我需要左连接SQL查询的帮助

我需要帮助写一个简单的程序。让我解释一下我想做什么

我有三张桌子

tJobOffer

应用

应用状态

我想创建一个程序，返回一个tJobOffer列表，其中包含响应此tJobOffer的候选人数量。候选人是我表格申请状态的一个状态。此表链接到链接到tJobOffer的tApplication。申请可以是候选/接受/拒绝/忽略/

我创建了这个查询：

    SELECT 
        [T].[JobOfferId],
        [T].[JobOfferTitle],
        COUNT([A].[ApplicationId]) AS [CandidateCount]

    FROM        [tJobOffer] AS [T]
    LEFT JOIN   [tApplication] AS [A]
        ON      [A].[JobOfferId] = [T].[JobOfferId]
    LEFT JOIN   [tApplicationStatus] AS [S]
        ON      [S].[ApplicationStatusId] = [A].[ApplicationStatusId]
        AND     [S].[ApplicationStatusTechnicalName] = 'CANDIDATE'

    GROUP BY
            [T].[JobOfferId],
            [T].[JobOfferTitle]
            --[A].[ApplicationStatusId]

    ORDER BY [T].[JobOfferTitle]

结果是

> 52ED7C67-21E1-49BB-A1F8-0601E6EED1EA  Annonce 1   0
> F26B228D-0C81-4DA8-A287-F8F997CC1F9C  Annonce 1b  0
> 9DA60B23-F113-4C7F-9707-2B90C1556D5D  Announce 25 2
> 258E11A7-79C1-47B6-8C61-413AA54E2360  Announce 3  0
> DA582383-5DF4-4E1D-837C-382371BDEF57  Announce 6  2

这是不正确的，因为我只有一个候选人宣布6。如果有人把我的路线

    --AND     [S].[ApplicationStatusTechnicalName] = 'CANDIDATE'

在注释中，结果是相同的。我的问题似乎忽略了这一行。怎么了

编辑--

我的正确结果应该是

> 52ED7C67-21E1-49BB-A1F8-0601E6EED1EA  Annonce 1   0
> F26B228D-0C81-4DA8-A287-F8F997CC1F9C  Annonce 1b  0
> 9DA60B23-F113-4C7F-9707-2B90C1556D5D  Announce 25 2
> 258E11A7-79C1-47B6-8C61-413AA54E2360  Announce 3  0
> DA582383-5DF4-4E1D-837C-382371BDEF57  Announce 6  1

试试这个：

SELECT 
        [T].[JobOfferId],
        [T].[JobOfferTitle],
        COUNT([A].[ApplicationId]) AS [CandidateCount]

    FROM        [tJobOffer] AS [T]
    LEFT JOIN   [tApplication] AS [A]
        ON      [A].[JobOfferId] = [T].[JobOfferId]
    LEFT JOIN   
         (SELECT ApplicationStatusId FROM [tApplicationStatus]
          WHERE  [ApplicationStatusTechnicalName] = 'CANDIDATE') [S]
        ON      [S].[ApplicationStatusId] = [A].[ApplicationStatusId]
    GROUP BY
            [T].[JobOfferId],
            [T].[JobOfferTitle]
    ORDER BY [T].[JobOfferTitle]

我猜你想要这个：

SELECT 
    [T].[JobOfferId],
    [T].[JobOfferTitle],
    COUNT([A].[ApplicationId]) AS [CandidateCount]

FROM        [tJobOffer] AS [T]
LEFT JOIN   [tApplication] AS [A]
    INNER JOIN  [tApplicationStatus] AS [S]
        ON      [S].[ApplicationStatusId] = [A].[ApplicationStatusId]
        AND     [S].[ApplicationStatusTechnicalName] = 'CANDIDATE'
    ON      [A].[JobOfferId] = [T].[JobOfferId]

GROUP BY
        [T].[JobOfferId],
        [T].[JobOfferTitle]

ORDER BY [T].[JobOfferTitle] ;

您还可以先连接2个“应用程序”表，分组，然后使用

JobOffer

连接派生表：

SELECT 
    [T].[JobOfferId],
    [T].[JobOfferTitle],
    COALESCE([G].[Cnt], 0) AS [CandidateCount]

FROM        [tJobOffer] AS [T]
LEFT JOIN   
            ( SELECT 
                  [A].[JobOfferId],
                  COUNT(*) AS [Cnt]

              FROM        [tApplication] AS [A]
              INNER JOIN  [tApplicationStatus] AS [S]
                  ON      [S].[ApplicationStatusId] = [A].[ApplicationStatusId]
                  AND     [S].[ApplicationStatusTechnicalName] = 'CANDIDATE'

              GROUP BY
                          [A].[JobOfferId]
            ) AS [G]
    ON      [G].[JobOfferId] = [T].[JobOfferId]

ORDER BY [T].[JobOfferTitle] ;

---

我不确定您是否在问，为什么使用第二个左连接的ON子句筛选的行仍然显示在最终结果中

如果这是你的问题，那么答案是：

基于on谓词的筛选器不是最终筛选器，即on谓词不确定该行是否显示在输出中，只确定它是否与其他表中的行匹配

另一方面，WHERE子句是final，在FROM子句之后处理，即在处理所有表运算符之后，以及（在外部联接的情况下）在生成所有外部行之后

---

因此，当您需要在生成外部行之后应用过滤器，并且希望过滤器是最终过滤器时，请在WHERE子句中指定谓词

根据您的要求，在必须使用内部联接之后，首先需要使用左联接

试试下面这个

SELECT 
    [T].[JobOfferId],
    [T].[JobOfferTitle],
    COUNT([A].[ApplicationId]) AS [CandidateCount]

FROM        [tJobOffer] AS [T]
LEFT JOIN   [tApplication] AS [A]
    INNER JOIN   [tApplicationStatus] AS [S]
        ON      [S].[ApplicationStatusId] = [A].[ApplicationStatusId]
        AND     [S].[ApplicationStatusTechnicalName] = 'CANDIDATE'
    ON      [A].[JobOfferId] = [T].[JobOfferId]

GROUP BY
        [T].[JobOfferId],
        [T].[JobOfferTitle]

ORDER BY [T].[JobOfferTitle] ;

---

如果您可以添加带有所需结果的示例记录，并回答您的问题

：D

如果您没有在任何地方使用它，为什么要以[S]的身份加入

[tApplicationStatus]呢？正确。非常感谢。我不知道可以加入加入加入。