mySQL 多个 INNER JOIN答案

【问题标题】：mySQL multiple INNER JOINmySQL 多个 INNER JOIN
【发布时间】：2011-05-29 00:44:21
【问题描述】：

我们要查询三个表，以了解每个企业的客户数量和每个企业的记录数

三个表分别是：

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

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

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

我们可以一次从两个表中选择一个计数（businessDetails 加上 clientDetails 或记录），这没有任何问题。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很好的结果：

--------------------------------
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 records 进行计数。我们尝试了以下查询，但发现由于某种原因它会乘以 totalClients 的数量：

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

这会返回类似的结果：

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

我希望我们只是犯了一些简单的错误。任何帮助将不胜感激。

【问题讨论】：

标签： mysql count inner-join

【解决方案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

【讨论】：

@direct00。请注意，由于子查询，这可能会对大型记录集产生一些性能问题。
+1 有趣：MySql 并不总是能很好地处理 COUNT DISTINCT；这可能就是为什么这种变化比我概述的更好。

【解决方案2】：

如果你像这样添加DISTINCT 怎么样：

SELECT  businessDetails.name AS businessName
        , COUNT(distinct clientDetails.ID) AS totalClients
        , COUNT(distinct records.ID) 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

这样，对于每组 businessDetails.name，您可以计算该“窗口”内不同的 clientDetails 和记录的数量。

【讨论】：

我认为这两项都会给你“1”。如果两个表都有自己的唯一 ID，则可以使用 distinct clientdetails.clientid 和 distinct records.recordid。
你是对的！感谢您指出疏忽。已相应地编辑了答案。
正确 - 只需添加 DISTINCT，两者都返回 1。
两个表都有自己的唯一 ID，所以我尝试使用 DISTINCT 代替，但查询超时。

【解决方案3】：

Count(Distinct [FieldName]) 在 MYSQL 中工作。试一试。

SELECT  businessDetails.name AS businessName
        , COUNT(Distinct 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

【讨论】：