Question

t1

DocNo                         Date            Description
Doc-12345-67890               15Jan19         QC-Passed            

t2
DocNo                         Date            Description
Doc-09876-21347               16Jan19         QC-Passed      

t3
DocNo                         Date            Description
12345                         15Jan19         QC-Passed                   
09876                         16Jan19         QC-Passed  

Select *
from
(
select left(right(docno,len(docno)-3),5) as DocNo,Date,Description 
from t1
union all
select *
from t2) as RC    
inner join    
t3    
on
rc.docno=t3.docno

它不起作用，任何人都可以帮忙吗？

Answer 1

我想你想要

select *
from ((select DocNo, Date, Description 
       from t1
      ) union all
      (select DocNo, Date, Description
       from t2
      )
     ) RC inner join    
     t3    
     on left(right(rc.docno, len(rc.docno) - 3), 5) = t3.docno;

或者：

select *
from ((select left(right(rc.docno, len(rc.docno) - 3), 5) as DocNo, Date, Description 
       from t1
      ) union all
      (select left(right(rc.docno, len(rc.docno) - 3), 5) as DocNo, Date, Description
       from t2
      )
     ) RC inner join    
     t3    
     on rc.docno = t3.docno;

换句话说，select的第一个union all中的表达式不会“携带”到第二个select中。

Answer 2

看来您的数据是t3表是驱动程序，它的docno是'Doc-'文本后t1和t2的前五个字符的主要字符。下面的tsql通过使用SubString来获取“ t1”和“ t2” DocNo：

查询：

Select *
from
(
select SubString(DocNo,5,5) as DocNo,dt,Description 
from #tbl1
union all
select SubString(DocNo,5,5) as DocNo,dt,Description
from #tbl2) as RC    
inner join    
#tbl3    
on
rc.docno=#tbl3.docno

结果：

DocNo   dt      Description DocNo   dt      Description
12345   15Jan19 QC-Passed   12345   15Jan19 QC-Passed
09876   16Jan19 QC-Passed   09876   16Jan19 QC-Passed

合并两个表的所有内容，然后嵌套内部联接

2 个答案: