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:
Field | Type | Null | Key | Default | Extra |
---|---|---|---|---|---|
employee_id | int | NO | PRI | ||
emp_name | varchar(25) | YES | |||
hire_date | date | YES | |||
job_id | varchar(25) | YES | |||
salary | decimal(10,2) | YES | |||
manager_id | int | YES | |||
department_id | int | YES |
Data:
employee_id | emp_name | hire_date | job_id | salary | manager_id | department_id |
---|---|---|---|---|---|---|
100 | Steven | 1987-06-17 | AD_PRES | 24000.00 | 0 | 90 |
101 | Neena | 1987-06-18 | AD_VP | 17000.00 | 100 | 90 |
102 | Lex | 1987-06-19 | AD_VP | 17000.00 | 100 | 90 |
103 | Alexander | 1987-06-20 | IT_PROG | 9000.00 | 102 | 60 |
104 | Bruce | 1987-06-21 | IT_PROG | 6000.00 | 103 | 60 |
105 | David | 1987-06-22 | IT_PROG | 4800.00 | 103 | 60 |
106 | Valli | 1987-06-23 | IT_PROG | 4800.00 | 103 | 60 |
107 | Diana | 1987-06-24 | IT_PROG | 4200.00 | 103 | 60 |
108 | Nancy | 1987-06-25 | FI_MGR | 12000.00 | 101 | 100 |
109 | Daniel | 1987-06-26 | FI_ACCOUNT | 9000.00 | 108 | 100 |
110 | John | 1987-06-27 | FI_ACCOUNT | 8200.00 | 108 | 100 |
111 | Ismael | 1987-06-28 | FI_ACCOUNT | 7700.00 | 108 | 100 |
112 | Jose Manuel | 1987-06-29 | FI_ACCOUNT | 7800.00 | 108 | 100 |
113 | Luis | 1987-06-30 | FI_ACCOUNT | 6900.00 | 108 | 100 |
114 | Den | 1987-07-01 | PU_MAN | 11000.00 | 100 | 30 |
115 | Alexander | 1987-07-02 | PU_CLERK | 3100.00 | 114 | 30 |
116 | Shelli | 1987-07-03 | PU_CLERK | 2900.00 | 114 | 30 |
117 | Sigal | 1987-07-04 | PU_CLERK | 2800.00 | 114 | 30 |
133 | Jason | 1987-07-20 | ST_CLERK | 3300.00 | 122 | 50 |
134 | Michael | 1987-07-21 | ST_CLERK | 2900.00 | 122 | 50 |
135 | Ki | 1987-07-22 | ST_CLERK | 2400.00 | 122 | 50 |
136 | Hazel | 1987-07-23 | ST_CLERK | 2200.00 | 122 | 50 |
137 | Renske | 1987-07-24 | ST_CLERK | 3600.00 | 123 | 50 |
138 | Stephen | 1987-07-25 | ST_CLERK | 3200.00 | 123 | 50 |
139 | John | 1987-07-26 | ST_CLERK | 2700.00 | 123 | 50 |
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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Previous: Find the highest salaried employees for each designation.
Next: Find cost percent for each order of total order amount.
SQL: Tips of the Day
SQL Server SELECT into existing table.
INSERT INTO dbo.TABLETWO SELECT col1, col2 FROM dbo.TABLEONE WHERE col3 LIKE @search_key
This assumes there's only two columns in dbo.TABLETWO - you need to specify the columns otherwise:
INSERT INTO dbo.TABLETWO (col1, col2) SELECT col1, col2 FROM dbo.TABLEONE WHERE col3 LIKE @search_key
Database: SQL Server
Ref: https://bit.ly/3y6tpA3
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