SQL Inner Join Tutorial with Example: Managers with at Least 5 Direct Reports


Given the following SQL Schema:

Create table If Not Exists Employee (Id int, Name varchar(255), Department varchar(255), ManagerId int)
Truncate table Employee
insert into Employee (Id, Name, Department, ManagerId) values ('101', 'John', 'A', 'None')
insert into Employee (Id, Name, Department, ManagerId) values ('102', 'Dan', 'A', '101')
insert into Employee (Id, Name, Department, ManagerId) values ('103', 'James', 'A', '101')
insert into Employee (Id, Name, Department, ManagerId) values ('104', 'Amy', 'A', '101')
insert into Employee (Id, Name, Department, ManagerId) values ('105', 'Anne', 'A', '101')
insert into Employee (Id, Name, Department, ManagerId) values ('106', 'Ron', 'B', '101')

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 	  |Department |ManagerId |
+------+----------+-----------+----------+
|101   |John 	  |A 	      |null      |
|102   |Dan 	  |A 	      |101       |
|103   |James 	  |A 	      |101       |
|104   |Amy 	  |A 	      |101       |
|105   |Anne 	  |A 	      |101       |
|106   |Ron 	  |B 	      |101       |
+------+----------+-----------+----------+

Given the Employee table, write a SQL query that finds out managers with at least 5 direct report. For the above table, your SQL query should return:

+-------+
| Name  |
+-------+
| John  |
+-------+

Note:
No one would report to himself.

How does SQL Inner-Join Work?

The SQL Inner-Join (or often referred to as Join) can be easily illustrated using the following Venn Diagram:

sql-joins-venn-diagrams-inner-join SQL Inner Join Tutorial with Example:  Managers with at Least 5 Direct Reports database mysql sql tutorial

sql-joins-venn-diagrams-inner-join

Two tables are joined and the result is the intersection of two tables.

select Name from Employee as A
inner join (
    select ManagerId
    from Employee
    group by ManagerId
    having count(1) >= 5           
) as B on A.ID = B.ManagerId

As we can see, two tables (queries) are joined using the keyword inner join (case insensitive) or join in short. You also need to specify the columns that connect two tables using syntax On.

SQL Sub Query

As of this problem, we can also use the sub query to solve.

select Name from
Employee where Id in (
 select ManagerId from Employee
 group by ManagerId
 having count(ManagerId) >= 5
)

This works slightly differently, the IDs from a query are returned and are used as inputs to another query (more or less like the pipe).

–EOF (The Ultimate Computing & Technology Blog) —

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