175. Combine Two Tables
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
MYSQL语句：
select FirstName, LastName, City, State
from Person left join Address on Person.PersonId=Address.PersonId176. Second Highest Salary
Write a SQL query to get the second highest salary from the Employee table.
+----+--------+
| Id | Salary |
+----+--------+
| 1 | 100 |
| 2 | 200 |
| 3 | 300 |
+----+--------+
For example, given the above Employee table, the query should return 200 as the second highest salary. If there is no second highest salary, then the query should return null.
+---------------------+
| SecondHighestSalary |
+---------------------+
| 200 |
+---------------------+
MYSQL语句：
select MAX(Salary) as SecondHighestSalary
from Employee
where Salary < (select MAX(Salary) from Employee);181. Employees Earning More Than Their Managers
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 |
+----------+
这里ManageId 就是管理这个员工的 管理员id
MYSQL 代码：
Select Name as Employee from Employee A where salary > (Select salary from Employee B where B.Id = A.ManagerId);
182. Duplicate Emails
Write a SQL query to find all duplicate emails in a table named Person.
+----+---------+
| Id | Email |
+----+---------+
| 1 | [email protected] |
| 2 | [email protected] |
| 3 | [email protected] |
+----+---------+
For example, your query should return the following for the above table:
+---------+
| Email |
+---------+
| [email protected] |
+---------+
MYSQL 代码：
select Email from Person group by Email having count(Email)>1;183. Customers Who Never Order
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 |
+-----------+
sql之left join、right join、inner join的区别
left join(左联接) 返回包括左表中的所有记录和右表中联结字段相等的记录
right join(右联接) 返回包括右表中的所有记录和左表中联结字段相等的记录
inner join(等值连接) 只返回两个表中联结字段相等的行
MYSQL语句：
select Name as Customers from Customers
where Id not in (
select Customers.Id from Customers
inner join
Orders
on Customers.Id=Orders.CustomerId
)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 | [email protected] |
| 2 | [email protected] |
| 3 | [email protected] |
+----+------------------+
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 | [email protected] |
| 2 | [email protected] |
+----+------------------+
Note:
Your output is the whole Person table after executing your sql. Use delete statement.
# Write your MySQL query statement below
delete from
Person
where
Id not in (select Id
from
(select min(Id) as Id
from Person
group by Email
) p
);197. Rising Temperature
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 | +----+
SQL语句
SELECT wt1.Id as Id
FROM Weather wt1,Weather wt2
WHERE TO_DAYS(wt1.RecordDate)-TO_DAYS(wt2.RecordDate)=1
&& wt1.Temperature>wt2.Temperature;