我需要帮助来编写一个简单的程序。让我解释一下我要做的事情。
我有3张桌子
我想创建一个程序,它返回一个tJobOffer列表,其中包含响应此tJobOffer的候选者数量。候选人是我的表tApplicationStatus的状态。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1个候选人宣布6.如果设置我的行
--AND [S].[ApplicationStatusTechnicalName] = 'CANDIDATE'
在评论中结果是一样的。我的查询似乎忽略了这一行。有什么问题?
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
答案 0 :(得分:1)
我想你想要这个:
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] ;
您也可以首先加入2个“应用程序”表,分组,然后将派生表与JobOffer
一起加入:
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] ;
答案 1 :(得分:1)
根据您的要求,您必须在必须使用内连接后使用左连接。
尝试以下
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] ;
答案 2 :(得分:0)
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]
答案 3 :(得分:0)
我不确定您是否在问为什么使用第二个LEFT JOIN的ON子句过滤的行仍显示在最终结果中。
如果这是你的问题,那么答案是:
基于ON谓词的过滤器不是最终的,即ON谓词不确定该行是否会显示在输出中,只是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,在处理完所有表运算符之后(在外连接的情况下),在生成所有外部行之后。
因此,如果需要在生成外部行之后应用过滤器,并且希望过滤器为final,请在WHERE子句中指定谓词。