文章目錄

• • 1. 題目
  • 2. 解題

1. 題目

支出表: Spending

+-------------+---------+ | Column Name | Type | +-------------+---------+ | user_id | int | | spend_date | date | | platform | enum | | amount | int | +-------------+---------+ 這張表記錄了用戶在一個在線購物網站的支出歷史， 該在線購物平臺同時擁有桌面端（'desktop'）和手機端（'mobile'）的應用程序。 這張表的主鍵是 (user_id, spend_date, platform)。 平臺列 platform 是一種 ENUM ，類型為（'desktop', 'mobile'）。

寫一段 SQL 來查找每天 僅 使用手機端用戶、僅 使用桌面端用戶、 同時 使用桌面端和手機端的用戶人數和總支出金額。

查詢結果格式如下例所示：

Spending table: +---------+------------+----------+--------+ | user_id | spend_date | platform | amount | +---------+------------+----------+--------+ | 1 | 2019-07-01 | mobile | 100 | | 1 | 2019-07-01 | desktop | 100 | | 2 | 2019-07-01 | mobile | 100 | | 2 | 2019-07-02 | mobile | 100 | | 3 | 2019-07-01 | desktop | 100 | | 3 | 2019-07-02 | desktop | 100 | +---------+------------+----------+--------+Result table: +------------+----------+--------------+-------------+ | spend_date | platform | total_amount | total_users | +------------+----------+--------------+-------------+ | 2019-07-01 | desktop | 100 | 1 | | 2019-07-01 | mobile | 100 | 1 | | 2019-07-01 | both | 200 | 1 | | 2019-07-02 | desktop | 100 | 1 | | 2019-07-02 | mobile | 100 | 1 | | 2019-07-02 | both | 0 | 0 | +------------+----------+--------------+-------------+ 在 2019-07-01, 用戶1 同時 使用桌面端和手機端購買, 用戶2 僅 使用了手機端購買， 而用戶3 僅 使用了桌面端購買。在 2019-07-02, 用戶2 僅 使用了手機端購買, 用戶3 僅 使用了桌面端購買， 且沒有用戶 同時 使用桌面端和手機端購買。

來源：力扣（LeetCode）
鏈接：https://leetcode-cn.com/problems/user-purchase-platform
著作權歸領扣網絡所有。商業轉載請聯系官方授權，非商業轉載請注明出處。

2. 解題

• 先造出表的各種組合

select distinct spend_date, "desktop" platform from Spending union select distinct spend_date, "mobile" platform from Spending union select distinct spend_date, "both" platform from Spending {"headers": ["spend_date", "platform"], "values": [ ["2019-07-01", "desktop"], ["2019-07-02", "desktop"], ["2019-07-01", "mobile"], ["2019-07-02", "mobile"], ["2019-07-01", "both"], ["2019-07-02", "both"]]}

• 計算每天，某類屬下的總金額、人數

select spend_date, if(count(distinct platform)=1, platform, 'both') plat,sum(amount) total_am,count(distinct user_id) total_u # 1 total_u 這么寫也對，就1個人 from Spending group by spend_date, user_id {"headers": ["spend_date", "plat", "total_am", "total_u"], "values": [ ["2019-07-01", "both", 200, 1], ["2019-07-01", "mobile", 100, 1], ["2019-07-01", "desktop", 100, 1], ["2019-07-02", "mobile", 100, 1], ["2019-07-02", "desktop", 100, 1]]}

• 上面2表連接

select p.spend_date, p.platform, t.total_am, t.total_u from (select distinct spend_date, "desktop" platform from Spendingunionselect distinct spend_date, "mobile" platform from Spendingunionselect distinct spend_date, "both" platform from Spending ) p left join (select spend_date, if(count(distinct platform)=1, platform, 'both') plat,sum(amount) total_am,count(distinct user_id) total_ufrom Spendinggroup by spend_date, user_id ) t on p.platform = t.plat and p.spend_date = t.spend_date {"headers": ["spend_date", "platform", "total_am", "total_u"], 。 "values": [ ["2019-07-01", "desktop", 100, 1], ["2019-07-02", "desktop", 100, 1], ["2019-07-01", "mobile", 100, 1], ["2019-07-02", "mobile", 100, 1], ["2019-07-01", "both", 200, 1], ["2019-07-02", "both", null, null]]}

• 對連接后的表，求和

# Write your MySQL query statement below selectspend_date, platform,ifnull(sum(total_am),0) total_amount,ifnull(sum(total_u),0) total_users from (select p.spend_date, p.platform, t.total_am, t.total_ufrom(select distinct spend_date, "desktop" platform from Spendingunionselect distinct spend_date, "mobile" platform from Spendingunionselect distinct spend_date, "both" platform from Spending) pleft join(select spend_date, if(count(distinct platform)=1, platform, 'both') plat,sum(amount) total_am,count(distinct user_id) total_ufrom Spendinggroup by spend_date, user_id) ton p.platform = t.plat and p.spend_date = t.spend_date ) temp group by spend_date, platform

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