如何使用左連接的用戶聚合功能？ MySQL的

Question

我有這樣的查詢：

SELECT business.bussId, COUNT(invoices.userId) as invoices, COUNT(rating.bussId) as ratingCount , 
     FROM business 
     LEFT JOIN invoice ON (invoice.bussId = business.bussId AND invoice.userId = '3000') 
     LEFT JOIN rating ON (rating.bussId = business.bussId) 
     WHERE business.bussId=100 

的COUNT（invoices.userId）用戶應該在發票表，返回的行數其中userid = 3000 AND bussId = 100。 換言之此查詢：

SELECT COUNT(*) as invoice FROM `invoices` WHERE bussId = '100' AND userId = '30000'. 

第二查詢返回COUNT（*）= 3，（當我用戶LEFT JOIN），它的返回15的第一個查詢，如何解決呢？

Answer 1

試試下面的查詢：

SELECT business.bussId, (select count(invoices.userId) from invoice where invoice.bussId = a.bussId AND invoice.userId = '3000') as invoices, (select COUNT(rating.bussId) from rating where rating.bussId = a.bussId) as ratingCount , 
     FROM business a 
       WHERE business.bussId=100 

Answer 2

您還沒有發佈任何測試數據或任何表聲明所以這正在對這些假設。

您當前的查詢將獲得發票和評級的每項業務組合。因此，如果有3個發票和5個評級，將導致這15個組合。

通常可以使用COUNT（DISTINCT）來修復此問題，但是您將計算髮票的用戶ID數量，當它是常量時。

假設您的發票表和評分表都具有名爲id的唯一密鑰，那麼你應該能夠計數更改爲那些視場中不同： -

SELECT business.bussId, COUNT(DISTINCT invoices.id) as invoices, COUNT(DISTINCT rating.id) as ratingCount 
FROM business 
LEFT JOIN invoice ON (invoice.bussId = business.bussId AND invoice.userId = '3000') 
LEFT JOIN rating ON (rating.bussId = business.bussId) 
WHERE business.bussId = 100 

這避免了使用子查詢，這取決於數據通常可能表現不佳。雖然在這種情況下，它看起來像你可能會返回只有一行，所以這是不太可能是

編輯

要獲得價格以及總和，這裏採用的子查詢的問題以獲得發票數量和價格總和，然後將其加入到業務表中： -

SELECT business.bussId, invoice_count as invoices, invoice_price, COUNT(DISTINCT rating.id) as ratingCount 
FROM business 
LEFT JOIN 
(
    SELECT bussId, COUNT(id) AS invoice_count, SUM(price) AS invoice_price 
    FROM invoice 
    WHERE userId = '3000' 
    GROUP BY bussId 
) invoice_sub 
ON invoice_sub.bussId = business.bussId 
LEFT JOIN rating ON (rating.bussId = business.bussId) 
WHERE business.bussId = 100 
GROUP BY business.bussId, invoices, invoice_price