我试图显示交易付款清单，并在每次付款后显示当前余额。
修改 这是我的架构和示例数据 http://sqlfiddle.com/#!9/77dcc6/1
以下是预期结果的示例
char这是我最初的尝试，但总付款是根据所有交易付款的总和计算的。我想要的是相对于当前number date total paid balance
1 2018-01-01 1000 1000 0
2 2018-01-02 1250 1000 250
2 2018-01-05 1250 250 0
3 2018-01-03 2500 2000 500
3 2018-01-07 2500 300 200
3 2018-01-08 2500 200 0
5 2018-01-07 2149 2149 0
付款行
p然后我想我可以在select
t.number,
DATE(p.date) 'date',
ti.total 'total',
SUM(p.amount) 'paid',
ti.total - paid.total 'balance'
from payments p
left join transactions t
on p.transaction_id = t.id
left join (
select inner_ti.transaction_id, sum((inner_ti.price - inner_ti.discount) * inner_ti.quantity) 'total'
from transaction_items inner_ti
group by inner_ti.transaction_id
) ti on t.id = ti.transaction_id
left join (
select inner_p.transaction_id, sum(inner_p.amount) 'total'
from payments inner_p
group by inner_p.transaction_id
) paid on t.id = paid.transaction_id
group by t.number, DATE(p.date), ti.total, paid.total
order by t.number, DATE(p.date) ASC
中放置where子句，仅将与paid相关的付款相加，但我收到错误p.date
unknown column p.date 请注意我select
t.number,
DATE(p.date),
ti.total 'total',
SUM(p.amount) 'paid',
ti.total - paid.total 'balance'
from payments p
left join transactions t
on p.transaction_id = t.id
left join (
select inner_ti.transaction_id, sum((inner_ti.price - inner_ti.discount) * inner_ti.quantity) 'total'
from transaction_items inner_ti
group by inner_ti.transaction_id
) ti on t.id = ti.transaction_id
left join (
select inner_p.transaction_id, sum(inner_p.amount) 'total'
from payments inner_p
where inner_p.date <= p.date -- error unknown column p.date
group by inner_p.transaction_id
) paid on t.id = paid.transaction_id
group by t.number, DATE(p.date), ti.total, paid.total
order by DATE(p.date) ASC
分组，因为我们担心当天付款。
有人可以告诉我为什么我会收到这个错误吗？是否有任何解决方法可以达到预期效果？
谢谢！
答案 0 :(得分：2)
也许你应该重新开始
drop table if exists p,t,ti;
create table t(id int,number int);
create table ti(id int,tid int, price int);
create table p(id int,tid int,dt date, paid int);
insert into t values (1,1355),(2,1359),(3,1361);
insert into ti values (1,1,400),(2,1,200),(3,1,299),(4,2,4045),(5,3,1500),(6,3,40),(7,3,8);
insert into p values (1,1,'2018-01-01',200),(2,1,'2018-01-01',250),(3,1,'2018-02-01',449),
(4,2,'2018-01-01',1515),(5,2,'2018-02-01',35),
(6,3,'2018-01-06',1548);
select t.id,number,p.dt paiddate,(select sum(price) from ti where ti.tid = t.id) due,
sum(p.paid) paid,
ifnull((Select sum(p1.paid) from p p1 where p1.dt < p.dt and p1.tid = p.tid),0) + sum(p.paid) aggsumpaid,
(select sum(price) from ti where ti.tid = t.id) -
(ifnull((Select sum(p1.paid) from p p1 where p1.dt < p.dt and p1.tid = p.tid),0) + sum(p.paid)) runningbal
from t
left join p on p.tid = t.id
group by t.id,number,p.dt
结果
+------+--------+------------+------+------+------------+------------+
| id | number | paiddate | due | paid | aggsumpaid | runningbal |
+------+--------+------------+------+------+------------+------------+
| 1 | 1355 | 2018-01-01 | 899 | 450 | 450 | 449 |
| 1 | 1355 | 2018-02-01 | 899 | 449 | 899 | 0 |
| 2 | 1359 | 2018-01-01 | 4045 | 1515 | 1515 | 2530 |
| 2 | 1359 | 2018-02-01 | 4045 | 35 | 1550 | 2495 |
| 3 | 1361 | 2018-01-06 | 1548 | 1548 | 1548 | 0 |
+------+--------+------------+------+------+------------+------------+
5 rows in set (0.02 sec)