mySQL 多重 INNER JOIN

Question

我们要查询三个表以找出每个企业的客户数量和每个企业的记录数

三个表是:

businessDetails
-------------------
businessDetails.ID
businessDetails.name

clientDetails
-------------------
clientDetails.ID
clientDetails.businessDetailsID

records
-------------------
records.ID
records.businessDetailsID

我们能够同时从两个表中选择计数(businessDetails 加上 clientDetails 或记录)没有任何问题。例如

SELECT  businessDetails.name AS businessName
        , COUNT(clientDetails.businessDetailsID) AS totalClients
FROM `businessDetails`
INNER JOIN clientDetails 
        ON clientDetails.businessDetailsID = businessDetails.businessDetailsID
GROUP BY 
businessDetails.name
ORDER BY
totalClients DESC

这个查询给了我们预期的不错的结果:

--------------------------------
businessName    |   totalClients
--------------------------------
Initech         |   23
Cylon Inc       |   148
The Dude Ltd    |   71

我们遇到的问题是如何同时对 clientDetails 和记录进行计数。我们尝试了以下查询，但发现它出于某种原因使 totalClients 的数量成倍增加:

SELECT  businessDetails.name AS businessName
        , COUNT(clientDetails.businessDetailsID) AS totalClients
        , COUNT(records.businessDetailsID) AS totalRecords
FROM `businessDetails`
INNER JOIN clientDetails 
        ON clientDetails.businessDetailsID = businessDetails.businessDetailsID
INNER JOIN records ON records.businessDetailsID = businessDetails.ID
GROUP BY 
businessDetails.name
ORDER BY
totalClients DESC

这会返回类似这样的结果:

--------------------------------------------------------
businessName    |   totalClients    |   totalRecords
--------------------------------------------------------
Initech         |   93              |    93
Cylon Inc       |   398             |    398
The Dude Ltd    |   215             |    215

我想我们只是犯了一些简单的错误。任何帮助将不胜感激。

Answer 1

你需要这样的东西:

SELECT  
    businessDetails.name AS businessName,
    (SELECT count('x') FROM clientDetails WHERE clientDetails.businessDetailsID = businessDetails.businessDetailsID) as totalclients,
    COUNT(records.businessDetailsID) AS totalRecords
FROM 
    `businessDetails`
    INNER JOIN records ON records.businessDetailsID = businessDetails.ID
GROUP BY 
    businessDetails.name
ORDER BY
    totalClients DESC