SQL练习03

1.每月交易

SQL

Create table If Not Exists Transactions (id int, country varchar(4), state enum('approved', 'declined'), amount int, trans_date date);
Truncate table Transactions;
insert into Transactions (id, country, state, amount, trans_date) values ('121', 'US', 'approved', '1000', '2018-12-18');
insert into Transactions (id, country, state, amount, trans_date) values ('122', 'US', 'declined', '2000', '2018-12-19');
insert into Transactions (id, country, state, amount, trans_date) values ('123', 'US', 'approved', '2000', '2019-01-01');
insert into Transactions (id, country, state, amount, trans_date) values ('124', 'DE', 'approved', '2000', '2019-01-07');

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。

以 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                  |
+----------+---------+-------------+----------------+--------------------+-----------------------+

思路

1.查找每个月国家/地区。利用group by DATE_FORMAT(trans_date, '%Y-%m'),country
2.查找总的事务数。第一步已经将数据按月和国家聚合，只需要使用count函数
3.查找总金额。使用sum函数计算总金额
4.查找已批准的事物数。
5.查找已批准的事物的总金额。

题解

SELECT DATE_FORMAT(trans_date, '%Y-%m') AS month,country,COUNT(*) AS trans_count,COUNT(IF(state = 'approved', 1, NULL)) AS approved_count,SUM(amount) AS trans_total_amount,SUM(IF(state = 'approved', amount, 0)) AS approved_total_amount
FROM Transactions
GROUP BY month, country

2.最后一个能进入巴士的人

SQL

Create table If Not Exists Queue (person_id int, person_name varchar(30), weight int, turn int);
Truncate table Queue;
insert into Queue (person_id, person_name, weight, turn) values ('5', 'Alice', '250', '1');
insert into Queue (person_id, person_name, weight, turn) values ('4', 'Bob', '175', '5');
insert into Queue (person_id, person_name, weight, turn) values ('3', 'Alex', '350', '2');
insert into Queue (person_id, person_name, weight, turn) values ('6', 'John Cena', '400', '3');
insert into Queue (person_id, person_name, weight, turn) values ('1', 'Winston', '500', '6');
insert into Queue (person_id, person_name, weight, turn) values ('2', 'Marie', '200', '4');

表: Queue

+-------------+---------+
| Column Name | Type    |
+-------------+---------+
| person_id   | int     |
| person_name | varchar |
| weight      | int     |
| turn        | int     |
+-------------+---------+
person_id 是这个表具有唯一值的列。
该表展示了所有候车乘客的信息。
表中 person_id 和 turn 列将包含从 1 到 n 的所有数字，其中 n 是表中的行数。
turn 决定了候车乘客上巴士的顺序，其中 turn=1 表示第一个上巴士，turn=n 表示最后一个上巴士。
weight 表示候车乘客的体重，以千克为单位。

有一队乘客在等着上巴士。然而，巴士有1000 千克的重量限制，所以其中一部分乘客可能无法上巴士。

编写解决方案找出 最后一个 上巴士且不超过重量限制的乘客，并报告 person_name 。题目测试用例确保顺位第一的人可以上巴士且不会超重。

返回结果格式如下所示。

示例 1：

输入：
Queue 表
+-----------+-------------+--------+------+
| person_id | person_name | weight | turn |
+-----------+-------------+--------+------+
| 5         | Alice       | 250    | 1    |
| 4         | Bob         | 175    | 5    |
| 3         | Alex        | 350    | 2    |
| 6         | John Cena   | 400    | 3    |
| 1         | Winston     | 500    | 6    |
| 2         | Marie       | 200    | 4    |
+-----------+-------------+--------+------+
输出：
+-------------+
| person_name |
+-------------+
| John Cena   |
+-------------+
解释：
为了简化，Queue 表按 turn 列由小到大排序。
+------+----+-----------+--------+--------------+
| Turn | ID | Name      | Weight | Total Weight |
+------+----+-----------+--------+--------------+
| 1    | 5  | Alice     | 250    | 250          |
| 2    | 3  | Alex      | 350    | 600          |
| 3    | 6  | John Cena | 400    | 1000         | (最后一个上巴士)
| 4    | 2  | Marie     | 200    | 1200         | (无法上巴士)
| 5    | 4  | Bob       | 175    | ___          |
| 6    | 1  | Winston   | 500    | ___          |
+------+----+-----------+--------+--------------+

思路

1.获取下一位上车的人，计算总体重 
2.根据turn进行降序排序
3.判断当前上车的人总体重是否超过1000
4.保留最后一个能进入巴士的人

题解

方式一：
select *
from Queue t1,Queue t2
where t1.turn>=t2.turn
group by t1.person_id
having sum(t2.weight) <=1000
order by t1.turn desc
limit 1方式二：
select person_name
from
(
select *,sum(weight) over(order by turn)as Total_Weight
from Queue
)t
where Total_Weight<=1000
order by Total_Weight desc
limit 1

3.餐馆营业额变化增长

SQL

Create table If Not Exists Customer (customer_id int, name varchar(20), visited_on date, amount int);
Truncate table Customer;
insert into Customer (customer_id, name, visited_on, amount) values ('1', 'Jhon', '2019-01-01', '100');
insert into Customer (customer_id, name, visited_on, amount) values ('2', 'Daniel', '2019-01-02', '110');
insert into Customer (customer_id, name, visited_on, amount) values ('3', 'Jade', '2019-01-03', '120');
insert into Customer (customer_id, name, visited_on, amount) values ('4', 'Khaled', '2019-01-04', '130');
insert into Customer (customer_id, name, visited_on, amount) values ('5', 'Winston', '2019-01-05', '110');
insert into Customer (customer_id, name, visited_on, amount) values ('6', 'Elvis', '2019-01-06', '140');
insert into Customer (customer_id, name, visited_on, amount) values ('7', 'Anna', '2019-01-07', '150');
insert into Customer (customer_id, name, visited_on, amount) values ('8', 'Maria', '2019-01-08', '80');
insert into Customer (customer_id, name, visited_on, amount) values ('9', 'Jaze', '2019-01-09', '110');
insert into Customer (customer_id, name, visited_on, amount) values ('1', 'Jhon', '2019-01-10', '130');
insert into Customer (customer_id, name, visited_on, amount) values ('3', 'Jade', '2019-01-10', '150');

表: Customer

+---------------+---------+
| Column Name   | Type    |
+---------------+---------+
| customer_id   | int     |
| name          | varchar |
| visited_on    | date    |
| amount        | int     |
+---------------+---------+
在 SQL 中，(customer_id, visited_on) 是该表的主键。
该表包含一家餐馆的顾客交易数据。
visited_on 表示 (customer_id) 的顾客在 visited_on 那天访问了餐馆。
amount 是一个顾客某一天的消费总额。

你是餐馆的老板，现在你想分析一下可能的营业额变化增长（每天至少有一位顾客）。

计算以 7 天（某日期 + 该日期前的 6 天）为一个时间段的顾客消费平均值。average_amount 要 保留两位小数。

结果按 visited_on 升序排序。

返回结果格式的例子如下。

示例 1:

输入：
Customer 表:
+-------------+--------------+--------------+-------------+
| customer_id | name         | visited_on   | amount      |
+-------------+--------------+--------------+-------------+
| 1           | Jhon         | 2019-01-01   | 100         |
| 2           | Daniel       | 2019-01-02   | 110         |
| 3           | Jade         | 2019-01-03   | 120         |
| 4           | Khaled       | 2019-01-04   | 130         |
| 5           | Winston      | 2019-01-05   | 110         | 
| 6           | Elvis        | 2019-01-06   | 140         | 
| 7           | Anna         | 2019-01-07   | 150         |
| 8           | Maria        | 2019-01-08   | 80          |
| 9           | Jaze         | 2019-01-09   | 110         | 
| 1           | Jhon         | 2019-01-10   | 130         | 
| 3           | Jade         | 2019-01-10   | 150         | 
+-------------+--------------+--------------+-------------+
输出：
+--------------+--------------+----------------+
| visited_on   | amount       | average_amount |
+--------------+--------------+----------------+
| 2019-01-07   | 860          | 122.86         |
| 2019-01-08   | 840          | 120            |
| 2019-01-09   | 840          | 120            |
| 2019-01-10   | 1000         | 142.86         |
+--------------+--------------+----------------+
解释：
第一个七天消费平均值从 2019-01-01 到 2019-01-07 是restaurant-growth/restaurant-growth/ (100 + 110 + 120 + 130 + 110 + 140 + 150)/7 = 122.86
第二个七天消费平均值从 2019-01-02 到 2019-01-08 是 (110 + 120 + 130 + 110 + 140 + 150 + 80)/7 = 120
第三个七天消费平均值从 2019-01-03 到 2019-01-09 是 (120 + 130 + 110 + 140 + 150 + 80 + 110)/7 = 120
第四个七天消费平均值从 2019-01-04 到 2019-01-10 是 (130 + 110 + 140 + 150 + 80 + 110 + 130 + 150)/7 = 142.86

思路

第一个七天消费平均值从 2019-01-01 到 2019-01-07 (获取2019-01-07 营业额，作为7天的营业额)
第二个七天消费平均值从 2019-01-02 到 2019-01-08 以此类推，
1.根据visited_on 分组，计算每天的营业额
2.在计算从第一天到第七天的累计营业额 sum()over()
3.将累计营业额进行排序 rank()over() 
4.筛选累计不到7天的营业额，where rk>=7 
5.再根据visited_on 分组，计算每个7天营业额的平均  round(sum(amount)/7,2)

题解

select visited_on,amount,round(sum(amount)/7,2) average_amount
from 
(-- 获取日期、排名、累计的营业额select visited_on,rank()over(order by visited_on) as rk,sum(sum(amount))over(order by visited_on range interval 6 day preceding) as amountfrom Customergroup by visited_on
)AS tep
where rk>=7 
group by visited_on

4.电影评分

SQL

Create table If Not Exists Movies (movie_id int, title varchar(30));
Create table If Not Exists Users (user_id int, name varchar(30));
Create table If Not Exists MovieRating (movie_id int, user_id int, rating int, created_at date);
Truncate table Movies;
insert into Movies (movie_id, title) values ('1', 'Avengers');
insert into Movies (movie_id, title) values ('2', 'Frozen 2');
insert into Movies (movie_id, title) values ('3', 'Joker');
Truncate table Users;
insert into Users (user_id, name) values ('1', 'Daniel');
insert into Users (user_id, name) values ('2', 'Monica');
insert into Users (user_id, name) values ('3', 'Maria');
insert into Users (user_id, name) values ('4', 'James');
Truncate table MovieRating;
insert into MovieRating (movie_id, user_id, rating, created_at) values ('1', '1', '3', '2020-01-12');
insert into MovieRating (movie_id, user_id, rating, created_at) values ('1', '2', '4', '2020-02-11');
insert into MovieRating (movie_id, user_id, rating, created_at) values ('1', '3', '2', '2020-02-12');
insert into MovieRating (movie_id, user_id, rating, created_at) values ('1', '4', '1', '2020-01-01');
insert into MovieRating (movie_id, user_id, rating, created_at) values ('2', '1', '5', '2020-02-17');
insert into MovieRating (movie_id, user_id, rating, created_at) values ('2', '2', '2', '2020-02-01');
insert into MovieRating (movie_id, user_id, rating, created_at) values ('2', '3', '2', '2020-03-01');
insert into MovieRating (movie_id, user_id, rating, created_at) values ('3', '1', '3', '2020-02-22');
insert into MovieRating (movie_id, user_id, rating, created_at) values ('3', '2', '4', '2020-02-25');

表：Movies

+---------------+---------+
| Column Name   | Type    |
+---------------+---------+
| movie_id      | int     |
| title         | varchar |
+---------------+---------+
movie_id 是这个表的主键(具有唯一值的列)。
title 是电影的名字。

表：Users

+---------------+---------+
| Column Name   | Type    |
+---------------+---------+
| user_id       | int     |
| name          | varchar |
+---------------+---------+
user_id 是表的主键(具有唯一值的列)。

表：MovieRating

+---------------+---------+
| Column Name   | Type    |
+---------------+---------+
| movie_id      | int     |
| user_id       | int     |
| rating        | int     |
| created_at    | date    |
+---------------+---------+
(movie_id, user_id) 是这个表的主键(具有唯一值的列的组合)。
这个表包含用户在其评论中对电影的评分 rating 。
created_at 是用户的点评日期。

请你编写一个解决方案：

查找评论电影数量最多的用户名。如果出现平局，返回字典序较小的用户名。
查找在 February 2020 平均评分最高 的电影名称。如果出现平局，返回字典序较小的电影名称。

字典序 ，即按字母在字典中出现顺序对字符串排序，字典序较小则意味着排序靠前。

返回结果格式如下例所示。

示例 1：

输入：
Movies 表：
+-------------+--------------+
| movie_id    |  title       |
+-------------+--------------+
| 1           | Avengers     |
| 2           | Frozen 2     |
| 3           | Joker        |
+-------------+--------------+
Users 表：
+-------------+--------------+
| user_id     |  name        |
+-------------+--------------+
| 1           | Daniel       |
| 2           | Monica       |
| 3           | Maria        |
| 4           | James        |
+-------------+--------------+
MovieRating 表：
+-------------+--------------+--------------+-------------+
| movie_id    | user_id      | rating       | created_at  |
+-------------+--------------+--------------+-------------+
| 1           | 1            | 3            | 2020-01-12  |
| 1           | 2            | 4            | 2020-02-11  |
| 1           | 3            | 2            | 2020-02-12  |
| 1           | 4            | 1            | 2020-01-01  |
| 2           | 1            | 5            | 2020-02-17  | 
| 2           | 2            | 2            | 2020-02-01  | 
| 2           | 3            | 2            | 2020-03-01  |
| 3           | 1            | 3            | 2020-02-22  | 
| 3           | 2            | 4            | 2020-02-25  | 
+-------------+--------------+--------------+-------------+
输出：
Result 表：
+--------------+
| results      |
+--------------+
| Daniel       |
| Frozen 2     |
+--------------+
解释：
Daniel 和 Monica 都点评了 3 部电影（"Avengers", "Frozen 2" 和 "Joker"） 但是 Daniel 字典序比较小。
Frozen 2 和 Joker 在 2 月的评分都是 3.5，但是 Frozen 2 的字典序比较小。

题解

select results
from 
(
-- 查找评论电影数量最多的用户名。如果出现平局，返回字典序较小的用户名
select t2.name results   
from MovieRating t1
left join Users  t2
on t1.user_id=t2.user_id
group by t2.user_id
order by count(t2.user_id) desc,t2.name
limit 1
)t1union allselect results
from (
-- 查找在 February 2020 平均评分最高 的电影名称。如果出现平局，返回字典序较小的电影名称。
select m.title results
from MovieRating mr 
left join Movies m using(movie_id)
where date_format(mr.created_at,'%Y-%m') = '2020-02'
group by m.movie_id
order by avg(rating) desc, m.title asc
limit 1
)t2

5.股票的资本损益

SQL

Create Table If Not Exists Stocks (stock_name varchar(15), operation ENUM('Sell', 'Buy'), operation_day int, price int);
Truncate table Stocks;
insert into Stocks (stock_name, operation, operation_day, price) values ('Leetcode', 'Buy', '1', '1000');
insert into Stocks (stock_name, operation, operation_day, price) values ('Corona Masks', 'Buy', '2', '10');
insert into Stocks (stock_name, operation, operation_day, price) values ('Leetcode', 'Sell', '5', '9000');
insert into Stocks (stock_name, operation, operation_day, price) values ('Handbags', 'Buy', '17', '30000');
insert into Stocks (stock_name, operation, operation_day, price) values ('Corona Masks', 'Sell', '3', '1010');
insert into Stocks (stock_name, operation, operation_day, price) values ('Corona Masks', 'Buy', '4', '1000');
insert into Stocks (stock_name, operation, operation_day, price) values ('Corona Masks', 'Sell', '5', '500');
insert into Stocks (stock_name, operation, operation_day, price) values ('Corona Masks', 'Buy', '6', '1000');
insert into Stocks (stock_name, operation, operation_day, price) values ('Handbags', 'Sell', '29', '7000');
insert into Stocks (stock_name, operation, operation_day, price) values ('Corona Masks', 'Sell', '10', '10000');

Stocks 表：

+---------------+---------+
| Column Name   | Type    |
+---------------+---------+
| stock_name    | varchar |
| operation     | enum    |
| operation_day | int     |
| price         | int     |
+---------------+---------+
(stock_name, day) 是这张表的主键(具有唯一值的列的组合)
operation 列使用的是一种枚举类型，包括：('Sell','Buy')
此表的每一行代表了名为 stock_name 的某支股票在 operation_day 这一天的操作价格。
此表可以保证，股票的每个“卖出”操作在前一天都有相应的“买入”操作。并且，股票的每个“买入”操作在即将到来的一天都有相应的“卖出”操作。

编写解决方案报告每只股票的 资本损益。

股票的 资本利得/损失 是指一次或多次买卖该股票后的总收益或损失。

以 任意顺序 返回结果表。

结果格式如下所示。

示例 1：

输入：
Stocks 表:
+---------------+-----------+---------------+--------+
| stock_name    | operation | operation_day | price  |
+---------------+-----------+---------------+--------+
| Leetcode      | Buy       | 1             | 1000   |
| Corona Masks  | Buy       | 2             | 10     |
| Leetcode      | Sell      | 5             | 9000   |
| Handbags      | Buy       | 17            | 30000  |
| Corona Masks  | Sell      | 3             | 1010   |
| Corona Masks  | Buy       | 4             | 1000   |
| Corona Masks  | Sell      | 5             | 500    |
| Corona Masks  | Buy       | 6             | 1000   |
| Handbags      | Sell      | 29            | 7000   |
| Corona Masks  | Sell      | 10            | 10000  |
+---------------+-----------+---------------+--------+
输出：
+---------------+-------------------+
| stock_name    | capital_gain_loss |
+---------------+-------------------+
| Corona Masks  | 9500              |
| Leetcode      | 8000              |
| Handbags      | -23000            |
+---------------+-------------------+
解释：
Leetcode 股票在第一天以1000美元的价格买入，在第五天以9000美元的价格卖出。资本收益=9000-1000=8000美元。
Handbags 股票在第17天以30000美元的价格买入，在第29天以7000美元的价格卖出。资本损失=7000-30000=-23000美元。
Corona Masks 股票在第1天以10美元的价格买入，在第3天以1010美元的价格卖出。在第4天以1000美元的价格再次购买，在第5天以500美元的价格出售。最后，它在第6天以1000美元的价格被买走，在第10天以10000美元的价格被卖掉。资本损益是每次（’Buy'->'Sell'）操作资本收益或损失的和=（1010-10）+（500-1000）+（10000-1000）=1000-500+9000=9500美元。

题解

方式一：
select stock_name, sum(if(operation='Sell',price,-1*price)) capital_gain_loss 
from Stocks 
group by stock_name
order by capital_gain_loss desc方式二：
selectstock_name,sum(casewhen operation = 'Buy' then -pricewhen operation = 'Sell' then priceend) as capital_gain_loss
from stocks
group by stock_name
order by capital_gain_loss desc