任何人都可以幫助解決這個錯誤嗎？其他人向我展示了如何創建連接，據我所知，我完全復制了它，但是使用另一個表，但我無法讓它作業：
SELECT [ITEMID]
,[ITEMDATAAREAID]
,ITEM.[ITEMGROUPID]
,[ITEMGROUPDATAAREAID]
FROM [ELG_DynamicsAX_DWH].[dbo].[INVENTITEMGROUPITEM] AS ITEM
Join(Select[NAME] AS [ITEMGROUPNAME]
,[ITEMGROUPID]
FROM[ELG_DynamicsAX_DWH].[dbo].[INVENTITEMGROUP] AS INVENTGROUP ON INVENTGROUP.ITEMGROUPID =
ITEM.ITEMGROUPID
謝謝您的幫助！
uj5u.com熱心網友回復：
SELECT
[ITEMID]
,[ITEMDATAAREAID]
,ITEM.[ITEMGROUPID]
,INVENTGROUP.[NAME] AS [ITEMGROUPNAME]
,[ITEMGROUPDATAAREAID]
FROM
[ELG_DynamicsAX_DWH].[dbo].[INVENTITEMGROUPITEM] AS ITEM
Join [ELG_DynamicsAX_DWH].[dbo].[INVENTITEMGROUP] AS INVENTGROUP
ON INVENTGROUP.ITEMGROUPID = ITEM.ITEMGROUPID
uj5u.com熱心網友回復：
SELECT [ITEMID]
,[ITEMDATAAREAID]
,ITEM.[ITEMGROUPID]
,[ITEMGROUPDATAAREAID]
X.[ITEMGROUPNAME]
FROM [ELG_DynamicsAX_DWH].[dbo].[INVENTITEMGROUPITEM] AS ITEM
Join
(
Select[NAME] AS [ITEMGROUPNAME]
,[ITEMGROUPID]
FROM[ELG_DynamicsAX_DWH].[dbo].[INVENTITEMGROUP] AS INVENTGROUP
)X ON X.ITEMGROUPID =ITEM.ITEMGROUPID
像這樣的東西
轉載請註明出處，本文鏈接：https://www.uj5u.com/yidong/510795.html
標籤：sql加入