SQL Challenges-1: Employees who are earning more salary than their managers

SQL Challenges-1: Exercise-68 with Solution

From the following table write a query in SQL to find the employees who are earning more salary than their managers. Return employee ID, name, and salary, manager ID, manager name and the salary of the manager.

Table: employees

Structure:

Data:

Sample Solution:

SQL Code(MySQL):

create table employees (
employee_id integer(4) not null unique,
emp_name varchar(25),
hire_date date,
job_id varchar(25),
salary decimal(10,2),
manager_id integer(4),
department_id integer(4)
);



insert into employees values( 100,'Steven     ','1987-06-17','AD_PRES   ',24000.00,         0,   90);
insert into employees values( 101,'Neena      ','1987-06-18','AD_VP     ',17000.00,       100,   90);
insert into employees values( 102,'Lex        ','1987-06-19','AD_VP     ',17000.00,       100,   90);
insert into employees values( 103,'Alexander  ','1987-06-20','IT_PROG   ', 9000.00,       102,   60);
insert into employees values( 104,'Bruce      ','1987-06-21','IT_PROG   ',11000.00,       103,   60);
insert into employees values( 105,'David      ','1987-06-22','IT_PROG   ', 4800.00,       103,   60);
insert into employees values( 106,'Valli      ','1987-06-23','IT_PROG   ', 4800.00,       103,   60);
insert into employees values( 107,'Diana      ','1987-06-24','IT_PROG   ', 4200.00,       103,   60);
insert into employees values( 114,'Den        ','1987-07-01','PU_MAN    ',11000.00,       100,   30);
insert into employees values( 115,'Alexander  ','1987-07-02','PU_CLERK  ', 3100.00,       114,   30);
insert into employees values( 116,'Shelli     ','1987-07-03','PU_CLERK  ', 2900.00,       114,   30);
insert into employees values( 117,'Sigal      ','1987-07-04','PU_CLERK  ', 2800.00,       114,   30);
insert into employees values( 108,'Nancy      ','1987-06-25','FI_MGR    ',12000.00,       101,  100);
insert into employees values( 109,'Daniel     ','1987-06-26','FI_ACCOUNT', 9000.00,       108,  100);
insert into employees values( 110,'John       ','1987-06-27','FI_ACCOUNT', 8200.00,       108,  100);
insert into employees values( 111,'Ismael     ','1987-06-28','FI_ACCOUNT', 7700.00,       108,  100);
insert into employees values( 112,'Jose Manuel','1987-06-29','FI_ACCOUNT', 7800.00,       108,  100);
insert into employees values( 113,'Luis       ','1987-06-30','FI_ACCOUNT', 6900.00,       108,  100);
insert into employees values( 133,'Jason      ','1987-07-20','ST_CLERK  ', 3300.00,       122,   50);
insert into employees values( 134,'Michael    ','1987-07-21','ST_CLERK  ', 2900.00,       122,   50);
insert into employees values( 135,'Ki         ','1987-07-22','ST_CLERK  ', 2400.00,       122,   50);
insert into employees values( 136,'Hazel      ','1987-07-23','ST_CLERK  ', 2200.00,       122,   50);
insert into employees values( 137,'Renske     ','1987-07-24','ST_CLERK  ', 3600.00,       123,   50);
insert into employees values( 138,'Stephen    ','1987-07-25','ST_CLERK  ', 3200.00,       123,   50);
insert into employees values( 139,'John       ','1987-07-26','ST_CLERK  ', 2700.00,       123,   50);



SELECT a.employee_id,
    a.emp_name,
    a.salary,
    a.manager_id,
    b.emp_name AS "Manager Name",
    b.salary AS "Manager Salary"
    FROM
    employees a INNER JOIN employees b
    ON a.manager_id = b.employee_id
WHERE
    a.salary > b.salary;

Sample Output:

mployee_id|emp_name   |salary  |manager_id|Manager Name|Manager Salary|
----------+-----------+--------+----------+------------+--------------+
       104|Bruce      |11000.00|       103|Alexander   |       9000.00|

SQL Code Editor:

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