# 1831.(Medium)Maximum Transaction Each Day

表: Transactions

```
+----------------+----------+
| Column Name    | Type     |
+----------------+----------+
| transaction_id | int      |
| day            | datetime |
| amount         | int      |
+----------------+----------+
transaction_id 是此表的主键。
每行包括了该次交易的信息。
```

写一条 SQL 返回每天交易金额 amount 最大的交易 ID 。如果某天有多个这样的交易，返回这些交易的 ID 。

返回结果根据 transaction\_id 升序排列。

查询结果样例如下：

```
Transactions table:
+----------------+--------------------+--------+
| transaction_id | day                | amount |
+----------------+--------------------+--------+
| 8              | 2021-4-3 15:57:28  | 57     |
| 9              | 2021-4-28 08:47:25 | 21     |
| 1              | 2021-4-29 13:28:30 | 58     |
| 5              | 2021-4-28 16:39:59 | 40     |
| 6              | 2021-4-29 23:39:28 | 58     |
+----------------+--------------------+--------+

Result table:
+----------------+
| transaction_id |
+----------------+
| 1              |
| 5              |
| 6              |
| 8              |
+----------------+
"2021-4-3"  --> 有一个 id 是 8 的交易，因此，把它加入结果表。 
"2021-4-28" --> 有两个交易，id 是 5 和 9 ，交易 5 的金额是 40 ，而交易 9 的数量是 21 。只需要将交易 5 加入结果表，因为它是当天金额最大的交易。
"2021-4-29" --> 有两个交易，id 是 1 和 6 ，这两个交易的金额都是 58 ，因此需要把它们都写入结果表。
最后，把交易 id 按照升序排列。
```

进阶：你可以不使用 `MAX()` 函数解决这道题目吗?

来源：力扣（LeetCode）

链接：<https://leetcode-cn.com/problems/maximum-transaction-each-day>

著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。

## Solution

* 用 `date()` 函数提取日期
* 以每日为窗口，取rank=1&#x20;

```sql
SELECT transaction_id
FROM (
    SELECT *, rank() over (PARTITION BY date(day) order by amount desc) rk
    FROM Transactions 
) a
where rk = 1
order by 1
```


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