如何在sql中对具有不同日期值的行进行分组和排序

wfsdck30  于 2021-07-29  发布在  Java
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我有这样一个结果:

+---------------------------+-----------+------------+
|           Date            | InvoiceID |  Amount    |
+---------------------------+-----------+------------+
| 2020-06-09 12:36:37.433   | AF-1      | 189,876996 |
| 2020-06-09 12:36:37.483   | AF-1      | 59,4       |
| 2020-06-09 12:36:37.490   | AF-1      | 15,8544    |
| 2020-06-09 12:37:42.790   | AF-2      | 20,2       |
| 2020-06-09 12:39:29.453   | AF-4      | 70,6596    |
| 2020-06-09 12:43:30.553   | SF-1      | 47,1064    |
| 2020-06-09 12:43:30.577   | SF-1      | 12,96      |
| 2020-06-09 12:43:30.583   | SF-1      | 17,3664    |
| 2020-06-09 12:44:51.963   | SF-3      | 34,3905    |
| 2020-06-09 12:49:34.147   | TM-1      | 500        |
| 2020-06-09 12:50:26.040   | TM-2      | 150        |
| 2020-06-09 12:50:26.063   | TM-2      | 600        |
| 2020-06-09 12:51:29.817   | GH-1      | 500        |
| 2020-06-09 12:51:29.823   | GH-1      | 313,68     |
+---------------------------+-----------+------------+

查询非常简单:

Select Date, InvoiceID, Amount from TableName order by Date

我需要把他们分组,得到总金额和订单日期。请注意,日期值不同。因为这个我不能把结果分组。重要的是我需要按日期列订购。这就是我想要的结果:

+-----------+-------------+
| InvoiceID |   Amount    |
+-----------+-------------+
| AF-1      | 265,131396  |
| AF-2      | 20,2        |
| AF-4      | 70,6596     |
| SF-1      | 77,4328     |
| SF-3      | 34,3905     |
| TM-1      | 500         |
| TM-2      | 750         |
| GH-1      | 813,68      |
+-----------+-------------+

我正在尝试此代码,但sql给了我一个错误,我需要在GROUPBY子句中使用date列。

SELECT InvoiceID, sum(Amount) as Amount from TableName group by InvoiceID order by Date

我对sql非常陌生,有人能建议我如何解决这个问题吗?

06odsfpq

06odsfpq1#

您需要聚合,但也需要用于排序的聚合函数:

SELECT InvoiceID, sum(Amount) as Amount 
FROM TableName 
GROUP BY InvoiceID 
ORDER BY MIN(Date);

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