId = 1 
       THEN 1 
      ELSE 0 
     END 
    ) QuestionsDown, 
    SUM(
     CASE 
      WHEN VoteTypeId = 2 AND Posts.PostTypeId = 2 
       THEN 1 
      ELSE 0 
     END 
    ) AnswersUp, 
    SUM(
     CASE 
      WHEN VoteTypeId = 3 AND Posts.PostTypeId = 2 
       THEN 1 
      ELSE 0 
     END 
    ) AnswersDown 
FROM Tags 
    INNER JOIN PostTags ON PostTags.TagId = Tags.id 
    INNER JOIN Posts ON Posts.ParentId = PostTags.PostId 
    INNER JOIN Votes ON Votes.PostId = Posts.Id 
WHERE 
    Posts.OwnerUserId = @UserId and 
    Tags.TagName = @Tag 

編輯：

你可以使用一個CTE和然後使用您的計算中的列。

喜歡的東西

;WITH Vals AS (
     SELECT 
      SUM(
       CASE 
        WHEN VoteTypeId = 2 AND Posts.PostTypeId = 1 
         THEN 1 
        ELSE 0 
       END 
      ) QuestionsUp, 
      SUM(
       CASE 
        WHEN VoteTypeId = 3 AND Posts.PostTypeId = 1 
         THEN 1 
        ELSE 0 
       END 
      ) QuestionsDown, 
      SUM(
       CASE 
        WHEN VoteTypeId = 2 AND Posts.PostTypeId = 2 
         THEN 1 
        ELSE 0 
       END 
      ) AnswersUp, 
      SUM(
       CASE 
        WHEN VoteTypeId = 3 AND Posts.PostTypeId = 2 
         THEN 1 
        ELSE 0 
       END 
      ) AnswersDown 
     FROM Tags 
      INNER JOIN PostTags ON PostTags.TagId = Tags.id 
      INNER JOIN Posts ON Posts.ParentId = PostTags.PostId 
      INNER JOIN Votes ON Votes.PostId = Posts.Id 
     WHERE 
      Posts.OwnerUserId = @UserId and 
      Tags.TagName = @Tag 
     ) 
SELECT QuestionsUp * 5 + 
     AnswersUp * 10 + 
     (QuestionsDown + AnswersDown) * -2 
FROM Vals