標籤：multiple-tablessumsqlmysqlsql-server

所以…我們有3個與比賽有關的表格,其中DB跟蹤他們為每個比賽獲得的積分.競賽1、2和3.每當使用者獲得成就時,就會為該使用者建立一個新行,並附加得分.因此,為了計算使用者獲得的所有積分,我使用選擇總和

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

因為我還有另外兩個比賽,所以我也要查詢每個比賽…

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
 

我基本上需要將每個使用者從每個競賽中獲得的所有積分加到1個基本上顯示結果的列中
USERID,共TOTALS(競賽1競賽2競賽3)

或至少像

USER,共1個競賽,共2個競賽,共3個競賽

到目前為止,我這樣做的方法是將每個結果複製/貼上到excel中,然後我使用VLOOKUP將它們相互匹配,這有點麻煩,並且肯定有一種方法可以在SQL中完成.我對SQL來說還很陌生,我曾嘗試加入並使用usig ON來匹配userid,但是我的語法以及我瞭解到查詢都可以插入其中的方式出了點問題.

解決方法:

您需要UNION結果：

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;
 

編輯：
要獲得三個單獨的列,您只需使用三個SUM而不是一個：

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;

當然,每個userDI /類別只有一行,因此MIN或MAX將返回相同的結果.
如果想要0,則將為不存在的資料返回NULL,而應使用COALESCE(“ Category x Points”,0).

您也可以加入結果集,但是除非保證每個使用者都參加了每個比賽,否則您需要使用COALESCE進行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;

標籤：multiple-tables,sum,sql,mysql,sql-server
來源： https://codeday.me/bug/20191122/2062214.html