3

那么这个问题的标题可能实际上并不能反映我想要的，我想不出更好的了。也许我应该在末尾附加一个“然后加入以这种方式获得的两个表”。

假设我有这张桌子。场景是，我们用 id 支付贷款loan_id，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，我们更改paid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 |
+---------+-------+-----------+

该paid列基本上是amountpay = 1 的所有字段的remaining总和。并且是 pay = 0 的行的总和。amount它们应该按 分组loan_id。

我尝试了什么：

我尝试了不同的连接、产品等，但我能从查询中得到的最好的结果是这个：

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。

4

2 回答 2

4 
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

为了代码清晰，我会避免使用列名作为别名（付费 => 付费）

于 2012-09-14T08:56:16.927 回答
2

这个怎么样：

SELECT Loan_ID,
SUM(Amount*Paid) Paid,
SUM(Amount*(1-Paid)) Remaining
FROM Payment
GROUP BY Loan_ID
于 2012-09-14T09:36:01.860 回答