bank-account-summary-ii

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Database

Table: <code>Users</code>

<pre>
+--------------+---------+
| Column Name | Type |
+--------------+---------+
| account | int |
| name | varchar |
+--------------+---------+
account is the primary key (column with unique values) for this table.
Each row of this table contains the account number of each user in the bank.
There will be no two users having the same name in the table.
</pre>

&nbsp;

Table: <code>Transactions</code>

<pre>
+---------------+---------+
| Column Name | Type |
+---------------+---------+
| trans_id | int |
| account | int |
| amount | int |
| transacted_on | date |
+---------------+---------+
trans_id is the primary key (column with unique values) for this table.
Each row of this table contains all changes made to all accounts.
amount is positive if the user received money and negative if they transferred money.
All accounts start with a balance of 0.
</pre>

&nbsp;

Write a solution to report the name and balance of users with a balance higher than <code>10000</code>. The balance of an account is equal to the sum of the amounts of all transactions involving that account.

Return the result table in any order.

The result format is in the following example.

&nbsp;
Example 1:

<pre>
Input: 
Users table:
+------------+--------------+
| account | name |
+------------+--------------+
| 900001 | Alice |
| 900002 | Bob |
| 900003 | Charlie |
+------------+--------------+
Transactions table:
+------------+------------+------------+---------------+
| trans_id | account | amount | transacted_on |
+------------+------------+------------+---------------+
| 1 | 900001 | 7000 | 2020-08-01 |
| 2 | 900001 | 7000 | 2020-09-01 |
| 3 | 900001 | -3000 | 2020-09-02 |
| 4 | 900002 | 1000 | 2020-09-12 |
| 5 | 900003 | 6000 | 2020-08-07 |
| 6 | 900003 | 6000 | 2020-09-07 |
| 7 | 900003 | -4000 | 2020-09-11 |
+------------+------------+------------+---------------+
Output: 
+------------+------------+
| name | balance |
+------------+------------+
| Alice | 11000 |
+------------+------------+
Explanation: 
Alice&#39;s balance is (7000 + 7000 - 3000) = 11000.
Bob&#39;s balance is 1000.
Charlie&#39;s balance is (6000 + 6000 - 4000) = 8000.
</pre>

account_summary

Bank Account Summary II

表: <code>Users</code>

<pre>
+--------------+---------+
| Column Name | Type |
+--------------+---------+
| account | int |
| name | varchar |
+--------------+---------+
account 是该表的主键(具有唯一值的列)。
该表的每一行都包含银行中每个用户的帐号。
表中不会有两个用户具有相同的名称。
</pre>

&nbsp;

表: <code>Transactions</code>

<pre>
+---------------+---------+
| Column Name | Type |
+---------------+---------+
| trans_id | int |
| account | int |
| amount | int |
| transacted_on | date |
+---------------+---------+
trans_id 是该表主键(具有唯一值的列)。
该表的每一行包含了所有账户的交易改变情况。
如果用户收到了钱, 那么金额是正的; 如果用户转了钱, 那么金额是负的。
所有账户的起始余额为 0。
</pre>

&nbsp;

编写解决方案,&nbsp;&nbsp;报告余额高于 10000 的所有用户的名字和余额.&nbsp;账户的余额等于包含该账户的所有交易的总和。

返回结果表单 无顺序要求 。

查询结果格式如下例所示。

&nbsp;

示例 1：

<pre>
<code>输入：
Users</code> table:
+------------+--------------+
| account | name |
+------------+--------------+
| 900001 | Alice |
| 900002 | Bob |
| 900003 | Charlie |
+------------+--------------+

<code>Transactions</code> table:
+------------+------------+------------+---------------+
| trans_id | account | amount | transacted_on |
+------------+------------+------------+---------------+
| 1 | 900001 | 7000 | 2020-08-01 |
| 2 | 900001 | 7000 | 2020-09-01 |
| 3 | 900001 | -3000 | 2020-09-02 |
| 4 | 900002 | 1000 | 2020-09-12 |
| 5 | 900003 | 6000 | 2020-08-07 |
| 6 | 900003 | 6000 | 2020-09-07 |
| 7 | 900003 | -4000 | 2020-09-11 |
+------------+------------+------------+---------------+
输出：
+------------+------------+
| <code>name </code> | <code>balance </code> |
+------------+------------+
| Alice | 11000 |
+------------+------------+
解释：
Alice 的余额为(7000 + 7000 - 3000) = 11000.
Bob 的余额为1000.
Charlie 的余额为(6000 + 6000 - 4000) = 8000.
</pre>