Question

这是我的表

 PayerID PaymentID Amount

    1   8386    5827.14
    1   8386    5827.14
    1   8386    5827.14
    1   8387    4464.68
    1   8387    4464.68
    1   8387    4464.68
    2   8388    1482.57
    2   8388    1482.57
    2   8388    1482.57
    2   8388    1482.57
    2   8388    1482.57
    2   8388    1482.57
    2   8388    1482.57
    2   8388    1482.57
    2   8388    1482.57

我希望通过PayerID支付不同PaymentID组的金额，如下所示

payerID AmountSum
1        5827.14 + 4464.68
2         1482.57

Answer 1

使用SUM(DISTINCT)

SELECT PayerId, SUM(DISTINCT Amount) AS AmountSum
FROM tab
GROUP BY PayerId;

此解决方案假定Amount对于不同的PaymentId是唯一的。

更一般的方法：

SELECT PayerId, SUM(Amount) AS AmountSum
FROM (SELECT DISTINCT PayerID, PaymentID, Amount
      FROM tab) sub
GROUP BY PayerId

Answer 2

您可以使用row_number进行查询，如下所示：

Select PayerId, sum(amount) from (
    Select *, RowN = Row_Number() over(partition by Payerid, PaymentId, Amount order by Amount) from #payerdata
) a
where a.RowN = 1
group by PayerId

输出如下：

+---------+-----------+
| PayerId | AmountSum |
+---------+-----------+
|       1 |  10291.82 |
|       2 |   1482.57 |
+---------+-----------+

获取SQL Server中不同字段的总和

2 个答案: