Question

我有两张桌子客户表

+--------+---------+
| refno  | deposit |
+--------+---------+
| 1/13   |     -10 |
| 10/13  |     500 |
| 100/13 |       0 |
| 101/13 |     250 |
| 102/13 |    1000 |
+--------+---------+

Ledger Table

+--------+----------+------+----------+
| refno  | quantity | rate | recieved |
+--------+----------+------+----------+
| 1/13   |        2 |   70 |        0 |
| 10/13  |        3 |   80 |        0 |
| 100/13 |        3 |   60 |        0 |
| 101/13 |        4 |   60 |        0 |
| 102/13 |       10 |   65 |        0 |
+--------+----------+------+----------+

我想在分类帐栏中添加客户列（存款）（总计）我不想创建另一个表。我想要

refno | total = customer.deposit+(ledger.quantity*ledger.rate-ledger.received)
1/13  | -200
10/13 | 4210
100/13| 625
101/13| 280
102/13| 1000

此致

Answer 1

select c.refno,  c.deposit + l.total
from customer c join ledger l on c.refno = l.refno

Answer 2

这是你想要的吗？

select l.refno, l.total, c.deposit, l.total + coalesce(c.deposit, 0) as TotalWithDeposit
from ledger l left outer join
     customer c
     on l.refno = c.refno;

您的问题提到了聚合函数，但我没有看到任何聚合。

编辑：

select l.refno, l.total, c.deposit,
       c.deposit + (l.quantity * l.rate - l.received) as TotalWithDeposit
from ledger l left outer join
     customer c
     on l.refno = c.refno;

Answer 3

如果您在Ledger中有重复的refno，您可以尝试这种方式：

select b.refno, b.total + IFNULL(a.deposit, 0) as total
from
    (select refno, sum(total) as total
    from Ledger
    group by refno) b
    inner join Customer a on a.refno = b.refno

SQLFiddle demo

Answer 4

我认为作者是指直接选择以外的更新，但不确定哪个表用作基本表。例如，如果我们想在Customer中将值添加到Ledger中，它应该如下所示：

更新Ledger l内部联接Cusomter c on l.refno = c.refno set l.total = l.total + c.deposit。

MySQL Query：我想在两个表之间使用聚合函数？

4 个答案: