mysql - 将 SUM 的多个查询合并为一个结果

Question

所以....我们有三个不同的表，它们与比赛有关，数据库在其中跟踪他们在每次比赛中获得的分数。比赛 1、2 和 3。每次用户取得成就时，都会为该用户创建一个新行，其中包含额外的分数。因此，为了计算用户收到的所有积分，我使用了一个选择总和

SELECT userID, SUM(amount1) as "Contest 1 Points"
FROM [Company].[dbo].[Contest1]
WHERE userid not in (0,1)
GROUP BY userId
ORDER BY userid

因为我还有两场比赛，所以我也对每场比赛都进行了查询...

SELECT userId, SUM(amount2)/.65 AS "Category 2 Points"
FROM [Company].[dbo].[Contest2]
WHERE dateGiven >=201301 AND dateGiven <= 201305
GROUP BY userId
ORDER BY userid



SELECT userid, SUM(amount3) AS "Category 3 Points"
FROM [Company].[dbo].[Contest3]
where userid not in (1,2)
GROUP BY userid
ORDER BY userid

我基本上需要将每个用户从每次比赛中获得的所有积分加到 1 列中，基本上显示结果 USERID，TOTAL OF TOTALS(Contest1 + Contest2 + Contest3)

或者至少让它喜欢，

USER，比赛 1 总计，比赛 2 总计，比赛 3 总计

到目前为止，我这样做的方法是将这些结果中的每一个复制/粘贴到 excel 中，然后我使用 VLOOKUP 将它们相互匹配，这有点麻烦，我相信有一种方法可以在 SQL 中做到这一点。我是 SQL 的新手，我尝试加入并使用 ON 来匹配用户 ID，但我的语法有问题，我理解它全部插入自身以进行查询。

Answer 1

您需要对结果进行 UNION:

SELECT userID, SUM(Points) AS total
FROM
 ( 
   SELECT userID, SUM(amount1) AS "Points"
   FROM [Company].[dbo].[Contest1]
   WHERE userid NOT IN (0,1)
   GROUP BY userId

   UNION ALL       

   SELECT userId, SUM(amount2)/.65 AS "Category 2 Points"
   FROM [Company].[dbo].[Contest2]
   WHERE dateGiven >=201301 AND dateGiven <= 201305
   GROUP BY userId

   UNION ALL       

   SELECT userid, SUM(amount3) AS "Category 3 Points"
   FROM [Company].[dbo].[Contest3]
   WHERE userid NOT IN (1,2)
   GROUP BY userid
 ) AS dt
GROUP BY userID
ORDER BY 2 DESC;

编辑: 要获得三个单独的列，您只需使用三个 SUM 而不是一个:

SELECT userID, SUM("Category 1 Points"), SUM("Category 2 Points"), SUM("Category 3 Points") 
FROM
 ( 
   SELECT userID, SUM(amount1) AS "Category 1 Points"
   FROM [Company].[dbo].[Contest1]
   WHERE userid NOT IN (0,1)
   GROUP BY userId

   UNION ALL       

   SELECT userId, SUM(amount2)/.65 AS "Category 2 Points"
   FROM [Company].[dbo].[Contest2]
   WHERE dateGiven >=201301 AND dateGiven <= 201305
   GROUP BY userId

   UNION ALL       

   SELECT userid, SUM(amount3) AS "Category 3 Points"
   FROM [Company].[dbo].[Contest3]
   WHERE userid NOT IN (1,2)
   GROUP BY userid
 ) AS dt
GROUP BY userID
ORDER BY 2 DESC;

当然，每个 userDI/类别只有一行，因此 MIN 或 MAX 会返回相同的结果。 这将为不存在的数据返回 NULL，如果您想要 0 而不是使用 COALESCE("Category x Points", 0)。

您也可以加入结果集，但除非保证每个用户都参加了每次比赛，否则您需要使用 COALESCE 的 FULL OUTER JOIN:

SELECT userID, "Category 1 Points", "Category 2 Points", "Category 3 Points"
FROM
 ( 
   SELECT userID, SUM(amount1) AS "Category 1 Points"
   FROM [Company].[dbo].[Contest1]
   WHERE userid NOT IN (0,1)
   GROUP BY userId
 ) AS t1
FULL JOIN
ON t1.userID = t2.userID
 (
   SELECT userId, SUM(amount2)/.65 AS "Category 2 Points"
   FROM [Company].[dbo].[Contest2]
   WHERE dateGiven >=201301 AND dateGiven <= 201305
   GROUP BY userId
 ) AS t2
FULL JOIN
 (
   SELECT userid, SUM(amount3) AS "Category 3 Points"
   FROM [Company].[dbo].[Contest3]
   WHERE userid NOT IN (1,2)
   GROUP BY userid
 ) AS t3
ON COALESCE(t1.userID, t2.userID) = t3.userID
ORDER BY 2 DESC;