一题：Combine Two Tables
中文：
Table: Person
+-------------+---------+
| Column Name | Type |
+-------------+---------+
| PersonId | int |
| FirstName | varchar |
| LastName | varchar |
+-------------+---------+
PersonId is the primary key column for this table.
Table: Address
+-------------+---------+
| Column Name | Type |
+-------------+---------+
| AddressId | int |
| PersonId | int |
| City | varchar |
| State | varchar |
+-------------+---------+
AddressId is the primary key column for this table.
为报告编写SQL查询，该报告为Person表中的每个人提供以下信息，无论这个人是否有一个地址：
FirstName, LastName, City, State英文：
Table: Person
+-------------+---------+
| Column Name | Type |
+-------------+---------+
| PersonId | int |
| FirstName | varchar |
| LastName | varchar |
+-------------+---------+
PersonId is the primary key column for this table.
Table: Address
+-------------+---------+
| Column Name | Type |
+-------------+---------+
| AddressId | int |
| PersonId | int |
| City | varchar |
| State | varchar |
+-------------+---------+
AddressId is the primary key column for this table.
Write a SQL query for a report that provides the following information for each person in the Person table, regardless if there is an address for each of those people:
FirstName, LastName, City, State
解题思路：
因为不管用户有没有对应的地址，一直要保证有用户数据，所以以person为左表，进行左关联。
# 创建Person表
Create table Person (PersonId int, FirstName varchar(255), LastName varchar(255));
# 创建Address表
Create table Address (AddressId int, PersonId int, City varchar(255), State varchar(255));
# 删除Person表中的所有数据，类似于delete from Person
Truncate table Person;
# 往Person表中插入一条数据
insert into Person (PersonId, LastName, FirstName) values ('1', 'Wang', 'Allen');
# 删除Address表中的所有数据
Truncate table Address;
# 往Address表中插入一条数据
insert into Address (AddressId, PersonId, City, State) values ('1', '2', 'New York City', 'New York');
# Date：2019.02.15
# num：195
# 解题答案：
# ==============================================================================
# 使用左关联查询表中的数据
SELECT p.FirstName,p.LastName,a.City,a.State from Person p left join Address a on p.PersonId = a.PersonId;二题：Employees Earning More Than Their Managers
中文：
Employee表包含所有员工，包括他们的经理。每个员工都有一个Id，并且还有一个经理Id列。
+----+-------+--------+-----------+
| 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，查找收入高于其经理的员工。对于上表，Joe是唯一一位收入超过其经理的员工。
+----------+
| Employee |
+----------+
| Joe |
+----------+英文：
The Employee table holds all employees including their managers. Every employee has an Id, and there is also a column for the manager Id.
+----+-------+--------+-----------+
| Id | Name | Salary | ManagerId |
+----+-------+--------+-----------+
| 1 | Joe | 70000 | 3 |
| 2 | Henry | 80000 | 4 |
| 3 | Sam | 60000 | NULL |
| 4 | Max | 90000 | NULL |
+----+-------+--------+-----------+
Given the Employee table, write a SQL query that finds out employees who earn more than their managers. For the above table, Joe is the only employee who earns more than his manager.
+----------+
| Employee |
+----------+
| Joe |
+----------+解题思路：
将表自身通过ManagerId进行关联，前半部分为自身信息，后半部分为经理信息，查询自身薪水大于经理薪水。
并将查询到的name重命名为Employee。
# 创建员工表，有id，名字，薪水，经理Id
CREATE TABLE
IF
NOT EXISTS Employee ( Id INT, NAME VARCHAR ( 255 ), Salary INT, ManagerId INT );# 清空员工表中所有数据
TRUNCATE TABLE Employee;# 向员工表中插入4条数据
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', '0' );
INSERT INTO Employee ( Id, NAME, Salary, ManagerId )
VALUES
( '4', 'Max', '90000', '0' );
DROP TABLE Employee;# date：2019.02.15
# num：181
# 解题答案：
# =====================================
# 将表自身通过ManagerId进行关联，前半部分为自身信息，后半部分为经理信息，查询自身薪水大于经理薪水
# 并将查询到的name重命名为Employee
SELECT
A.`Name` AS Employee
FROM
Employee AS A,
Employee AS B
WHERE
A.ManagerId = B.Id
AND A.Salary > B.Salary;三题：Customers Who Never Order
中文：
假设一个网站包含两个表，Customers表和Orders表。编写SQL查询以查找从未订购任何内容的所有客户。
Table: Customers.
+----+-------+
| Id | Name |
+----+-------+
| 1 | Joe |
| 2 | Henry |
| 3 | Sam |
| 4 | Max |
+----+-------+
Table: Orders.
+----+------------+
| Id | CustomerId |
+----+------------+
| 1 | 3 |
| 2 | 1 |
+----+------------+使用上面的表作为示例，返回以下内容：
+-----------+
| Customers |
+-----------+
| Henry |
| Max |
+-----------+
英文：
Suppose that a website contains two tables, the Customers table and the Orders table. Write a SQL query to find all customers who never order anything.
Table: Customers.
+----+-------+
| Id | Name |
+----+-------+
| 1 | Joe |
| 2 | Henry |
| 3 | Sam |
| 4 | Max |
+----+-------+
Table: Orders.
+----+------------+
| Id | CustomerId |
+----+------------+
| 1 | 3 |
| 2 | 1 |
+----+------------+
Using the above tables as example, return the following:
+-----------+
| Customers |
+-----------+
| Henry |
| Max |
+-----------+解题思路：
第一种：
使用子查询。
查询在订单表中有的customerId作为查询条件，查询Customer表中Id不在其中的用户名称。
第二种：
使用左关联，将Customers和Orders表使用customerId进行关联，如果该用户不存在订单表中，则customerId为空。
# 创建客户表
CREATE TABLE
IF
NOT EXISTS Customers ( Id INT, NAME VARCHAR ( 255 ) );
# 创建订单表
CREATE TABLE
IF
NOT EXISTS Orders ( Id INT, CustomerId INT );
# 清空客户表原有数据，并插入数据
TRUNCATE TABLE Customers;
INSERT INTO Customers ( Id, NAME )
VALUES
( '1', 'Joe' );
INSERT INTO Customers ( Id, NAME )
VALUES
( '2', 'Henry' );
INSERT INTO Customers ( Id, NAME )
VALUES
( '3', 'Sam' );
INSERT INTO Customers ( Id, NAME )
VALUES
( '4', 'Max' );
# 清空订单表原有数据，并插入数据
TRUNCATE TABLE Orders;
INSERT INTO Orders ( Id, CustomerId )
VALUES
( '1', '3' );
INSERT INTO Orders ( Id, CustomerId )
VALUES
( '2', '1' );
# date: 2039.02.15
# num: 183
# 解题答案：
# =====================================
# 方法一：使用子查询。查询在订单表中有的customerId作为查询条件，查询Customer表中Id不在其中的用户名称。
SELECT NAME AS
Customers
FROM
Customers
WHERE
Id NOT IN ( SELECT CustomerId FROM Orders );
# 方法二：使用左关联，将Customers和Orders表使用customerId进行关联，如果该用户不存在订单表中，则customerId为空。
SELECT
*
FROM
Customers C
LEFT JOIN Orders O ON C.Id = O.CustomerId
WHERE
O.CustomerId IS NULL;