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.
# Write your MySQL query statement below
SELECT A.Name Employee
FROM Employee A,Employee B
WHERE A.ManagerId=B.Id AND A.Salary >B.Salary;
182. 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 | +---------+
Note: All emails are in lowercase.
# Write your MySQL query statement below
SELECT Distinct(a.Email)
FROM Person a,Person b
WHERE a.Id<>b.Id and a.Email=b.Email
or
select Email
from Person
group by Email
having count(*) > 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 | +-----------+
SELECT A.Name from Customers A LEFT JOIN Orders B on a.Id = B.CustomerId WHERE b.CustomerId is NULL;
select c.Name from Customers c
where (select count(*) from Orders o where o.customerId=c.id)=0
select c.Name from Customers c
where not exists (select * from Orders o where o.customerId=c.id)