select kmb.km, sum(a.srje),sum(a.zcje), sum(a.cy)
from kmb
leftjoin wztz a on kmb.kmbh = a.kmbh and SUBSTR(a.pzh,1,6)='200808'groupby kmb.km orderby kmb.kmbh
SQL code
select kmb.km, sum(a.srje),sum(a.zcje), sum(a.cy) , (sum(b.srje)-sum(b.zcje))
from kmb
leftjoin wztz a on kmb.kmbh = a.kmbh and SUBSTR(a.pzh,1,6)='200808'leftjoin wztz b on kmb.kmbh = b.kmbh and SUBSTR(b.pzh,1,6)<='200808'groupby kmb.km orderby kmb.kmbh
呵呵, 是对比 select kmb.km, a.srje, a.zcje, a.cy from kmb left join wztz a on kmb.kmbh = a.kmbh and SUBSTR(a.pzh,1,6)='200808' order by kmb.kmbh
与
select kmb.km, sum(a.srje),sum(a.zcje), sum(a.cy) , (sum(b.srje)-sum(b.zcje)) from kmb left join wztz a on kmb.kmbh = a.kmbh and SUBSTR(a.pzh,1,6)='200808' left join wztz b on kmb.kmbh = b.kmbh and SUBSTR(b.pzh,1,6) <='200808' group by kmb.km order by kmb.kmbh 记录有什么不同 找到原因,就知道为什么结果不同了
select kmb.km, sum(a.srje),sum(a.zcje)
from kmb
leftjoin wztz a on kmb.kmbh = a.kmbh and SUBSTR(a.pzh,1,6)='200808'groupby kmb.km orderby kmb.kmbh
select kmb.km, sum(b.srje),sum(b.zcje)
from kmb
leftjoin wztz b on kmb.kmbh = b.kmbh and SUBSTR(b.pzh,1,6)<='200808'groupby kmb.km orderby kmb.kmbh
我的目的是想把以上两个结果合并起来,但是
SQL code
select kmb.km, sum(a.srje),sum(a.zcje), sum(b.srje),sum(b.zcje)
from kmb
leftjoin wztz a on kmb.kmbh = a.kmbh and SUBSTR(a.pzh,1,6)='200808'leftjoin wztz b on kmb.kmbh = b.kmbh and SUBSTR(b.pzh,1,6)<='200808'groupby kmb.km orderby kmb.kmbh