数据库

leetcode

175.

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

答案：
左连接： 从左表那里返回所有的行，即使在右表中没有匹配的行。

SELECT FirstName, LastName, City, State FROM Person LEFT JOIN Address ON Person.PersonId = Address.PersonId

176.

编写一个 SQL 查询语句，获取Employee表中第二高的薪水（Salary）。

答案：

此题可以分为两块考虑，第一块是最高的薪水，第二个是比最高薪水小的最高的薪水

 select Max(Salary)  SecondHighestSalary from Employee where (select Max(Salary)from Employee) > Salary

Max(Salary)  SecondHighestSalary(在结果中，用SecondHighestSalary替代Max(Salary)作为COLUMN NAME)

181.

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      |
+----------+

 答案：

select E.Name Employee from Employee E, Employee H where E.ManagerId = H.Id and E.Salary > H.Salary

182.

Write a SQL query to find all duplicate emails in a table named Person.

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

For example, your query should return the following for the above table:

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

Note: All emails are in lowercase.

答案：

使用group by 和having子句

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

183.

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       |
+-----------+

答案：

select Name Customers from Customers left join Orders on Customers.Id = Orders.CustomerId where Orders.Id is null