「SQL面试题库」 No_77 每月交易II

🍅 今日真题

题目介绍： 每月交易II monthly-transactions-ii

难度中等

SQL架构

Transactions

记录表

+----------------+---------+
| Column Name    | Type    |
+----------------+---------+
| id             | int     |
| country        | varchar |
| state          | enum    |
| amount         | int     |
| trans_date     | date    |
+----------------+---------+
id 是这个表的主键。
该表包含有关传入事务的信息。
状态列是类型为 [approved（已批准）、declined（已拒绝）] 的枚举。

Chargebacks

表

+----------------+---------+
| Column Name    | Type    |
+----------------+---------+
| trans_id       | int     |
| charge_date    | date    |
+----------------+---------+
退单包含有关放置在事务表中的某些事务的传入退单的基本信息。
trans_id 是 transactions 表的 id 列的外键。
每项退单都对应于之前进行的交易，即使未经批准。

编写一个 SQL 查询，以查找每个月和每个国家/地区的已批准交易的数量及其总金额、退单的数量及其总金额。

注意：在您的查询中，给定月份和国家，忽略所有为零的行。

查询结果格式如下所示：

``` Transactions 表： +------+---------+----------+--------+------------+ | id | country | state | amount | trans_date | +------+---------+----------+--------+------------+ | 101 | US | approved | 1000 | 2019-05-18 | | 102 | US | declined | 2000 | 2019-05-19 | | 103 | US | approved | 3000 | 2019-06-10 | | 104 | US | declined | 4000 | 2019-06-13 | | 105 | US | approved | 5000 | 2019-06-15 | +------+---------+----------+--------+------------+

Chargebacks 表： +------------+------------+ | trans_id | trans_date | +------------+------------+ | 102 | 2019-05-29 | | 101 | 2019-06-30 | | 105 | 2019-09-18 | +------------+------------+

Result 表： +----------+---------+----------------+-----------------+-------------------+--------------------+ | month | country | approved_count | approved_amount | chargeback_count | chargeback_amount | +----------+---------+----------------+-----------------+-------------------+--------------------+ | 2019-05 | US | 1 | 1000 | 1 | 2000 | | 2019-06 | US | 2 | 8000 | 1 | 1000 | | 2019-09 | US | 0 | 0 | 1 | 5000 | +----------+---------+----------------+-----------------+-------------------+--------------------+ ```

sql
select 
    date_format(trans_date, '%Y-%m') month, 
    country,
    sum(state = 'approved') approved_count,
    sum(if(state = 'approved', amount, 0)) approved_amount,
    sum(state = 'chargeback') chargeback_count,
    sum(if(state = 'chargeback', amount, 0)) chargeback_amount
from (
    select * from transactions
    union all
    select id, country, 'chargeback' state, amount, c.trans_date
    from chargebacks c left join transactions t 
    on c.trans_id = t.id
) tmp 
group by month, country
having approved_amount or chargeback_amount

• 已经有灵感了？在评论区写下你的思路吧！