175. Combine Two Tables
SQL架构
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


175. 组合两个表
SQL架构
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

# Write your MySQL query statement below
select FirstName, LastName, City, State from Person left join Address on Person.PersonId = Address.PersonId;

183. Customers Who Never Order
SQL架构
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       |
+-----------+

# Write your MySQL query statement below
select Name  as "Customers" from Customers where id not in (select CustomerId from Orders);

196. Delete Duplicate Emails
Write a SQL query to delete all duplicate email entries in a table named Person, keeping only unique emails based on its smallest Id.

+----+------------------+
| Id | Email            |
+----+------------------+
| 1  | john@example.com |
| 2  | bob@example.com  |
| 3  | john@example.com |
+----+------------------+
Id is the primary key column for this table.
For example, after running your query, the above Person table should have the following rows:

+----+------------------+
| Id | Email            |
+----+------------------+
| 1  | john@example.com |
| 2  | bob@example.com  |
+----+------------------+
Note:

Your output is the whole Person table after executing your sql. Use delete statement.

196. 删除重复的电子邮箱
编写一个 SQL 查询，来删除 Person 表中所有重复的电子邮箱，重复的邮箱里只保留 Id 最小 的那个。

+----+------------------+
| Id | Email            |
+----+------------------+
| 1  | john@example.com |
| 2  | bob@example.com  |
| 3  | john@example.com |
+----+------------------+
Id 是这个表的主键。
例如，在运行你的查询语句之后，上面的 Person 表应返回以下几行:

+----+------------------+
| Id | Email            |
+----+------------------+
| 1  | john@example.com |
| 2  | bob@example.com  |
+----+------------------+
 

提示：

执行 SQL 之后，输出是整个 Person 表。
使用 delete 语句。

# Write your MySQL query statement below
DELETE p1 from Person p1,Person p2 where p1.Id > p2.Id and p1.Email=p2.Email;

197. Rising Temperature
SQL架构
Given a Weather table, write a SQL query to find all dates' Ids with higher temperature compared to its previous (yesterday's) dates.

+---------+------------------+------------------+
| Id(INT) | RecordDate(DATE) | Temperature(INT) |
+---------+------------------+------------------+
|       1 |       2015-01-01 |               10 |
|       2 |       2015-01-02 |               25 |
|       3 |       2015-01-03 |               20 |
|       4 |       2015-01-04 |               30 |
+---------+------------------+------------------+
For example, return the following Ids for the above Weather table:

+----+
| Id |
+----+
|  2 |
|  4 |
+----+
197. 上升的温度
SQL架构
给定一个 Weather 表，编写一个 SQL 查询，来查找与之前（昨天的）日期相比温度更高的所有日期的 Id。

+---------+------------------+------------------+
| Id(INT) | RecordDate(DATE) | Temperature(INT) |
+---------+------------------+------------------+
|       1 |       2015-01-01 |               10 |
|       2 |       2015-01-02 |               25 |
|       3 |       2015-01-03 |               20 |
|       4 |       2015-01-04 |               30 |
+---------+------------------+------------------+
例如，根据上述给定的 Weather 表格，返回如下 Id:

+----+
| Id |
+----+
|  2 |
|  4 |
+----+

select w2.Id from Weather w1,Weather w2 where DATEDIFF(w2.RecordDate, w1.RecordDate) = 1 and w1.Temperature < w2.Temperature;