Skip to content

Latest commit

 

History

History
99 lines (74 loc) · 3 KB

File metadata and controls

99 lines (74 loc) · 3 KB
comments difficulty edit_url tags
true
中等
数据库

English Version

题目描述

表:Transactions

+---------------+---------+
| 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。
state 列类型为 ["approved", "declined"] 之一。

 

编写一个 sql 查询来查找每个月和每个国家/地区的事务数及其总金额、已批准的事务数及其总金额。

任意顺序 返回结果表。

查询结果格式如下所示。

 

示例 1:

输入:
Transactions table:
+------+---------+----------+--------+------------+
| id   | country | state    | amount | trans_date |
+------+---------+----------+--------+------------+
| 121  | US      | approved | 1000   | 2018-12-18 |
| 122  | US      | declined | 2000   | 2018-12-19 |
| 123  | US      | approved | 2000   | 2019-01-01 |
| 124  | DE      | approved | 2000   | 2019-01-07 |
+------+---------+----------+--------+------------+
输出:
+----------+---------+-------------+----------------+--------------------+-----------------------+
| month    | country | trans_count | approved_count | trans_total_amount | approved_total_amount |
+----------+---------+-------------+----------------+--------------------+-----------------------+
| 2018-12  | US      | 2           | 1              | 3000               | 1000                  |
| 2019-01  | US      | 1           | 1              | 2000               | 2000                  |
| 2019-01  | DE      | 1           | 1              | 2000               | 2000                  |
+----------+---------+-------------+----------------+--------------------+-----------------------+

解法

方法一:分组求和

我们可以先按照月份和国家分组,然后利用 COUNTSUM 函数分别求出每个分组的事务数、已批准的事务数、总金额和已批准的总金额。

MySQL

# Write your MySQL query statement below
SELECT
    DATE_FORMAT(trans_date, '%Y-%m') AS month,
    country,
    COUNT(1) AS trans_count,
    SUM(state = 'approved') AS approved_count,
    SUM(amount) AS trans_total_amount,
    SUM(IF(state = 'approved', amount, 0)) AS approved_total_amount
FROM Transactions
GROUP BY 1, 2;