總和在MySQL

Question

我有以下簡單的表

FROM_CATEGORY  TO_CATEGORY  LENGTH 
    A     D    5 
    B     C    6 
    B     B    2 
    C     D    4 
    D     D    1 

我只想得到以下結果

CATEGORY  COUNT_REPEATED  SUM_LENGTH 
    A    1    5 
    B    2    8 
    C    2    10 
    D    2    10 

我曾嘗試使用下面的查詢，但在對（BB）和對（DD）我會計算和總結兩次

SELECT t1.CATEGORY , COUNT(*) , SUM(t1.LENGTH) FROM 
(SELECT FROM_CATEGORY AS CATEGORY , SUM(LENGTH) AS LENGTH FROM categoryTable GROUP BY FROM_CATEGORY 
UNION ALL 
SELECT TO_CATEGORY AS CATEGORY, SUM(LENGTH) AS LENGTH FROM categoryTable GROUP BY TO_CATEGORY) AS t1 GROUP BY t1.CATEGORY 

任何想法如何做到這一點？

Answer 1

只過濾那些兩類都是相同的行：

SELECT t1.CATEGORY , COUNT(*) , SUM(t1.LENGTH) 
FROM 
(SELECT FROM_CATEGORY AS CATEGORY , SUM(LENGTH) AS LENGTH 
    FROM categoryTable 
    WHERE FROM_CATEGORY <> TO_CATEGORY 
    GROUP BY FROM_CATEGORY 

    UNION ALL 

    SELECT TO_CATEGORY AS CATEGORY, SUM(LENGTH) AS LENGTH 
    FROM categoryTable 
    GROUP BY TO_CATEGORY) AS t1 
GROUP BY t1.CATEGORY 

編輯：

根據您希望包括其中兩個FROM_CATEGORY和TO_CATEGORY在你數相同的行您的評論，但不是總和。然後，你需要在WHERE條件進入一個CASE，做條件聚集：

SELECT t1.CATEGORY , Count(*) , Sum(t1.LENGTH) 
FROM 
(SELECT FROM_CATEGORY AS CATEGORY , 
     Sum(CASE WHEN FROM_CATEGORY <> TO_CATEGORY THEN LENGTH ELSE 0 end) AS LENGTH 
    FROM categoryTable 
    GROUP BY FROM_CATEGORY 

    UNION ALL 

    SELECT TO_CATEGORY AS CATEGORY, Sum(LENGTH) AS LENGTH 
    FROM categoryTable 
    GROUP BY TO_CATEGORY) AS t1 
GROUP BY t1.CATEGORY 

Answer 2

修改dnoeth的回答如下

SELECT t1.CATEGORY , SUM(t1.counter), SUM(t1.LENGTH) FROM 
(SELECT FROM_CATEGORY AS CATEGORY , COUNT(*) AS counter , SUM(LENGTH) AS LENGTH FROM categoryTable 
WHERE FROM_CATEGORY <> TO_CATEGORY 
GROUP BY FROM_CATEGORY 
UNION ALL 
SELECT TO_CATEGORY AS CATEGORY, COUNT(*) AS counter , SUM(LENGTH) AS LENGTH FROM categoryTable GROUP BY TO_CATEGORY) AS t1 
GROUP BY t1.CATEGORY