cust table:
cust_id, name, etc
bill table:
bill_id, cust_id, amt, due_date, status (open/closed)
payment table:
payment_id, bill_id, amt etc
客户可以通过分期付款来结算单笔账单。因此,一个bill_id可能与payment_ids有关。
cust_id |到期了
'due amt'是所有bill.amts的总和 - 所有payment.amts和已打开状态的总和。
比尔表
bill_id cust_id amt status
1 18 200 open
2 18 200 open
3 17 200 open
4 17 200 open
5 17 200 open
6 17 200 closed
付款表
payment_id bill_id cust_id amt
1 1 18 50
2 2 18 40
3 3 17 10
预期输出
cust_id due_amt hint/how
17 590 (600-10) .. 600 -> because one is closed
18 310 (400-(50+40))
答案 0 :(得分:2)
select c.cust_id, sum(b.amt) - sum(p.amt) as due_amt
from cust c
left join bill b on b.cust_id = c.cust_id and b.status = 'open'
left join payment p on p.bill_id = b.bill_id
group by c.cust_id