我有一个问题，获取一些数据来显示我想要它 . 这是一种称为Cabinet Vision的软件中的某种形式的SQL .

这是应该做的：表1应该显示5 CabQty，对于CabJob1和3 CabQty用于CabJob2表1目前显示5为CabJob1和CabJob2

DoorCount工作正常，但CabCount没有看到表1（每周时间表）Weekly Schedule Table

下面是我的两个DoorCount和CabCount表的截图，当前正在正确添加它们，但是在计划表的某个地方，它几乎选择了TOP CabQty

Door Count Table Cab Count Table

下面是每个表的SQL ------------------ DoorCount SQL

SELECT Doors.[Door Num], Doors.[Material Schedule], Doors.[Cabinet ID], 
Count(Doors.[Door Num]) As DoorCount, Doors.[Door Name]

FROM Doors

WHERE Doors.[Door Num] > 0

GROUP BY Doors.[Door Num], Doors.[Material Schedule], Doors.[Cabinet ID], 
Doors.[Door Name]
----------------------

------------------ CabCount SQL
SELECT SUM(Cabinets.Quantity) AS CabQty, [Job Info].[Job Name]

FROM ([Job Info] 
INNER JOIN Rooms ON [Job Info].ID=Rooms.[Job ID]) 
INNER JOIN Cabinets ON Rooms.RoomNumber=Cabinets.[Room ID]

WHERE Cabinets.[Cabinet Name] LIKE 'W1D%' OR Cabinets.[Cabinet Name] LIKE 
't1D%' OR Cabinets.[Cabinet Name] LIKE 'SB%'

GROUP BY [Job Info].[Job Name]
------------------

---------------Weekly Schedule SQL
SELECT  [Job Info].[Job Number], [Job Info].[Customer Name], [Job Info].[Job 
Name], [Job Info].[Customer Comments], DoorCount.[Material Schedule], 
Sum(DoorCount.[DoorCount]) AS DoorCount,  [Job Info].[Customer Email], [Job 
Info].[Customer Fax], [Job Info].[Customer Mobile], [Job Info].[Customer 
Phone], COUNT(CabCount.CabQty) AS CabQty, CabCount.[Job name], Rooms.[Job 
ID]

FROM ((([Job Info]
INNER JOIN Rooms ON [Job Info].ID=Rooms.[Job ID])
INNER JOIN Cabinets ON Rooms.RoomNumber=Cabinets.[Room ID])  
INNER JOIN DoorCount ON Cabinets.[Cabinet ID]=DoorCount.[Cabinet ID])
INNER JOIN CabCount ON [Job Info].[Job Name]=CabCount.[Job Name]

GROUP BY [Job Info].[Job Name], [Job Info].[Job Number], [Job Info]. 
[Customer Name], DoorCount.[Material Schedule],  [Job Info].[Customer 
Comments], [Job Info].[Customer Email], [Job Info].[Customer Fax], [Job 
Info].[Customer Mobile], [Job Info].[Customer Phone], CabCount.[Job name], 
Rooms.[Job ID]

ORDER BY [Job Info].[Customer Email] ASC
---------------------------

感谢您的时间和意见

sql sum