odd-and-even-transactions

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Database

Table: <code>transactions</code>

<pre>
+------------------+------+
| Column Name | Type | 
+------------------+------+
| transaction_id | int |
| amount | int |
| transaction_date | date |
+------------------+------+
The transactions_id column uniquely identifies each row in this table.
Each row of this table contains the transaction id, amount and transaction date.
</pre>

Write a solution to find the sum of amounts for odd and even transactions for each day. If there are no odd or even transactions for a specific date, display as <code>0</code>.

Return the result table ordered by <code>transaction_date</code> in ascending order.

The result format is in the following example.

&nbsp;
Example:

<div class="example-block">
Input:

<code>transactions</code> table:

<pre class="example-io">
+----------------+--------+------------------+
| transaction_id | amount | transaction_date |
+----------------+--------+------------------+
| 1 | 150 | 2024-07-01 |
| 2 | 200 | 2024-07-01 |
| 3 | 75 | 2024-07-01 |
| 4 | 300 | 2024-07-02 |
| 5 | 50 | 2024-07-02 |
| 6 | 120 | 2024-07-03 |
+----------------+--------+------------------+
 </pre>

Output:

<pre class="example-io">
+------------------+---------+----------+
| transaction_date | odd_sum | even_sum |
+------------------+---------+----------+
| 2024-07-01 | 75 | 350 |
| 2024-07-02 | 0 | 350 |
| 2024-07-03 | 0 | 120 |
+------------------+---------+----------+
 </pre>

Explanation:

<ul>
	<li>For transaction dates:
	<ul>
		<li>2024-07-01:
		<ul>
			<li>Sum of amounts for odd transactions: 75</li>
			<li>Sum of amounts for even transactions: 150 + 200 = 350</li>
		</ul>
		</li>
		<li>2024-07-02:
		<ul>
			<li>Sum of amounts for odd transactions: 0</li>
			<li>Sum of amounts for even transactions: 300 + 50 = 350</li>
		</ul>
		</li>
		<li>2024-07-03:
		<ul>
			<li>Sum of amounts for odd transactions: 0</li>
			<li>Sum of amounts for even transactions: 120</li>
		</ul>
		</li>
	</ul>
	</li>
</ul>

Note: The output table is ordered by <code>transaction_date</code> in ascending order.
</div>

sum_daily_odd_even

Odd and Even Transactions

表：<code>transactions</code>

<pre>
+------------------+------+
| Column Name | Type | 
+------------------+------+
| transaction_id | int |
| amount | int |
| transaction_date | date |
+------------------+------+
transactions_id 列唯一标识了表中的每一行。
这张表的每一行包含交易 id，金额总和和交易日期。
</pre>

编写一个解决方案来查找每天 奇数 交易金额和 偶数 交易金额的 总和。如果某天没有奇数或偶数交易，显示为&nbsp;<code>0</code>。

返回结果表以&nbsp;<code>transaction_date</code> 升序&nbsp;排序。

结果格式如下所示。

&nbsp;

示例：

<div class="example-block">
输入：

<code>transactions</code> 表：

<pre class="example-io">
+----------------+--------+------------------+
| transaction_id | amount | transaction_date |
+----------------+--------+------------------+
| 1 | 150 | 2024-07-01 |
| 2 | 200 | 2024-07-01 |
| 3 | 75 | 2024-07-01 |
| 4 | 300 | 2024-07-02 |
| 5 | 50 | 2024-07-02 |
| 6 | 120 | 2024-07-03 |
+----------------+--------+------------------+
 </pre>

输出：

<pre class="example-io">
+------------------+---------+----------+
| transaction_date | odd_sum | even_sum |
+------------------+---------+----------+
| 2024-07-01 | 75 | 350 |
| 2024-07-02 | 0 | 350 |
| 2024-07-03 | 0 | 120 |
+------------------+---------+----------+
 </pre>

解释：

<ul>
	<li>对于交易日期：
	<ul>
		<li>2024-07-01:
		<ul>
			<li>奇数交易金额总和：75</li>
			<li>偶数交易金额总和：150 + 200 = 350</li>
		</ul>
		</li>
		<li>2024-07-02:
		<ul>
			<li>奇数交易金额总和：0</li>
			<li>偶数交易金额总和：300 + 50 = 350</li>
		</ul>
		</li>
		<li>2024-07-03:
		<ul>
			<li>奇数交易金额总和：0</li>
			<li>偶数交易金额总和：120</li>
		</ul>
		</li>
	</ul>
	</li>
</ul>

注意：输出表以&nbsp;<code>transaction_date</code>&nbsp;升序排序。
</div>