Question

这个问题的标题可能实际上并不能反映出我想要的东西，我想不出更好的东西。也许我应该附加一个'然后加入两个以这种方式获得的表格到最后。

假设我有这张桌子。情况是，我们支付身份loan_id的贷款，我们将付款存储在“付款”表中，每笔付款都有自己的id。当客户付款时，我们会将paid更改为1。

+---------+----+--------+------+
| loan_id | id | amount | paid |
+---------+----+--------+------+
|       1 |  1 |   1000 |    1 |
|       1 |  2 |   1000 |    1 |
|       1 |  3 |   1000 |    0 |
|       2 |  4 |  10000 |    0 |
|       3 |  5 |  20000 |    1 |
+---------+----+--------+------+

现在我们需要一份贷款报告。输出应如下所示：

+---------+-------+-----------+
| loan_id |  paid | remaining |
+---------+-------+-----------+
|       1 |  2000 |      1000 |
|       2 |     0 |     10000 |
|       3 | 20000 |         0 |
+---------+-------+-----------+

paid列基本上是所有amount字段的总和，其中pay = 1.而remaining是amount行的总和，其中pay = 0。应按loan_id分组。

我尝试了什么：

我尝试了不同的连接，产品等，但我可以从我的查询中获得的最好的事情是：

SELECT loan_id, SUM(paid), SUM(remaining)
FROM (
    SELECT loan_id, 0 AS paid, SUM(amount) AS remaining
    FROM Payment
    WHERE paid = 0
    GROUP BY loan_id

    UNION

    SELECT loan_id, SUM(amount) AS paid, 0 AS remaining
    FROM Payment
    WHERE paid = 1
    GROUP BY loan_id
)
GROUP BY loan_id

但我想应该有比我更好的方法。

Answer 1

SELECT 
loan_id, 
SUM(case when paid = 1 then amount else 0 end) as payed, 
SUM(case when paid = 0 then amount else 0 end) as remaining
FROM Payment
GROUP BY loan_id

为避免代码清晰，我会避免使用列名作为别名（付费=＆gt;付费）

Answer 2

这个怎么样：

SELECT Loan_ID,
SUM(Amount*Paid) Paid,
SUM(Amount*(1-Paid)) Remaining
FROM Payment
GROUP BY Loan_ID

如何有条件地将某列的不同值传递给聚合函数？

我尝试了什么：

2 个答案: