一统计sql语句

Question

--用户表
create table user_info 
(
  u_id int primary key,--用户id
  u_name varchar2(20),--姓名
  Department_id int ,--部门id
  Position_id int ,--职位id
)
--到账记录表
create table user_yes_info
(
  u_id int ,
  fee  float,--金额
  theMonth varchar(10)--月份
)
--待发记录表
create table user_waiting_info
(
  u_id int ,
  fee  float,--金额
 theMonth varchar(10)--月份)
--部门表
create table Department_info 
(
 Department_id int primary key ,
 Department_name varchar2(50)
)
--职位表
create table Position_info 
(
 Position_id int primary key ,
 Position_name varchar2(50)
)最后要查出来的字段是：
(查出所有用户的，没有的为0
姓名，部门 ,职位 ,总金额累计，到账金额累计，待发金额累计,本月总金额，本月到账金额，本月待发金额

Answer 1

select ui.u_name,
(select Department_name from Department_info where Department_id=ui.Department_id) as Department_name,
(select Position_name from Position_info where Position_id=ui.Position_id) as Position_name,
(select sum(fee) from user_yes_info where u_id=ui.u_id)+(select sum(fee) from user_waiting_info where u_id=ui.u_id) as sum_fee,
(select sum(fee) from user_yes_info where u_id=ui.u_id) as yes_fee,
(select sum(fee) from user_waiting_info where u_id=ui.u_id) as waiting_fee,
(select sum(fee) from user_yes_info where u_id=ui.u_id and theMonth='11')+(select sum(fee) from user_waiting_info where u_id=ui.u_id and theMonth='11') as thismonth_sum_fee,
(select sum(fee) from user_yes_info where u_id=ui.u_id and theMonth='11') as thismonth_yes_fee,
(select sum(fee) from user_waiting_info where u_id=ui.u_id and theMonth='11') as thismonth_waiting_fee,
from user_info ui

Answer 2

select u.u_name,
(select d.department_name from Department_info d where d.department_id=u.department_id) department_name,
(select p.position_name from Position_info p where p.position_id=u.position_id) position_id,
(select nvl(sum(nvl(uy.fee,0)),0)+nvl(sum(nvl(uw.fee,0)),0) from user_yes_info uy,user_waiting_info uw where uy.u_id=uw.u_id and uy.u_id=u.u_id) zje,
(select nvl(sum(nvl(uy.fee,0)),0) from user_yes_info uy,user_waiting_info uw where uy.u_id=uw.u_id and uy.u_id=u.u_id) dzje,
(select nvl(sum(nvl(uw.fee,0)),0) from user_yes_info uy,user_waiting_info uw where uy.u_id=uw.u_id and uy.u_id=u.u_id) dfje,
(select nvl(sum(nvl(uy.fee,0)),0)+nvl(sum(nvl(uw.fee,0)),0) from user_yes_info uy,user_waiting_info uw where uy.u_id=uw.u_id and uy.u_id=u.u_id and uy.themonth=uw.themonth and to_char(uy.themonth,'yyyy-mm') like '11') byzje,
(select nvl(sum(nvl(uy.fee,0)),0) from user_yes_info uy,user_waiting_info uw where uy.u_id=uw.u_id and uy.u_id=u.u_id and to_char(uy.themonth,'yyyy-mm') like '11') bydzje,
(select nvl(sum(nvl(uw.fee,0)),0) from user_yes_info uy,user_waiting_info uw where uy.u_id=uw.u_id and uy.u_id=u.u_id and to_char(uw.themonth,'yyyy-mm') like '11') bydfje
from user_info u
已经过测试，无误。
注：nvl(col,0)函数，当col字段值为空时，则取0。oracle里sum函数里，如果有null值，则全为计算结果为空。

Answer 3

这个语句有问题把，要考虑到如果user_yes_info表中的用户没有在user_waiting_info中，会导致总金额累计和本月总金额为0

Answer 4

select sum(fee) from user_yes_info where u_id=ui.u_id and theMonth='11')+(select sum(fee) from user_waiting_info where u_id=ui.u_id and theMonth='11'这种写法，如果user_yes_info 有而 user_waiting_info 没有的用户，会得到为0吗

Answer 5

select ui.u_name,
(select Department_name from Department_info where Department_id=ui.Department_id) as Department_name,
(select Position_name from Position_info where Position_id=ui.Position_id) as Position_name,
(select sum(nvl(fee,0)) from user_yes_info where u_id=ui.u_id)+(select sum(nvl(fee,0)) from user_waiting_info where u_id=ui.u_id) as sum_fee,
(select sum(nvl(fee,0)) from user_yes_info where u_id=ui.u_id) as yes_fee,
(select sum(nvl(fee,0)) from user_waiting_info where u_id=ui.u_id) as waiting_fee,
(select sum(nvl(fee,0)) from user_yes_info where u_id=ui.u_id and theMonth='11')+(select sum(nvl(fee,0)) from user_waiting_info where u_id=ui.u_id and theMonth='11') as thismonth_sum_fee,
(select sum(nvl(fee,0)) from user_yes_info where u_id=ui.u_id and theMonth='11') as thismonth_yes_fee,
(select sum(nvl(fee,0)) from user_waiting_info where u_id=ui.u_id and theMonth='11') as thismonth_waiting_fee,
from user_info ui

Answer 6

谢谢 lxpbs8851
我想问下 ，这样写执行速度怎么样，会不会比较慢呢
而且在一条语句里面，多出出现 (select sum(nvl(fee,0)) from user_yes_info where u_id=ui.u_id)
是不是这句话在出现的每一处，都要重新计算一次呢

Answer 7

select t2.u_name,d.Department_name ,p.Position_name, sum(nvl(t1.fee,0))+sum(nvl(w.fee,0)),sum(nvl(t1.fee,0)) ,sum(nvl(w.fee,0)),nvl(t1.fee,0)+nvl(w.fee,0),nvl(t1.fee,0) ,nvl(w.fee,0)from (select t1.uid,t1.u_name,t1.Department_id ,t1.Position_id, nvl(t1.fee,0) ,nvl(w.fee,0) from (select u.uid,u.u_name,u.Department_id ,u.Position_id ,y.fee from user_info u,(select * from user_yes_info  where theMonth='11')y  where u_id=y.u_id(+))t1 ,(select * user_waiting_info where theMonth='11')w where t1.uid=w.uid(+))t2,
 Department_info d,Position_info p where t2.uid=d.uid and t2.uid=p.uid 
左连接 + nvl做的 

Answer 8

试试下面的sql，测试过：select 
t.u_name,--姓名
t1.department_name,--部门
t2.position_name,--职位
t3.fee_yes, --到账金额累计
t4.fee_waiting,--待发金额累计
t5.fee_yes_month,--本月到账金额
t4.fee_waiting,--本月待发金额
t4.fee_waiting+t5.fee_yes_month as fee_month --本月总金额
from user_info t 
left join Department_info t1 on t.department_id=t1.department_id
left join Position_info t2 on t.position_id=t2.position_id
left join (
     select t.u_id,sum(nvl(t1.fee,0)) fee_yes from user_info t 
     left join user_yes_info t1 on t.u_id=t1.u_id
     group by t.u_id
) t3 on t.u_id=t3.u_id
left join (
     select t.u_id,sum(nvl(t1.fee,0)) fee_waiting from user_info t 
     left join user_waiting_info t1 on t.u_id=t1.u_id
     group by t.u_id
) t4 on t.u_id=t4.u_id
left join (
     select t.u_id,sum(nvl(t1.fee,0)) fee_yes_month from user_info t 
     left join user_yes_info t1 on t.u_id=t1.u_id
     where t1.themonth='11' --'11'为本月
     group by t.u_id
) t5 on t.u_id=t5.u_id
left join (
     select t.u_id,sum(nvl(t1.fee,0)) fee_waiting_month from user_info t 
     left join user_waiting_info t1 on t.u_id=t1.u_id
     where t1.themonth='11' --'11'为本月
     group by t.u_id
) t6 on t.u_id=t6.u_id
where 1=1 
order by t.u_name,t1.department_name,t2.position_name

Answer 9

select ui.u_name,
(select Department_name from Department_info where Department_id=ui.Department_id) as Department_name,
(select Position_name from Position_info where Position_id=ui.Position_id) as Position_name,
sum(nvl(uyi.fee,0))+sum(nvl(uw.fee,0)) as sum_fee,
sum(nvl(uyi.fee,0) as yes_fee,
sum(nvl(uw.fee,0)) as  waiting_fee,
sum(nvl(muyi.fee,0))+sum(nvl(muw.fee,0)) as thismonth_sum_fee,
sum(nvl(muyi.fee,0)) as thismonth_yes_fee,
sum(nvl(muw.fee,0)) as thismonth_waiting_fee,
from user_info ui left join user_yes_info uyi on uyi.u_id=ui.u_id
                  left join user_waiting_info uw on uw.u_id=ui.u_id
  left join (select fee,u_id from user_yes_info where theMonth='11') muyi on muyi.u_id=ui.u_id
  left join (select fee,u_id from user_waiting_info where theMonth='11') muw on muw.u_id=ui.u_id

Answer 10

貌似我写的简单一点select u_name as 姓名，department_name as 部门 ,position_name as 职位 ,
       '请指明计算公式' as 总金额累计，incomeTotal as 到账金额累计，payTotal as 待发金额累计,
       '请指明计算公式1' as 本月总金额，incomeTotalThis as 本月到账金额，payTotalThis as 本月待发金额
from 
(
select u.u_id,u.u_name, d.department_name, p.position_name
  from user_info u, Department_info d, Position_info p
 where u.Department_id = d.Department_id
   and u.Position_id = p.Position_id
) employee left join       
(
select u_id,
       sum(fee) as incomeTotal,
       sum(decode(theMonth,to_char(sysdate, 'mm'), fee, 0)) as incomeTotalThis,
       sum(decode(theMonth,to_char(add_months(sysdate,-1), 'mm'), fee, 0)) as incomeTotalLast
from user_yes_info
 group by u_id
) income on employee.u_id = income.u_id left join 
(
select u_id,
       sum(fee) as payTotal,
       sum(decode(theMonth,to_char(sysdate, 'mm'), fee, 0)) as payTotalThis,
       sum(decode(theMonth,to_char(add_months(sysdate,-1), 'mm'), fee, 0)) as payTotalLast
from user_waiting_info
 group by u_id
) pay on employee.u_id = pay.u_id
group by u_name, department_name, position_name