SQL Challenges-1: Department where highest salaried employee(s) are working
SQL Challenges-1: Exercise-75 with Solution
From the following tables write a query in SQL to find the department where the highest salaried employee(s) are working. Return department name and highest salary to this department.
Table: departments
Structure:
Field | Type | Null | Key | Default | Extra |
---|---|---|---|---|---|
department_id | int | NO | PRI | ||
department_name | varchar(30) | YES |
Data:
department_id | department_name |
---|---|
30 | Sales |
50 | Export |
60 | Marketing |
80 | Audit |
90 | Production |
100 | Administration |
Table: employees
Structure:
Field | Type | Null | Key | Default | Extra |
---|---|---|---|---|---|
emp_id | int | NO | PRI | ||
emp_name | varchar(30) | YES | |||
emp_sex | varchar(1) | YES | |||
emp_salary | int | YES | |||
emp_department | int | YES | MUL |
Data:
emp_id | emp_name | emp_sex | emp_salary | emp_department |
---|---|---|---|---|
100 | Steven | M | 24000 | 90 |
101 | Neena | F | 17000 | 90 |
102 | Lex | M | 17000 | 80 |
103 | Alexander | M | 9000 | 60 |
104 | Bruce | M | 6000 | 60 |
105 | David | M | 4800 | 80 |
106 | Valli | F | 4800 | 60 |
107 | Diana | F | 4200 | 60 |
108 | Nancy | M | 12000 | 100 |
109 | Daniel | F | 9000 | 100 |
110 | John | M | 8200 | 100 |
111 | Ismael | M | 7700 | 100 |
112 | Jose Manuel | M | 7800 | 100 |
113 | Luis | F | 6900 | 100 |
114 | Den | M | 11000 | 30 |
115 | Alexander | M | 3100 | 30 |
116 | Shelli | F | 2900 | 30 |
117 | Sigal | F | 2800 | 30 |
133 | Jason | M | 3300 | 50 |
134 | Michael | F | 2900 | 50 |
135 | Ki | F | 2400 | 50 |
Sample Solution:
SQL Code(MySQL):
create table departments (
department_id integer(4) not null unique,
department_name varchar(30));
insert into departments values(30,'Sales');
insert into departments values( 50,'Export ');
insert into departments values( 60,'Marketing ');
insert into departments values( 80,'Audit ');
insert into departments values( 90,'Production ');
insert into departments values(100,'Administration');
create table employees (
emp_id integer(4) not null unique,
emp_name varchar(30),
emp_sex varchar(1),
emp_salary int(6),
emp_department int(3),
foreign key(emp_department) references departments(department_id));
insert into employees values(100,'Steven ','M',24000, 90);
insert into employees values(101,'Neena ','F',17000, 90);
insert into employees values(102,'Lex ','M',17000, 80);
insert into employees values(103,'Alexander ','M',9000, 60);
insert into employees values(104,'Bruce ','M',6000, 60);
insert into employees values(105,'David ','M',4800, 80);
insert into employees values(106,'Valli ','F',4800, 60);
insert into employees values(107,'Diana ','F',4200, 60);
insert into employees values(114,'Den ','M',11000, 30);
insert into employees values(115,'Alexander ','M',3100, 30);
insert into employees values(116,'Shelli ','F',2900, 30);
insert into employees values(117,'Sigal ','F',2800, 30);
insert into employees values(108,'Nancy ','M',12000, 100);
insert into employees values(109,'Daniel ','F',9000, 100);
insert into employees values(110,'John ','M',8200, 100);
insert into employees values(111,'Ismael ','M',7700, 100);
insert into employees values(112,'Jose Manuel','M',7800, 100);
insert into employees values(113,'Luis ','F',6900, 100);
insert into employees values(133,'Jason ','M',3300, 50);
insert into employees values(134,'Michael ','F',2900, 50);
insert into employees values(135,'Ki ','F',2400, 50);
SELECT departments.department_name,employees.emp_salary
FROM employees
JOIN departments
ON employees.emp_department = departments.department_id
WHERE employees.emp_salary IN
(SELECT MAX(employees.emp_salary) FROM employees);
Sample Output:
department_name|emp_salary| ---------------+----------+ Production | 24000|
SQL Code Editor:
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