leetcode_sql_3,181,182,183
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)
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