1. 交换性别
Given a table salary, such as the one below, that has m=male and f=female values. Swap all f and m values (i.e., change all f values to m and vice versa) with a single update query and no intermediate temp table.
For example:
| id | name | sex | salary | |----|------|-----|--------| | 1 | A | m | 2500 | | 2 | B | f | 1500 | | 3 | C | m | 5500 | | 4 | D | f | 500 |
After running your query, the above salary table should have the following rows:
| id | name | sex | salary | |----|------|-----|--------| | 1 | A | f | 2500 | | 2 | B | m | 1500 | | 3 | C | f | 5500 | | 4 | D | m | 500 |
Answer：
update salary
set sex = case sex
when 'm' then 'f'
else 'm'
end;2. 不boring的电影
X city opened a new cinema, many people would like to go to this cinema. The cinema also gives out a poster indicating the movies’ ratings and descriptions.
Please write a SQL query to output movies with an odd numbered ID and a description that is not ‘boring’. Order the result by rating.
For example, table cinema:
+---------+-----------+--------------+-----------+ | id | movie | description | rating | +---------+-----------+--------------+-----------+ | 1 | War | great 3D | 8.9 | | 2 | Science | fiction | 8.5 | | 3 | irish | boring | 6.2 | | 4 | Ice song | Fantacy | 8.6 | | 5 | House card| Interesting| 9.1 | +---------+-----------+--------------+-----------+
For the example above, the output should be:
+---------+-----------+--------------+-----------+ | id | movie | description | rating | +---------+-----------+--------------+-----------+ | 5 | House card| Interesting| 9.1 | | 1 | War | great 3D | 8.9 | +---------+-----------+--------------+-----------+
Answer:
select *
from cinema
where id%2=1 and description <> 'boring'
order by rating desc
3. Duplicate Emails
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 | +---------+
Answer:
select Email
from Person
group by Email
having count(*) > 14. 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
Answer:
select P.FirstName,P.LastName,A.City,A.State
from Person as P left join Address as A
on P.PersonId = A.PersonId
5. 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 | +----------+
Answer:
select a.Name as Employee
from Employee as a inner join Employee as b
on a.ManagerId = b.Id
where a.Salary > b.Salary6. 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 | +-----------+
Answer:
select Name as Customers
from Customers
where Id not in (select CustomerId from Orders)7. 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 | +----+
Answer:
select a.Id
from Weather as a inner join Weather as b
on datediff(a.Date,b.Date) = 1 and a.Temperature > b.Temperature8. Classes More Than 5 Students
There is a table courses with columns: student and class
Please list out all classes which have more than or equal to 5 students.
For example, the table:
+---------+------------+ | student | class | +---------+------------+ | A | Math | | B | English | | C | Math | | D | Biology | | E | Math | | F | Computer | | G | Math | | H | Math | | I | Math | +---------+------------+
Should output:
+---------+ | class | +---------+ | Math | +---------+
Note:
The students should not be counted duplicate in each course.
Answer:
select class
from courses
group by class
having count(distinct student) >= 59. 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 | +----+------------------+
Answer:
DELETE p1
FROM Person p1 inner join Person p2
on p1.Email = p2.Email AND p1.Id > p2.Id10. 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 | +---------------------+
Answer:
select max(Salary) as SecondHighestSalary
from Employee
where Salary != (select max(Salary) from Employee)11. 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 | +-------+------+
Answer:
select s1.Score,count(distinct s2.Score) as Rank
from Scores as s1 inner join Scores as s2
on s1.Score <= s2.Score
group by s1.Id
order by s1.Score desc12. Concecutive 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.
+-----------------+ | ConsecutiveNums | +-----------------+ | 1 | +-----------------+
Answer:
select distinct a.Num ConsecutiveNums
from logs a, logs b, logs c
where a.Id = b.Id - 1 and b.Id = C.Id - 1 and a.Num = b.Num and b.Num = c.num13. 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 | +------------+----------+--------+
Answer:
select d.Name as Department,e.Name as Employee, e.Salary
from Employee as e inner join Department as d
on e.DepartmentId = d.Id
where (e.DepartmentId,e.salary) in (select DepartmentId,max(Salary) from Employee group by DepartmentId)
