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Question

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

Answer 1

sum(decode(a.cbxz,'估抄',(decode(c.sffs,'托收',nvl(a.ql,0),0)),0))2*/1000 尾部")2*/1000"是不是写错了?

Answer 2

看看是不是下面两个地方的问题
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

Answer 3

15 * (sum(decode(a.cbxz,'正常',sum(nvl(a.ql,0)),0))  --意义何在
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