15 * (sum(decode(a.cbxz,'正常',sum(nvl(a.ql,0)),0))+sum(decode(a.cbxz,'换表',sum(nvl(a.ql,0)),0))+sum(decode(a.cbxz,'坏表',sum(nvl(a.ql,0)),0))+sum(decode(a.cbxz,'纠错',sum(nvl(a.ql,0)),0)))/1000
- sum(decode(a.cbxz,'估抄',1,0))) + sum(decode(a.cbxz,'估抄',(decode(c.sffs,'托收',nvl(a.ql,0),0)),0))2*/1000 tc这种情况一般是括号的问题,数一下吧
- sum(decode(a.cbxz,'估抄',1,0))) + sum(decode(a.cbxz,'估抄',(decode(c.sffs,'托收',nvl(a.ql,0),0)),0))2*/1000 tc这种情况一般是括号的问题,数一下吧
解决方案 »
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select b.xm cby,
sum(decode(a.cbxz,'正常',sum(nvl(a.ql,0)),0))
+sum(decode(a.cbxz,'换表',sum(nvl(a.ql,0)),0))
+sum(decode(a.cbxz,'坏表',sum(nvl(a.ql,0)),0))
+sum(decode(a.cbxz,'纠错',sum(nvl(a.ql,0)),0)) scql,
sum(decode(a.cbxz,'估抄',(decode(c.sffs,'托收',nvl(a.ql,0),0)),0)) gcqlt,
sum(decode(a.cbxz,'估抄',(decode(c.sffs,'托收',0,nvl(a.ql,0))),0)) gcqly,
sum(nvl(a.ql,0)),15 * (sum(decode(a.cbxz,'正常',sum(nvl(a.ql,0)),0))
+sum(decode(a.cbxz,'换表',sum(nvl(a.ql,0)),0))
+sum(decode(a.cbxz,'坏表',sum(nvl(a.ql,0)),0))
+sum(decode(a.cbxz,'纠错',sum(nvl(a.ql,0)),0)))/1000--这里有点问题吧,多个)
- sum(decode(a.cbxz,'估抄',1,0)))-- 这里有点问题吧,多个)
+ sum(decode(a.cbxz,'估抄',(decode(c.sffs,'托收',nvl(a.ql,0),0)),0))2*/1000 tc
from t_yh_yqjl a,t_dm_ry b,t_yh_jcxx c
where a.cby=b.gh and a.cbyf=200402 group by b.xm
这句意思cbxz='正常',就作一个sum(ql),否则,0,根本就重复了sum(decode(a.cbxz,'正常',a.ql,0)) --是否这样意思