oracle统计比例,oracle实现累加,累计百分比计算

这几天一直在搞报表,发现Oracle真的很强大,需要用到累加功能,发现强大的oracle还真有,用over(order by field)解决的

例子:

0883bb65f257301d5d2df66f7567bb32.png

数据表中最后一列就是累加的效果

累加sql:

select substr(hn.fd_sjsxsj_time,1,7) 日期,

sum(hn.fd_sh_workload) 工作量,

sum(sum(hn.fd_sh_workload)) over(order by substr(hn.fd_sjsxsj_time,1,7)) 累加

from tjstat.his_need hn

where substr(hn.fd_sjsxsj_time,1,4) = '2017'

group by substr(hn.fd_sjsxsj_time,1,7)

order by substr(hn.fd_sjsxsj_time,1,7);

趁热打铁,根据累计求和,进一步求占总和的百分比

例子:

2441863c1d89402e8e1b58911cc1dd3f.png

​sql:

思路:计算累计百分比,先求列和,然后嵌套求百分比

select t1.*,round(t1.accu_sum/t2.allsum*100,2)||'%' from (select substr(hn.fd_sjsxsj_time,1,7) 日期,

sum(hn.fd_sh_workload) 工作量,

sum(sum(hn.fd_sh_workload)) over(order by substr(hn.fd_sjsxsj_time,1,7)) accu_sum

from tjstat.his_need hn

where substr(hn.fd_sjsxsj_time,1,4) = '2017'

group by substr(hn.fd_sjsxsj_time,1,7)

order by substr(hn.fd_sjsxsj_time,1,7))t1,(select sum(hn.fd_sh_workload) allsum from tjstat.his_need hn where substr(hn.fd_sjsxsj_time,1,4) = '2017') t2