leetcode中Database题

Posted 2020-09-10 爱姿病

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

题目意思就是找出人的信息，即使AddressId不存在。这样我们就要使用左连接(left join)

select FirstName,LastName,City,state from Person left join Address on Person.PersonId=Address.PersonId

 

补充：

A right join B 是根据表B来匹配，表A中只能查询出和表B匹配的内容。

A inner joinB 是内连接查询，只能查询出指定字段相同的记录。

 

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 second highest salary is 200. If there is no second highest salary, then the query should return null.

要注意两个特例一个只有一个数据行，还有一种情况是有两行数据，但工资相等，这样也要返回null 所以要用group by，还有一种情况是有多个第二高的工资。

select (
  select distinct Salary from Employee order by Salary Desc limit 1 offset 1
)

1. limit numA offset numB 表示读取numA行数据，从numB+1行读取。

   limit numA,numB 表示返回的记录，从numA行以后开始读取，读取numB行数据。

2. distinct 去除表中从复记录，只保留一条。去除多条第二高的Salary。

3. order by 对记录进行排序，可以采用Asc和Desc。

 

 

Rank Scores

 

Write a SQL query to rank scores. If there is a tie between two scores, both should have the same ranking. Note that after a tie, the next ranking number should be the next consecutive integer value. In other words, there should be no "holes" between ranks.

+----+-------+
| Id | Score |
+----+-------+
| 1  | 3.50  |
| 2  | 3.65  |
| 3  | 4.00  |
| 4  | 3.85  |
| 5  | 4.00  |
| 6  | 3.65  |
+----+-------+

For example, given the above Scores table, your query should generate the following report (order by highest score):

+-------+------+
| Score | Rank |
+-------+------+
| 4.00  | 1    |
| 4.00  | 1    |
| 3.85  | 2    |
| 3.65  | 3    |
| 3.65  | 3    |
| 3.50  | 4    |
+-------+------+

SELECT
  Score,
  (SELECT count(*) FROM (SELECT distinct Score s FROM Scores) tmp WHERE s >= Score) Rank
FROM Scores
ORDER BY Score desc

 

Consecutive Numbers

Write a SQL query to find all numbers that appear at least three times consecutively.

+----+-----+
| Id | Num |
+----+-----+
| 1  |  1  |
| 2  |  1  |
| 3  |  1  |
| 4  |  2  |
| 5  |  1  |
| 6  |  2  |
| 7  |  2  |
+----+-----+

For example, given the above Logs table, 1 is the only number that appears consecutively for at least three times.

Select DISTINCT l1.Num ConsecutiveNums from Logs l1, Logs l2, Logs l3 
    where l1.Id=l2.Id-1 and l2.Id=l3.Id-1 
        and l1.Num=l2.Num and l2.Num=l3.Num;

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

select E1.Name Employee from Employee as E1,Employee as E2 where E1.ManagerId = E2.Id and E1.Salary > E2.Salary;

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] |
+---------+

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

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

select a.Name Customers from Customers a where a.Id not in (select b.CustomerId from Orders b);

Department Highest Salary

The Employee table holds all employees. Every employee has an Id, a salary, and there is also a column for the department Id.

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

The Department table holds all departments of the company.

+----+----------+
| Id | Name     |
+----+----------+
| 1  | IT       |
| 2  | Sales    |
+----+----------+

Write a SQL query to find employees who have the highest salary in each of the departments. For the above tables, Max has the highest salary in the IT department and Henry has the highest salary in the Sales department.

+------------+----------+--------+
| Department | Employee | Salary |
+------------+----------+--------+
| IT         | Max      | 90000  |
| Sales      | Henry    | 80000  |
+------------+----------+--------+

SELECT dep.Name as Department, emp.Name as Employee, emp.Salary 
    from Department dep, Employee emp 
        where emp.DepartmentId=dep.Id 
            and emp.Salary=(Select max(Salary) from Employee e2 where e2.DepartmentId=dep.Id)

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]  |
+----+------------------+

DELETE p from Person p, Person q where p.Id>q.Id AND q.Email=p.Email;

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) | Date(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 |
+----+

SELECT a.Id FROM Weather a INNER JOIN Weather b ON TO_DAYS(a.Date) = TO_DAYS(b.Date)+1 WHERE a.Temperature > b.Temperature;