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LeetCode 数据库专题

2020-03-23 04:03

业精于勤荒于嬉。

以下题目从易到难,持续更新中。

1378. 使用唯一标识码替换员工ID

SQL架构

Create table If Not Exists Employees (id int, name varchar(20))
Create table If Not Exists EmployeeUNI (id int, unique_id int)
Truncate table Employees
insert into Employees (id, name) values ('1', 'Alice')
insert into Employees (id, name) values ('7', 'Bob')
insert into Employees (id, name) values ('11', 'Meir')
insert into Employees (id, name) values ('90', 'Winston')
insert into Employees (id, name) values ('3', 'Jonathan')
Truncate table EmployeeUNI
insert into EmployeeUNI (id, unique_id) values ('3', '1')
insert into EmployeeUNI (id, unique_id) values ('11', '2')
insert into EmployeeUNI (id, unique_id) values ('90', '3')

Employees 表:

+---------------+---------+
| Column Name | Type |
+---------------+---------+
| id | int |
| name | varchar |
+---------------+---------+
id 是这张表的主键。
这张表的每一行分别代表了某公司其中一位员工的名字和 ID 。

EmployeeUNI 表:

+---------------+---------+
| Column Name | Type |
+---------------+---------+
| id | int |
| unique_id | int |
+---------------+---------+
(id, unique_id) 是这张表的主键。
这张表的每一行包含了该公司某位员工的 ID 和他的唯一标识码(unique ID)。

写一段SQL查询来展示每位用户的 唯一标识码(unique ID );如果某位员工没有唯一标识码,使用 null 填充即可。你可以以 任意 顺序返回结果表。查询结果的格式如下例所示:

Employees table:
+----+----------+
| id | name |
+----+----------+
| 1 | Alice |
| 7 | Bob |
| 11 | Meir |
| 90 | Winston |
| 3 | Jonathan |
+----+----------+

EmployeeUNI table:
+----+-----------+
| id | unique_id |
+----+-----------+
| 3 | 1 |
| 11 | 2 |
| 90 | 3 |
+----+-----------+

EmployeeUNI table:
+-----------+----------+
| unique_id | name |
+-----------+----------+
| null | Alice |
| null | Bob |
| 2 | Meir |
| 3 | Winston |
| 1 | Jonathan |
+-----------+----------+

Alice and Bob 没有唯一标识码, 因此我们使用 null 替代。
Meir 的唯一标识码是 2 。
Winston 的唯一标识码是 3 。
Jonathan 唯一标识码是 1 。

SQL 语句

select unique_id, name
from Employees
left join EmployeeUNI
on Employees.id = EmployeeUNI.id;

1350. 院系无效的学生

SQL架构

Create table If Not Exists Departments (id int, name varchar(30))
Create table If Not Exists Students (id int, name varchar(30), department_id int)
Truncate table Departments
insert into Departments (id, name) values ('1', 'Electrical Engineering')
insert into Departments (id, name) values ('7', 'Computer Engineering')
insert into Departments (id, name) values ('13', 'Bussiness Administration')
Truncate table Students
insert into Students (id, name, department_id) values ('23', 'Alice', '1')
insert into Students (id, name, department_id) values ('1', 'Bob', '7')
insert into Students (id, name, department_id) values ('5', 'Jennifer', '13')
insert into Students (id, name, department_id) values ('2', 'John', '14')
insert into Students (id, name, department_id) values ('4', 'Jasmine', '77')
insert into Students (id, name, department_id) values ('3', 'Steve', '74')
insert into Students (id, name, department_id) values ('6', 'Luis', '1')
insert into Students (id, name, department_id) values ('8', 'Jonathan', '7')
insert into Students (id, name, department_id) values ('7', 'Daiana', '33')
insert into Students (id, name, department_id) values ('11', 'Madelynn', '1')

院系表: Departments

+---------------+---------+
| Column Name | Type |
+---------------+---------+
| id | int |
| name | varchar |
+---------------+---------+
id 是该表的主键
该表包含一所大学每个院系的 id 信息

学生表: Students

+---------------+---------+
| Column Name | Type |
+---------------+---------+
| id | int |
| name | varchar |
| department_id | int |
+---------------+---------+
id 是该表的主键
该表包含一所大学每个学生的 id 和他/她就读的院系信息

写一条 SQL 语句以查询那些所在院系不存在的学生的 id 和姓名。

可以以任何顺序返回结果,下面是返回结果格式的例子

Departments 表:
+------+--------------------------+
| id | name |
+------+--------------------------+
| 1 | Electrical Engineering |
| 7 | Computer Engineering |
| 13 | Bussiness Administration |
+------+--------------------------+

Students 表:
+------+----------+---------------+
| id | name | department_id |
+------+----------+---------------+
| 23 | Alice | 1 |
| 1 | Bob | 7 |
| 5 | Jennifer | 13 |
| 2 | John | 14 |
| 4 | Jasmine | 77 |
| 3 | Steve | 74 |
| 6 | Luis | 1 |
| 8 | Jonathan | 7 |
| 7 | Daiana | 33 |
| 11 | Madelynn | 1 |
+------+----------+---------------+

结果表:
+------+----------+
| id | name |
+------+----------+
| 2 | John |
| 7 | Daiana |
| 4 | Jasmine |
| 3 | Steve |
+------+----------+

John, Daiana, Steve 和 Jasmine 所在的院系分别是 14, 33, 74 和 77, 其中 14, 33, 74 和 77 并不存在于院系表

SQL 语句

select id, `name`
from Students
where department_id not in ( select id from Departments)

1068. 产品销售分析 I

SQL架构

Create table Sales (sale_id int, product_id int, year int, quantity int, price int)
Create table Product (product_id int, product_name varchar(10))
Truncate table Sales
insert into Sales (sale_id, product_id, year, quantity, price) values ('1', '100', '2008', '10', '5000')
insert into Sales (sale_id, product_id, year, quantity, price) values ('2', '100', '2009', '12', '5000')
insert into Sales (sale_id, product_id, year, quantity, price) values ('7', '200', '2011', '15', '9000')
Truncate table Product
insert into Product (product_id, product_name) values ('100', 'Nokia')
insert into Product (product_id, product_name) values ('200', 'Apple')
insert into Product (product_id, product_name) values ('300', 'Samsung')

销售表 Sales

+-------------+-------+
| Column Name | Type |
+-------------+-------+
| sale_id | int |
| product_id | int |
| year | int |
| quantity | int |
| price | int |
+-------------+-------+
(sale_id, year) 是销售表 Sales 的主键.
product_id 是产品表 Product 的外键.
注意: price 表示每单位价格

产品表 Product

+--------------+---------+
| Column Name | Type |
+--------------+---------+
| product_id | int |
| product_name | varchar |
+--------------+---------+
product_id 是表的主键.

写一条SQL 查询语句获取产品表 Product 中所有的 产品名称 product name 以及 该产品在 Sales 表中相对应的 上市年份 year价格 price

示例:

Sales 表:
+---------+------------+------+----------+-------+
| sale_id | product_id | year | quantity | price |
+---------+------------+------+----------+-------+
| 1 | 100 | 2008 | 10 | 5000 |
| 2 | 100 | 2009 | 12 | 5000 |
| 7 | 200 | 2011 | 15 | 9000 |
+---------+------------+------+----------+-------+

Product 表:
+------------+--------------+
| product_id | product_name |
+------------+--------------+
| 100 | Nokia |
| 200 | Apple |
| 300 | Samsung |
+------------+--------------+

Result 表:
+--------------+-------+-------+
| product_name | year | price |
+--------------+-------+-------+
| Nokia | 2008 | 5000 |
| Nokia | 2009 | 5000 |
| Apple | 2011 | 9000 |
+--------------+-------+-------+

SQL 语句

select product_name, `year`, price
from Sales s
join Product p
on s.product_id = p.product_id;


select product_name, `year`, price
from Sales s
join Product using(product_id);

1303. 求团队人数

SQL架构

Create table If Not Exists Employee (employee_id int, team_id int)
Truncate table Employee
insert into Employee (employee_id, team_id) values ('1', '8')
insert into Employee (employee_id, team_id) values ('2', '8')
insert into Employee (employee_id, team_id) values ('3', '8')
insert into Employee (employee_id, team_id) values ('4', '7')
insert into Employee (employee_id, team_id) values ('5', '9')
insert into Employee (employee_id, team_id) values ('6', '9')

员工表:Employee

+---------------+---------+
| Column Name | Type |
+---------------+---------+
| employee_id | int |
| team_id | int |
+---------------+---------+
employee_id 字段是这张表的主键,表中的每一行都包含每个员工的 ID 和他们所属的团队。

编写一个 SQL 查询,以求得每个员工所在团队的总人数。

查询结果中的顺序无特定要求。查询结果格式示例如下:

Employee Table:
+-------------+------------+
| employee_id | team_id |
+-------------+------------+
| 1 | 8 |
| 2 | 8 |
| 3 | 8 |
| 4 | 7 |
| 5 | 9 |
| 6 | 9 |
+-------------+------------+
Result table:
+-------------+------------+
| employee_id | team_size |
+-------------+------------+
| 1 | 3 |
| 2 | 3 |
| 3 | 3 |
| 4 | 1 |
| 5 | 2 |
| 6 | 2 |
+-------------+------------+
ID 为 1、2、3 的员工是 team_id 为 8 的团队的成员,
ID 为 4 的员工是 team_id 为 7 的团队的成员,
ID 为 5、6 的员工是 team_id 为 9 的团队的成员。

SQL 语句

select a.employee_id, b.team_size
from Employee a
join (select team_id, count(employee_id) team_size from Employee group by team_id) b
on a.team_id = b.team_id

1069. 产品销售分析 II

SQL架构

Create table Sales (sale_id int, product_id int, year int, quantity int, price int)
Create table Product (product_id int, product_name varchar(10))
Truncate table Sales
insert into Sales (sale_id, product_id, year, quantity, price) values ('1', '100', '2008', '10', '5000')
insert into Sales (sale_id, product_id, year, quantity, price) values ('2', '100', '2009', '12', '5000')
insert into Sales (sale_id, product_id, year, quantity, price) values ('7', '200', '2011', '15', '9000')
Truncate table Product
insert into Product (product_id, product_name) values ('100', 'Nokia')
insert into Product (product_id, product_name) values ('200', 'Apple')
insert into Product (product_id, product_name) values ('300', 'Samsung')

销售表:Sales

+-------------+-------+
| Column Name | Type |
+-------------+-------+
| sale_id | int |
| product_id | int |
| year | int |
| quantity | int |
| price | int |
+-------------+-------+
sale_id 是这个表的主键。
product_id 是 Product 表的外键。
请注意价格是每单位的。

产品表:Product

+--------------+---------+
| Column Name | Type |
+--------------+---------+
| product_id | int |
| product_name | varchar |
+--------------+---------+
product_id 是这个表的主键。

编写一个 SQL 查询,按产品 id product_id 来统计每个产品的销售总量。

查询结果格式如下面例子所示:

Sales 表:
+---------+------------+------+----------+-------+
| sale_id | product_id | year | quantity | price |
+---------+------------+------+----------+-------+
| 1 | 100 | 2008 | 10 | 5000 |
| 2 | 100 | 2009 | 12 | 5000 |
| 7 | 200 | 2011 | 15 | 9000 |
+---------+------------+------+----------+-------+

Product 表:
+------------+--------------+
| product_id | product_name |
+------------+--------------+
| 100 | Nokia |
| 200 | Apple |
| 300 | Samsung |
+------------+--------------+

Result 表:
+--------------+----------------+
| product_id | total_quantity |
+--------------+----------------+
| 100 | 22 |
| 200 | 15 |
+--------------+----------------+

SQL 语句

select product_id, sum(quantity) total_quantity
from Sales
group by product_id

-- 显示 product_name
select product_id, product_name, sum(quantity) total_quantity
from Sales
join Product using(product_id)
group by product_id

613. 直线上的最近距离

SQL架构

CREATE TABLE If Not Exists point (x INT NOT NULL, UNIQUE INDEX x_UNIQUE (x ASC))
Truncate table point
insert into point (x) values ('-1')
insert into point (x) values ('0')
insert into point (x) values ('2')

point 保存了一些点在 x 轴上的坐标,这些坐标都是整数。

写一个查询语句,找到这些点中最近两个点之间的距离。

| x   |
|-----|
| -1 |
| 0 |
| 2 |

最近距离显然是 ‘1’ ,是点 ‘-1’ 和 ‘0’ 之间的距离。所以输出应该如下:

| shortest|
|---------|
| 1 |

注意:每个点都与其他点坐标不同,表 table 不会有重复坐标出现。

进阶:如果这些点在 x 轴上从左到右都有一个编号,输出结果时需要输出最近点对的编号呢?

SQL 语句

select min(abs(p1.x - p2.x)) shortest
from point p1, point p2
where p1.x != p2.x

182. 查找重复的电子邮箱

SQL架构

Create table If Not Exists Person (Id int, Email varchar(255))
Truncate table Person
insert into Person (Id, Email) values ('1', 'a@b.com')
insert into Person (Id, Email) values ('2', 'c@d.com')
insert into Person (Id, Email) values ('3', 'a@b.com')

编写一个 SQL 查询,查找 Person 表中所有重复的电子邮箱。

示例:

+----+---------+
| Id | Email |
+----+---------+
| 1 | a@b.com |
| 2 | c@d.com |
| 3 | a@b.com |
+----+---------+

根据以上输入,你的查询应返回以下结果:

+---------+
| Email |
+---------+
| a@b.com |
+---------+

说明:所有电子邮箱都是小写字母。

SQL 语句

select Email
from Person
group by Email
having count(Email) > 1;

595. 大的国家

SQL架构

Create table If Not Exists World (name varchar(255), continent varchar(255), area int, population int, gdp int)
Truncate table World
insert into World (name, continent, area, population, gdp) values ('Afghanistan', 'Asia', '652230', '25500100', '20343000000')
insert into World (name, continent, area, population, gdp) values ('Albania', 'Europe', '28748', '2831741', '12960000000')
insert into World (name, continent, area, population, gdp) values ('Algeria', 'Africa', '2381741', '37100000', '188681000000')
insert into World (name, continent, area, population, gdp) values ('Andorra', 'Europe', '468', '78115', '3712000000')
insert into World (name, continent, area, population, gdp) values ('Angola', 'Africa', '1246700', '20609294', '100990000000')

这里有张 World

+-----------------+------------+------------+--------------+---------------+
| name | continent | area | population | gdp |
+-----------------+------------+------------+--------------+---------------+
| Afghanistan | Asia | 652230 | 25500100 | 20343000 |
| Albania | Europe | 28748 | 2831741 | 12960000 |
| Algeria | Africa | 2381741 | 37100000 | 188681000 |
| Andorra | Europe | 468 | 78115 | 3712000 |
| Angola | Africa | 1246700 | 20609294 | 100990000 |
+-----------------+------------+------------+--------------+---------------+

如果一个国家的面积超过300万平方公里,或者人口超过2500万,那么这个国家就是大国家。

编写一个SQL查询,输出表中所有大国家的名称、人口和面积。

例如,根据上表,我们应该输出:

+--------------+-------------+--------------+
| name | population | area |
+--------------+-------------+--------------+
| Afghanistan | 25500100 | 652230 |
| Algeria | 37100000 | 2381741 |
+--------------+-------------+--------------+

SQL 语句

select `name`, population, `area`
from World
where population > 25000000 or `area` > 3000000;

1251. 平均售价

SQL架构

Create table If Not Exists Prices (product_id int, start_date date, end_date date, price int)
Create table If Not Exists UnitsSold (product_id int, purchase_date date, units int)
Truncate table Prices
insert into Prices (product_id, start_date, end_date, price) values ('1', '2019-02-17', '2019-02-28', '5')
insert into Prices (product_id, start_date, end_date, price) values ('1', '2019-03-01', '2019-03-22', '20')
insert into Prices (product_id, start_date, end_date, price) values ('2', '2019-02-01', '2019-02-20', '15')
insert into Prices (product_id, start_date, end_date, price) values ('2', '2019-02-21', '2019-03-31', '30')
Truncate table UnitsSold
insert into UnitsSold (product_id, purchase_date, units) values ('1', '2019-02-25', '100')
insert into UnitsSold (product_id, purchase_date, units) values ('1', '2019-03-01', '15')
insert into UnitsSold (product_id, purchase_date, units) values ('2', '2019-02-10', '200')
insert into UnitsSold (product_id, purchase_date, units) values ('2', '2019-03-22', '30')

Table: Prices

+---------------+---------+
| Column Name | Type |
+---------------+---------+
| product_id | int |
| start_date | date |
| end_date | date |
| price | int |
+---------------+---------+
(product_id,start_date,end_date) 是 Prices 表的主键。
Prices 表的每一行表示的是某个产品在一段时期内的价格。
每个产品的对应时间段是不会重叠的,这也意味着同一个产品的价格时段不会出现交叉。

Table: UnitsSold

+---------------+---------+
| Column Name | Type |
+---------------+---------+
| product_id | int |
| purchase_date | date |
| units | int |
+---------------+---------+
UnitsSold 表没有主键,它可能包含重复项。
UnitsSold 表的每一行表示的是每种产品的出售日期,单位和产品 id。

编写SQL查询以查找每种产品的平均售价。 average_price 应该四舍五入到小数点后两位。

查询结果格式如下例所示:

Prices table:
+------------+------------+------------+--------+
| product_id | start_date | end_date | price |
+------------+------------+------------+--------+
| 1 | 2019-02-17 | 2019-02-28 | 5 |
| 1 | 2019-03-01 | 2019-03-22 | 20 |
| 2 | 2019-02-01 | 2019-02-20 | 15 |
| 2 | 2019-02-21 | 2019-03-31 | 30 |
+------------+------------+------------+--------+

UnitsSold table:
+------------+---------------+-------+
| product_id | purchase_date | units |
+------------+---------------+-------+
| 1 | 2019-02-25 | 100 |
| 1 | 2019-03-01 | 15 |
| 2 | 2019-02-10 | 200 |
| 2 | 2019-03-22 | 30 |
+------------+---------------+-------+

Result table:
+------------+---------------+
| product_id | average_price |
+------------+---------------+
| 1 | 6.96 |
| 2 | 16.96 |
+------------+---------------+
平均售价 = 产品总价 / 销售的产品数量。
产品 1 的平均售价 = ((100 * 5)+(15 * 20) )/ 115 = 6.96
产品 2 的平均售价 = ((200 * 15)+(30 * 30) )/ 230 = 16.96

SQL 语句

先查出当时候的价格,再总的进行处理。

select
product_id, round(sum(price*units)/sum(units), 2) average_price
from (
select
u.product_id product_id, price, units
from UnitsSold u
left join Prices p
on p.product_id=u.product_id
and u.purchase_date >= p.start_date
and u.purchase_date <= p.end_date
) res group by product_id;

584. 寻找用户推荐人

SQL架构

CREATE TABLE IF NOT EXISTS customer (id INT,name VARCHAR(25),referee_id INT);
Truncate table customer
insert into customer (id, name, referee_id) values ('1', 'Will', 'None')
insert into customer (id, name, referee_id) values ('2', 'Jane', 'None')
insert into customer (id, name, referee_id) values ('3', 'Alex', '2')
insert into customer (id, name, referee_id) values ('4', 'Bill', 'None')
insert into customer (id, name, referee_id) values ('5', 'Zack', '1')
insert into customer (id, name, referee_id) values ('6', 'Mark', '2')

给定表 customer ,里面保存了所有客户信息和他们的推荐人。

+------+------+-----------+
| id | name | referee_id|
+------+------+-----------+
| 1 | Will | NULL |
| 2 | Jane | NULL |
| 3 | Alex | 2 |
| 4 | Bill | NULL |
| 5 | Zack | 1 |
| 6 | Mark | 2 |
+------+------+-----------+

写一个查询语句,返回一个编号列表,列表中编号的推荐人的编号都 不是 2。

对于上面的示例数据,结果为:

+------+
| name |
+------+
| Will |
| Jane |
| Bill |
| Zack |
+------+

SQL 语句

select name
from customer
where referee_id != 2 or referee_id is null;

627. 交换工资

SQL架构

create table if not exists salary(id int, name varchar(100), sex char(1), salary int)
Truncate table salary
insert into salary (id, name, sex, salary) values ('1', 'A', 'm', '2500')
insert into salary (id, name, sex, salary) values ('2', 'B', 'f', '1500')
insert into salary (id, name, sex, salary) values ('3', 'C', 'm', '5500')
insert into salary (id, name, sex, salary) values ('4', 'D', 'f', '500')

给定一个 salary 表,如下所示,有 m = 男性 和 f = 女性 的值。交换所有的 f 和 m 值(例如,将所有 f 值更改为 m,反之亦然)。要求只使用一个更新(Update)语句,并且没有中间的临时表。

注意,您必只能写一个 Update 语句,请不要编写任何 Select 语句。

例如:

| id | name | sex | salary |
|----|------|-----|--------|
| 1 | A | m | 2500 |
| 2 | B | f | 1500 |
| 3 | C | m | 5500 |
| 4 | D | f | 500 |

运行你所编写的更新语句之后,将会得到以下表:

| id | name | sex | salary |
|----|------|-----|--------|
| 1 | A | f | 2500 |
| 2 | B | m | 1500 |
| 3 | C | f | 5500 |
| 4 | D | m | 500 |

SQL 语句

update salary set sex = if(sex='m', 'f', 'm');

1173. 即时食物配送 I

SQL架构

Create table If Not Exists Delivery (delivery_id int, customer_id int, order_date date, customer_pref_delivery_date date)
Truncate table Delivery
insert into Delivery (delivery_id, customer_id, order_date, customer_pref_delivery_date) values ('1', '1', '2019-08-01', '2019-08-02')
insert into Delivery (delivery_id, customer_id, order_date, customer_pref_delivery_date) values ('2', '5', '2019-08-02', '2019-08-02')
insert into Delivery (delivery_id, customer_id, order_date, customer_pref_delivery_date) values ('3', '1', '2019-08-11', '2019-08-11')
insert into Delivery (delivery_id, customer_id, order_date, customer_pref_delivery_date) values ('4', '3', '2019-08-24', '2019-08-26')
insert into Delivery (delivery_id, customer_id, order_date, customer_pref_delivery_date) values ('5', '4', '2019-08-21', '2019-08-22')
insert into Delivery (delivery_id, customer_id, order_date, customer_pref_delivery_date) values ('6', '2', '2019-08-11', '2019-08-13')

配送表: Delivery

+-----------------------------+---------+
| Column Name | Type |
+-----------------------------+---------+
| delivery_id | int |
| customer_id | int |
| order_date | date |
| customer_pref_delivery_date | date |
+-----------------------------+---------+
delivery_id 是表的主键。
该表保存着顾客的食物配送信息,顾客在某个日期下了订单,并指定了一个期望的配送日期(和下单日期相同或者在那之后)。

如果顾客期望的配送日期和下单日期相同,则该订单称为 「即时订单」,否则称为「计划订单」。

写一条 SQL 查询语句获取即时订单所占的比例, 保留两位小数。

查询结果如下所示:

Delivery 表:
+-------------+-------------+------------+-----------------------------+
| delivery_id | customer_id | order_date | customer_pref_delivery_date |
+-------------+-------------+------------+-----------------------------+
| 1 | 1 | 2019-08-01 | 2019-08-02 |
| 2 | 5 | 2019-08-02 | 2019-08-02 |
| 3 | 1 | 2019-08-11 | 2019-08-11 |
| 4 | 3 | 2019-08-24 | 2019-08-26 |
| 5 | 4 | 2019-08-21 | 2019-08-22 |
| 6 | 2 | 2019-08-11 | 2019-08-13 |
+-------------+-------------+------------+-----------------------------+

Result 表:
+----------------------+
| immediate_percentage |
+----------------------+
| 33.33 |
+----------------------+
2 和 3 号订单为即时订单,其他的为计划订单。

SQL 语句

select round(sum(if(order_date = customer_pref_delivery_date, 1, 0))/count(delivery_id)*100, 2) immediate_percentage
from Delivery

620. 有趣的电影

SQL架构

Create table If Not Exists cinema (id int, movie varchar(255), description varchar(255), rating float(2, 1))
Truncate table cinema
insert into cinema (id, movie, description, rating) values ('1', 'War', 'great 3D', '8.9')
insert into cinema (id, movie, description, rating) values ('2', 'Science', 'fiction', '8.5')
insert into cinema (id, movie, description, rating) values ('3', 'irish', 'boring', '6.2')
insert into cinema (id, movie, description, rating) values ('4', 'Ice song', 'Fantacy', '8.6')
insert into cinema (id, movie, description, rating) values ('5', 'House card', 'Interesting', '9.1')

某城市开了一家新的电影院,吸引了很多人过来看电影。该电影院特别注意用户体验,专门有个 LED显示板做电影推荐,上面公布着影评和相关电影描述。

作为该电影院的信息部主管,您需要编写一个 SQL查询,找出所有影片描述为 boring (不无聊) 的并且 id 为奇数 的影片,结果请按等级 rating 排列。

例如,下表 cinema:

+---------+-----------+--------------+-----------+
| id | movie | description | rating |
+---------+-----------+--------------+-----------+
| 1 | War | great 3D | 8.9 |
| 2 | Science | fiction | 8.5 |
| 3 | irish | boring | 6.2 |
| 4 | Ice song | Fantacy | 8.6 |
| 5 | House card| Interesting| 9.1 |
+---------+-----------+--------------+-----------+

对于上面的例子,则正确的输出是为:

+---------+-----------+--------------+-----------+
| id | movie | description | rating |
+---------+-----------+--------------+-----------+
| 5 | House card| Interesting| 9.1 |
| 1 | War | great 3D | 8.9 |
+---------+-----------+--------------+-----------+

SQL 语句

select *
from cinema
where description != 'boring' and mod(id, 2) = 1 -- id%2 = 1 / id&1 = 1
order by rating desc

1148. 文章浏览 I

SQL架构

Create table If Not Exists Views (article_id int, author_id int, viewer_id int, view_date date)
Truncate table Views
insert into Views (article_id, author_id, viewer_id, view_date) values ('1', '3', '5', '2019-08-01')
insert into Views (article_id, author_id, viewer_id, view_date) values ('1', '3', '6', '2019-08-02')
insert into Views (article_id, author_id, viewer_id, view_date) values ('2', '7', '7', '2019-08-01')
insert into Views (article_id, author_id, viewer_id, view_date) values ('2', '7', '6', '2019-08-02')
insert into Views (article_id, author_id, viewer_id, view_date) values ('4', '7', '1', '2019-07-22')
insert into Views (article_id, author_id, viewer_id, view_date) values ('3', '4', '4', '2019-07-21')
insert into Views (article_id, author_id, viewer_id, view_date) values ('3', '4', '4', '2019-07-21')

Views 表:

+---------------+---------+
| Column Name | Type |
+---------------+---------+
| article_id | int |
| author_id | int |
| viewer_id | int |
| view_date | date |
+---------------+---------+
此表无主键,因此可能会存在重复行。
此表的每一行都表示某人在某天浏览了某位作者的某篇文章。
请注意,同一人的 author_id 和 viewer_id 是相同的。

请编写一条 SQL 查询以找出所有浏览过自己文章的作者,结果按照 id 升序排列。查询结果的格式如下所示:

Views 表:
+------------+-----------+-----------+------------+
| article_id | author_id | viewer_id | view_date |
+------------+-----------+-----------+------------+
| 1 | 3 | 5 | 2019-08-01 |
| 1 | 3 | 6 | 2019-08-02 |
| 2 | 7 | 7 | 2019-08-01 |
| 2 | 7 | 6 | 2019-08-02 |
| 4 | 7 | 1 | 2019-07-22 |
| 3 | 4 | 4 | 2019-07-21 |
| 3 | 4 | 4 | 2019-07-21 |
+------------+-----------+-----------+------------+

结果表:
+------+
| id |
+------+
| 4 |
| 7 |
+------+

SQL 语句

select distinct author_id id
from Views
where author_id = viewer_id
order by id

1082. 销售分析 I

SQL架构

Create table If Not Exists Product (product_id int, product_name varchar(10), unit_price int)
Create table If Not Exists Sales (seller_id int, product_id int, buyer_id int, sale_date date, quantity int, price int)
Truncate table Product
insert into Product (product_id, product_name, unit_price) values ('1', 'S8', '1000')
insert into Product (product_id, product_name, unit_price) values ('2', 'G4', '800')
insert into Product (product_id, product_name, unit_price) values ('3', 'iPhone', '1400')
Truncate table Sales
insert into Sales (seller_id, product_id, buyer_id, sale_date, quantity, price) values ('1', '1', '1', '2019-01-21', '2', '2000')
insert into Sales (seller_id, product_id, buyer_id, sale_date, quantity, price) values ('1', '2', '2', '2019-02-17', '1', '800')
insert into Sales (seller_id, product_id, buyer_id, sale_date, quantity, price) values ('2', '2', '3', '2019-06-02', '1', '800')
insert into Sales (seller_id, product_id, buyer_id, sale_date, quantity, price) values ('3', '3', '4', '2019-05-13', '2', '2800')

产品表:Product

+--------------+---------+
| Column Name | Type |
+--------------+---------+
| product_id | int |
| product_name | varchar |
| unit_price | int |
+--------------+---------+
product_id 是这个表的主键.

销售表:Sales

+-------------+---------+
| Column Name | Type |
+-------------+---------+
| seller_id | int |
| product_id | int |
| buyer_id | int |
| sale_date | date |
| quantity | int |
| price | int |
+------ ------+---------+
这个表没有主键,它可以有重复的行.
product_id 是 Product 表的外键.

编写一个 SQL 查询,查询总销售额最高的销售者,如果有并列的,就都展示出来。

查询结果格式如下所示:

Product 表:
+------------+--------------+------------+
| product_id | product_name | unit_price |
+------------+--------------+------------+
| 1 | S8 | 1000 |
| 2 | G4 | 800 |
| 3 | iPhone | 1400 |
+------------+--------------+------------+

Sales 表:
+-----------+------------+----------+------------+----------+-------+
| seller_id | product_id | buyer_id | sale_date | quantity | price |
+-----------+------------+----------+------------+----------+-------+
| 1 | 1 | 1 | 2019-01-21 | 2 | 2000 |
| 1 | 2 | 2 | 2019-02-17 | 1 | 800 |
| 2 | 2 | 3 | 2019-06-02 | 1 | 800 |
| 3 | 3 | 4 | 2019-05-13 | 2 | 2800 |
+-----------+------------+----------+------------+----------+-------+

Result 表:
+-------------+
| seller_id |
+-------------+
| 1 |
| 3 |
+-------------+
Id 为 1 和 3 的销售者,销售总金额都为最高的 2800。

SQL 语句

select seller_id
from Sales
group by seller_id
having sum(price) >= all(select sum(price) from sales group by seller_id)

1050. 合作过至少三次的演员和导演

SQL架构

Create table If Not Exists ActorDirector (actor_id int, director_id int, timestamp int)
Truncate table ActorDirector
insert into ActorDirector (actor_id, director_id, timestamp) values ('1', '1', '0')
insert into ActorDirector (actor_id, director_id, timestamp) values ('1', '1', '1')
insert into ActorDirector (actor_id, director_id, timestamp) values ('1', '1', '2')
insert into ActorDirector (actor_id, director_id, timestamp) values ('1', '2', '3')
insert into ActorDirector (actor_id, director_id, timestamp) values ('1', '2', '4')
insert into ActorDirector (actor_id, director_id, timestamp) values ('2', '1', '5')
insert into ActorDirector (actor_id, director_id, timestamp) values ('2', '1', '6')

ActorDirector 表:

+-------------+---------+
| Column Name | Type |
+-------------+---------+
| actor_id | int |
| director_id | int |
| timestamp | int |
+-------------+---------+
timestamp 是这张表的主键.

写一条SQL查询语句获取合作过至少三次的演员和导演的 id 对 (actor_id, director_id)

示例:

ActorDirector 表:
+-------------+-------------+-------------+
| actor_id | director_id | timestamp |
+-------------+-------------+-------------+
| 1 | 1 | 0 |
| 1 | 1 | 1 |
| 1 | 1 | 2 |
| 1 | 2 | 3 |
| 1 | 2 | 4 |
| 2 | 1 | 5 |
| 2 | 1 | 6 |
+-------------+-------------+-------------+

Result 表:
+-------------+-------------+
| actor_id | director_id |
+-------------+-------------+
| 1 | 1 |
+-------------+-------------+
唯一的 id 对是 (1, 1),他们恰好合作了 3 次。

SQL 语句

select actor_id, director_id
from ActorDirector
group by actor_id, director_id
having count(*) >= 3

586. 订单最多的客户

SQL架构

Create table If Not Exists orders (order_number int, customer_number int, order_date date, required_date date, shipped_date date, status char(15), comment char(200), key(order_number))
Truncate table orders
insert into orders (order_number, customer_number) values ('1', '1')
insert into orders (order_number, customer_number) values ('2', '2')
insert into orders (order_number, customer_number) values ('3', '3')
insert into orders (order_number, customer_number) values ('4', '3')

在表 orders 中找到订单数最多客户对应的 customer_number

数据保证订单数最多的顾客恰好只有一位。

orders 定义如下:

| Column            | Type      |
|-------------------|-----------|
| order_number (PK) | int |
| customer_number | int |
| order_date | date |
| required_date | date |
| shipped_date | date |
| status | char(15) |
| comment | char(200) |

样例输入

| order_number | customer_number | order_date | required_date | shipped_date | status | comment |
|--------------|-----------------|------------|---------------|--------------|--------|---------|
| 1 | 1 | 2017-04-09 | 2017-04-13 | 2017-04-12 | Closed | |
| 2 | 2 | 2017-04-15 | 2017-04-20 | 2017-04-18 | Closed | |
| 3 | 3 | 2017-04-16 | 2017-04-25 | 2017-04-20 | Closed | |
| 4 | 3 | 2017-04-18 | 2017-04-28 | 2017-04-25 | Closed | |

样例输出

| customer_number |
|-----------------|
| 3 |

解释

customer_number 为 '3' 的顾客有两个订单,比顾客 '1' 或者 '2' 都要多,因为他们只有一个订单
所以结果是该顾客的 customer_number ,也就是 3 。

进阶: 如果有多位顾客订单数并列最多,你能找到他们所有的 customer_number 吗?

SQL 语句

select customer_number
from orders
group by customer_number
order by count(*) desc
limit 1

-- 进阶
select customer_number
from orders
group by customer_number
having count(*) >= all(select count(*) from orders group by customer_number)

511. 游戏玩法分析 I

SQL架构

Create table If Not Exists Activity (player_id int, device_id int, event_date date, games_played int)
Truncate table Activity
insert into Activity (player_id, device_id, event_date, games_played) values ('1', '2', '2016-03-01', '5')
insert into Activity (player_id, device_id, event_date, games_played) values ('1', '2', '2016-05-02', '6')
insert into Activity (player_id, device_id, event_date, games_played) values ('2', '3', '2017-06-25', '1')
insert into Activity (player_id, device_id, event_date, games_played) values ('3', '1', '2016-03-02', '0')
insert into Activity (player_id, device_id, event_date, games_played) values ('3', '4', '2018-07-03', '5')

活动表 Activity

+--------------+---------+
| Column Name | Type |
+--------------+---------+
| player_id | int |
| device_id | int |
| event_date | date |
| games_played | int |
+--------------+---------+
表的主键是 (player_id, event_date)。
这张表展示了一些游戏玩家在游戏平台上的行为活动。
每行数据记录了一名玩家在退出平台之前,当天使用同一台设备登录平台后打开的游戏的数目(可能是 0 个)。

写一条 SQL 查询语句获取每位玩家 第一次登陆平台的日期

查询结果的格式如下所示:

Activity 表:
+-----------+-----------+------------+--------------+
| player_id | device_id | event_date | games_played |
+-----------+-----------+------------+--------------+
| 1 | 2 | 2016-03-01 | 5 |
| 1 | 2 | 2016-05-02 | 6 |
| 2 | 3 | 2017-06-25 | 1 |
| 3 | 1 | 2016-03-02 | 0 |
| 3 | 4 | 2018-07-03 | 5 |
+-----------+-----------+------------+--------------+

Result 表:
+-----------+-------------+
| player_id | first_login |
+-----------+-------------+
| 1 | 2016-03-01 |
| 2 | 2017-06-25 |
| 3 | 2016-03-02 |
+-----------+-------------+

SQL 语句

select player_id, min(event_date) first_login
from Activity
group by player_id

175. 组合两个表

SQL架构

Create table Person (PersonId int, FirstName varchar(255), LastName varchar(255))
Create table Address (AddressId int, PersonId int, City varchar(255), State varchar(255))
Truncate table Person
insert into Person (PersonId, LastName, FirstName) values ('1', 'Wang', 'Allen')
Truncate table Address
insert into Address (AddressId, PersonId, City, State) values ('1', '2', 'New York City', 'New York')

表1: Person

+-------------+---------+
| 列名 | 类型 |
+-------------+---------+
| PersonId | int |
| FirstName | varchar |
| LastName | varchar |
+-------------+---------+
PersonId 是上表主键

表2: Address

+-------------+---------+
| 列名 | 类型 |
+-------------+---------+
| AddressId | int |
| PersonId | int |
| City | varchar |
| State | varchar |
+-------------+---------+
AddressId 是上表主键

编写一个 SQL 查询,满足条件:无论 person 是否有地址信息,都需要基于上述两表提供 person 的以下信息:

FirstName, LastName, City, State

SQL 语句

select FirstName, LastName, City, State
from Person p
left join Address a
on p.PersonId = a. PersonId

1327. 列出指定时间段内所有的下单产品

SQL架构

Create table If Not Exists Products (product_id int, product_name varchar(40), product_category varchar(40))
Create table If Not Exists Orders (product_id int, order_date date, unit int)
Truncate table Products
insert into Products (product_id, product_name, product_category) values ('1', 'Leetcode Solutions', 'Book')
insert into Products (product_id, product_name, product_category) values ('2', 'Jewels of Stringology', 'Book')
insert into Products (product_id, product_name, product_category) values ('3', 'HP', 'Laptop')
insert into Products (product_id, product_name, product_category) values ('4', 'Lenovo', 'Laptop')
insert into Products (product_id, product_name, product_category) values ('5', 'Leetcode Kit', 'T-shirt')
Truncate table Orders
insert into Orders (product_id, order_date, unit) values ('1', '2020-02-05', '60')
insert into Orders (product_id, order_date, unit) values ('1', '2020-02-10', '70')
insert into Orders (product_id, order_date, unit) values ('2', '2020-01-18', '30')
insert into Orders (product_id, order_date, unit) values ('2', '2020-02-11', '80')
insert into Orders (product_id, order_date, unit) values ('3', '2020-02-17', '2')
insert into Orders (product_id, order_date, unit) values ('3', '2020-02-24', '3')
insert into Orders (product_id, order_date, unit) values ('4', '2020-03-01', '20')
insert into Orders (product_id, order_date, unit) values ('4', '2020-03-04', '30')
insert into Orders (product_id, order_date, unit) values ('4', '2020-03-04', '60')
insert into Orders (product_id, order_date, unit) values ('5', '2020-02-25', '50')
insert into Orders (product_id, order_date, unit) values ('5', '2020-02-27', '50')
insert into Orders (product_id, order_date, unit) values ('5', '2020-03-01', '50')

表: Products

+------------------+---------+
| Column Name | Type |
+------------------+---------+
| product_id | int |
| product_name | varchar |
| product_category | varchar |
+------------------+---------+
product_id 是该表主键。
该表包含该公司产品的数据。

表: Orders

+---------------+---------+
| Column Name | Type |
+---------------+---------+
| product_id | int |
| order_date | date |
| unit | int |
+---------------+---------+
该表无主键,可能包含重复行。
product_id 是表单 Products 的外键。
unit 是在日期 order_date 内下单产品的数目。

写一个 SQL 语句,要求获取在 2020 年 2 月份下单的数量不少于 100 的产品的名字和数目。

返回结果表单的顺序无要求。

查询结果的格式如下:

Products 表:
+-------------+-----------------------+------------------+
| product_id | product_name | product_category |
+-------------+-----------------------+------------------+
| 1 | Leetcode Solutions | Book |
| 2 | Jewels of Stringology | Book |
| 3 | HP | Laptop |
| 4 | Lenovo | Laptop |
| 5 | Leetcode Kit | T-shirt |
+-------------+-----------------------+------------------+

Orders 表:
+--------------+--------------+----------+
| product_id | order_date | unit |
+--------------+--------------+----------+
| 1 | 2020-02-05 | 60 |
| 1 | 2020-02-10 | 70 |
| 2 | 2020-01-18 | 30 |
| 2 | 2020-02-11 | 80 |
| 3 | 2020-02-17 | 2 |
| 3 | 2020-02-24 | 3 |
| 4 | 2020-03-01 | 20 |
| 4 | 2020-03-04 | 30 |
| 4 | 2020-03-04 | 60 |
| 5 | 2020-02-25 | 50 |
| 5 | 2020-02-27 | 50 |
| 5 | 2020-03-01 | 50 |
+--------------+--------------+----------+

Result 表:
+--------------------+---------+
| product_name | unit |
+--------------------+---------+
| Leetcode Solutions | 130 |
| Leetcode Kit | 100 |
+--------------------+---------+

2020 年 2 月份下单 product_id = 1 的产品的数目总和为 (60 + 70) = 130 。
2020 年 2 月份下单 product_id = 2 的产品的数目总和为 80 。
2020 年 2 月份下单 product_id = 3 的产品的数目总和为 (2 + 3) = 5 。
2020 年 2 月份 product_id = 4 的产品并没有下单。
2020 年 2 月份下单 product_id = 5 的产品的数目总和为 (50 + 50) = 100 。

SQL 语句

select product_name, sum(unit) unit
from Orders o
join Products p
on o.product_id = p.product_id
where order_date between '2020-02-01' and '2020-02-29'
group by o.product_id
having sum(o.unit) >= 100

577. 员工奖金

SQL架构

Create table If Not Exists Employee (EmpId int, Name varchar(255), Supervisor int, Salary int)
Create table If Not Exists Bonus (EmpId int, Bonus int)
Truncate table Employee
insert into Employee (EmpId, Name, Supervisor, Salary) values ('3', 'Brad', 'None', '4000')
insert into Employee (EmpId, Name, Supervisor, Salary) values ('1', 'John', '3', '1000')
insert into Employee (EmpId, Name, Supervisor, Salary) values ('2', 'Dan', '3', '2000')
insert into Employee (EmpId, Name, Supervisor, Salary) values ('4', 'Thomas', '3', '4000')
Truncate table Bonus
insert into Bonus (EmpId, Bonus) values ('2', '500')
insert into Bonus (EmpId, Bonus) values ('4', '2000')

选出所有 bonus < 1000 的员工的 name 及其 bonus。

Employee 表单

+-------+--------+-----------+--------+
| empId | name | supervisor| salary |
+-------+--------+-----------+--------+
| 1 | John | 3 | 1000 |
| 2 | Dan | 3 | 2000 |
| 3 | Brad | null | 4000 |
| 4 | Thomas | 3 | 4000 |
+-------+--------+-----------+--------+
empId 是这张表单的主关键字

Bonus 表单

+-------+-------+
| empId | bonus |
+-------+-------+
| 2 | 500 |
| 4 | 2000 |
+-------+-------+
empId 是这张表单的主关键字

输出示例:

+-------+-------+
| name | bonus |
+-------+-------+
| John | null |
| Dan | 500 |
| Brad | null |
+-------+-------+

SQL 语句

select `name`, bonus
from Employee e left join Bonus b
on e.empId = b.empId
where b.bonus < 1000 or b.bonus is null

181. 超过经理收入的员工

SQL架构

Create table If Not Exists Employee (Id int, Name varchar(255), Salary int, ManagerId int)
Truncate table Employee
insert into Employee (Id, Name, Salary, ManagerId) values ('1', 'Joe', '70000', '3')
insert into Employee (Id, Name, Salary, ManagerId) values ('2', 'Henry', '80000', '4')
insert into Employee (Id, Name, Salary, ManagerId) values ('3', 'Sam', '60000', 'None')
insert into Employee (Id, Name, Salary, ManagerId) values ('4', 'Max', '90000', 'None')

Employee 表包含所有员工,他们的经理也属于员工。每个员工都有一个 Id,此外还有一列对应员工的经理的 Id。

+----+-------+--------+-----------+
| Id | Name | Salary | ManagerId |
+----+-------+--------+-----------+
| 1 | Joe | 70000 | 3 |
| 2 | Henry | 80000 | 4 |
| 3 | Sam | 60000 | NULL |
| 4 | Max | 90000 | NULL |
+----+-------+--------+-----------+

给定 Employee 表,编写一个 SQL 查询,该查询可以获取收入超过他们经理的员工的姓名。在上面的表格中,Joe 是唯一一个收入超过他的经理的员工。

+----------+
| Employee |
+----------+
| Joe |
+----------+

SQL 语句

select e1.`Name` Employee
from Employee e1, Employee e2
where e1.ManagerId = e2.id and e1.Salary > e2.Salary

603. 连续空余座位

SQL架构

Create table If Not Exists cinema (seat_id int primary key auto_increment, free bool)
Truncate table cinema
insert into cinema (seat_id, free) values ('1', '1')
insert into cinema (seat_id, free) values ('2', '0')
insert into cinema (seat_id, free) values ('3', '1')
insert into cinema (seat_id, free) values ('4', '1')
insert into cinema (seat_id, free) values ('5', '1')

几个朋友来到电影院的售票处,准备预约连续空余座位。

你能利用表 cinema ,帮他们写一个查询语句,获取所有空余座位,并将它们按照 seat_id 排序后返回吗?

| seat_id | free |
|---------|------|
| 1 | 1 |
| 2 | 0 |
| 3 | 1 |
| 4 | 1 |
| 5 | 1 |

对于如上样例,你的查询语句应该返回如下结果。

| seat_id |
|---------|
| 3 |
| 4 |
| 5 |

注意:

  • seat_id 字段是一个自增的整数,free 字段是布尔类型(’1’ 表示空余, ‘0’ 表示已被占据)。
  • 连续空余座位的定义是大于等于 2 个连续空余的座位。

SQL 语句



SQL 语句

SQL 语句

SQL 语句

SQL 语句

SQL 语句

SQL 语句

SQL 语句

SQL 语句

SQL 语句

SQL 语句


知识来源: https://wywwzjj.top/2020/03/21/LeetCode-数据库专题/

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