SQL - 試圖找出1組的超過0的比率？

Question

我有兩個表像以下：

CouponTable

• ID
• IsAvailable
• COwnerId

CouponOwner

• COwnerId
• COwnerName

下面是情形：我需要計算IsAvailable的速率（1或0）對於每個優惠券所有者？我找不到我可以先由COwners對它進行分組，然後找到每個分組的速率。

我想這下面的SQL代碼，這是具體的只對業主，這個效果很好，但我不應該給每個擁有者的標識的，我需要做的更通用的方法。

SELECT 
    C.COwnerId, 
    O.COwnerName, 
    (select count(*) from CouponTable where COwnerId = 105 and IsAvailable = 1) * 100 /(select count(*) from CouponTable where COwnerId = 105) as Rate 
FROM 
    CouponTable AS C 
     JOIN 
    CouponOwner AS O ON C.COwnerId = O.COwnerId where C.COwnerId = 105 
GROUP BY C.COwnerId 
ORDER BY Rate DESC; 

這將返回如下：

105，加雷斯·威爾遜，21.9229

Answer 1

您可以使用相關子查詢：

SELECT 
    C.COwnerId, 
    O.COwnerName, 
    (SELECT COUNT(*) 
    FROM CouponTable 
    WHERE COwnerId = c.COwnerId AND IsAvailable = 1) * 100/
    (SELECT count(*) 
    FROM CouponTable 
    WHERE COwnerId = c.COwnerId) AS Rate 
FROM 
    CouponTable AS C 
INNER JOIN 
    CouponOwner AS O ON C.COwnerId = O.COwnerId 
GROUP BY C.COwnerId 
ORDER BY Rate DESC; 

這樣，你只需要替換特定COwnerId的值，即105，該字段的值來自外部查詢，即c.COwnerId。

Demo here

Answer 2

您可以用CouponOwner表加入CouponTable，並使用GROUP BY查詢用SUM和COUNT：

SELECT 
    ID, 
    CouponOwner, 
    (SUM(IsAvailable)/COUNT(*))*100 As Rate 
FROM 
    CouponTable ct INNER JOIN CouponOwner co 
    ON ct.CouponOwnner=co.CoOwnerID 
GROUP BY 
    ID, 
    CouponOwner 
ORDER BY 
    (SUM(IsAvailable)/COUNT(*))*100 DESC 

SUM（IsAvailabe）的優惠券供業主數目和COUNT（*）是爲業主所有優惠券的數量。