日期:2014-05-17  浏览次数:21086 次

左连接出现重复记录时取时间最小的那条记录匹配的SQL
SQL code
select p.rackingstarttime intoTime,s.partcode,s.partname,s.factory,s.house,
s.mlotno,s.vendorlot,pd.stockinqty intoQty,s.revisionno,'入库' intoType,to_char(U.Fullname) opearterName,
null orderNo,s.orderno sapOrderNo,s.reference sapreference,s.movementtype,wm.remark
from Pickingdetail pd
join Picking p on p.pickingid=pd.pickingid
left join sapmlot s on s.sapmlotid=pd.sapmlotid
left join "USER" U on U.Userid=pd.racker
left join wminventory wm on wm.sapmlotid=pd.sapmlotid [color=#FF0000]and wm.createdtime=min(wm.createdtime)[/color]
Where (pd.state=6)


上面and wm.createdtime=min(wm.createdtime)错误, left join wminventory会出现重复的记录,现在需要把wminventory表中去时间最小的哪条数据用来左连接,请问怎么做改动最小,谢谢。

------解决方案--------------------
SQL code

left join (select row_number() over(partition by sapmlotid order by createdtime) rn,sapmlotid,remark
    from wminventory) wm on wm.sapmlotid=pd.sapmlotid and wm.rn=1