Sql server 具有更多连接属性的表上的连接

我有一个带有应用程序（

AppID，RequestDt，OrgID

）的表

App

，一个带有分支机构名称（

Branch\u id，Branch\u name

）的表

Wspace

和员工（

branchID，NmbrEmp

）

这个select将返回：（我只显示了两行的示例，实际上有超过1000行）

现在我想用

INFO

添加下一列。此

INFO

位于另一个名为

App2

的表中，该表仅包含特定的应用程序（

AppID，INFO

）

我想计算表App2中信息为1的行，并将该数字放入每行的下一列

但每次它把我原来的小组搞砸了。。。这是因为App2中只有“OKApps”（还有一些应用根本不在App中），所以当我这样做时：

join App2 c on b.APPId = c.APPId

我的NOTOK应用程序将从群中消失，当我离开加入时，我的应用程序将比以前更多

---

感谢您的帮助

正如我在评论中所说的，您可以使用Like变量，也可以通过交叉连接添加：

DECLARE @someTable TABLE ([Date] VARCHAR(10), Branch VARCHAR(10), NmbrEmp INT, TotalApp INT, TotalAppOK INT, TotalAppNOK INT)

INSERT INTO @someTable SELECT '201402', 'Home', 5, 28, 20, 8
INSERT INTO @someTable SELECT '201402', 'Office', 4, 15, 8, 7
INSERT INTO @someTable SELECT '201403', 'Home', 5, 24, 7, 17
INSERT INTO @someTable SELECT '201403', 'Office', 4, 30, 9, 21



DECLARE @App2 TABLE (AppID INT, INFO INT)

INSERT INTO @App2 SELECT 2, 2
INSERT INTO @App2 SELECT 3, 1
INSERT INTO @App2 SELECT 4, 2

---FIRST CHOICE

DECLARE @App2Count INT = (SELECT COUNT(*) FROM @App2 WHERE INFO = 2)

SELECT Date, Branch, NmbrEmp, TotalApp, TotalAppOK, TotalAppNOK, @App2Count as App2
FROM @someTable

---SECOND CHOICE
SELECT Date, Branch, NmbrEmp, TotalApp, TotalAppOK, TotalAppNOK, A.App2 as App2
FROM @someTable sT
CROSS JOIN (SELECT COUNT(*) as App2 FROM @App2 WHERE INFO = 2) as A

---

您应该只计算子查询中的

Info

s，然后加入其中：

select  substring(b.RequestDt,1,6) as Date,
    a.branch_name,
    d.NmbrEmp,
    COUNT(b.AppId) as TotalApp,
    SUM(convert(int,b.OKFlg)) as TotalAppOK
    SUM(convert(int,b.NotOKFlg)) as TotalAppNOK, 
    e.C 
from    
    App b 
join Branch a on a.branch_id  = b.OrgId 
left join WSpace d on a.branch_id = d.branchID 
LEFT JOIN   
    (SELECT COUNT(*) AS C, AppID 
        FROM App2 
        WHERE Info = 1 
        GROUP BY AppId) e 
on e.AppId = b.AppId

where 
    substring(b.RequestDt,1,6) >= 201301
group by 
    substring(b.RequestDt,1,6), a.branch_name, d.NmbrEmp

---

为什么不在这之前只进行

COUNT

呢

选择

并将其作为变量输入？在我处理答案时，您能否显示表格的结构？请尝试按APPID分组，并在单独的表格中进行计数，并使用APPID加入此可用结果谢谢您的提示，我会在30分钟内试用，然后让你知道。我想我解释得很糟糕。我想知道在表App2中，每个原始版本中有多少应用程序的Info=1。这意味着在第行：“201402，Home，5，28，20，8”，我已经拒绝了28个应用程序，我想知道其中有多少在表App2中，INFO=1。和in-App table是根本不在App2表中的AppID，而in-App2表是不在App表中的AppID

DECLARE @someTable TABLE ([Date] VARCHAR(10), Branch VARCHAR(10), NmbrEmp INT, TotalApp INT, TotalAppOK INT, TotalAppNOK INT)

INSERT INTO @someTable SELECT '201402', 'Home', 5, 28, 20, 8
INSERT INTO @someTable SELECT '201402', 'Office', 4, 15, 8, 7
INSERT INTO @someTable SELECT '201403', 'Home', 5, 24, 7, 17
INSERT INTO @someTable SELECT '201403', 'Office', 4, 30, 9, 21



DECLARE @App2 TABLE (AppID INT, INFO INT)

INSERT INTO @App2 SELECT 2, 2
INSERT INTO @App2 SELECT 3, 1
INSERT INTO @App2 SELECT 4, 2

---FIRST CHOICE

DECLARE @App2Count INT = (SELECT COUNT(*) FROM @App2 WHERE INFO = 2)

SELECT Date, Branch, NmbrEmp, TotalApp, TotalAppOK, TotalAppNOK, @App2Count as App2
FROM @someTable

---SECOND CHOICE
SELECT Date, Branch, NmbrEmp, TotalApp, TotalAppOK, TotalAppNOK, A.App2 as App2
FROM @someTable sT
CROSS JOIN (SELECT COUNT(*) as App2 FROM @App2 WHERE INFO = 2) as A

select  substring(b.RequestDt,1,6) as Date,
    a.branch_name,
    d.NmbrEmp,
    COUNT(b.AppId) as TotalApp,
    SUM(convert(int,b.OKFlg)) as TotalAppOK
    SUM(convert(int,b.NotOKFlg)) as TotalAppNOK, 
    e.C 
from    
    App b 
join Branch a on a.branch_id  = b.OrgId 
left join WSpace d on a.branch_id = d.branchID 
LEFT JOIN   
    (SELECT COUNT(*) AS C, AppID 
        FROM App2 
        WHERE Info = 1 
        GROUP BY AppId) e 
on e.AppId = b.AppId

where 
    substring(b.RequestDt,1,6) >= 201301
group by 
    substring(b.RequestDt,1,6), a.branch_name, d.NmbrEmp