Sql 我有两个列名称相同的临时表。我想连接这两个表，但通过内部连接，它会发送如下结果

-DtI有两个具有相同列名的临时表。我想连接这两个表，但通过内部连接，它会发送如下结果

我希望结果的外观：

工会的问题是它没有比较这两个结果

使用的查询：

--Detox  

select distinct client.CL_CASENO , p.PR_NAME , a.AD_EFFDT,a.AD_EXPDT, a.CR_CODE
into #d
from MCCSADPF as sud
inner join  PCHDOCPF as doc on sud.SAP_DOCID = doc.DC_RCDID
inner join v_clients as client on client.CL_RCDID =  doc.DCF_CLTID
inner join v_Providers as p on sud.SAF_APRVID = p.PR_RCDID
left outer join CODCODPF as code on sud.SAF_TDTXTP = code.CO_RCDID
inner join v_Authorizations as a on a.CL_CASENO = client.CL_CASENO
where sud.SAF_APRVID  in ('10118','10115','191853' )
and sud.SAF_TDTXTP = '16774' --Detox  
and CR_CODE in ('H0010') 
and AD_EFFDT >= '2014-10-1' and  AD_EXPDT <= GETDATE()


--Ambulatory  

select distinct client.CL_CASENO , p.PR_NAME , a.AD_EFFDT , a.AD_EXPDT,a.CR_CODE 
into #amb
from MCCSADPF as sud
inner join PCHDOCPF as doc on sud.SAP_DOCID = doc.DC_RCDID
inner join v_clients as client on client.CL_RCDID =  doc.DCF_CLTID
inner join v_Providers as p on sud.SAF_APRVID = p.PR_RCDID
left outer join CODCODPF as code on sud.SAF_TDTXTP = code.CO_RCDID
inner join v_Authorizations as a on a.CL_CASENO = client.CL_CASENO
where sud.SAF_APRVID  in ('191847','10115','191853') 
and sud.SAF_TDTXTP in ('16778') --OutPatient
and CR_CODE in('H0020')
and AD_EFFDT >= '2014-10-1' and  AD_EXPDT <= GETDATE()  

----------------------------------------------------------------------
--JOIN--

Select * from #d as d
inner join #amb as a on d.CL_CASENO = a.CL_CASENO
where d.AD_EFFDT <= a.AD_EFFDT  

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如果您只需要两个表中都存在的患者，那么您的查询几乎是正确的。你只要抓住第一张桌子就行了

SELECT d.* 
FROM #d d
JOIN #amb a ON d.CL_CASENO = a.CL_CASENO
WHERE d.AD_EFFDT <= a.AD_EFFDT

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这是一个很好的起点。与您的声明联合比较，我认为您需要完全外部连接。感谢您的链接SeanAditya，完全外部连接将提供所有数据结果。我只需要经过排毒和门诊的客户