SQL Challenges-1: Find those salespersons whose commission is less than ten thousand
SQL Challenges-1: Exercise-26 with Solution
From the following table, write a SQL query to find those salespersons whose commission is less than ten thousand. Return salesperson name, commission.
Input:
Table: salemast
Structure:
Field | Type | Null | Key | Default | Extra |
---|---|---|---|---|---|
salesman_id | int(11) | YES | |||
salesman_name | varchar(255) | YES | |||
yearly_sale | int(11) | YES |
Data:
salesman_id | salesman_name | yearly_sale |
---|---|---|
101 | Adam | 250000 |
103 | Mark | 100000 |
104 | Liam | 200000 |
102 | Evan | 150000 |
105 | Blake | 275000 |
106 | Noah | 50000 |
Table: commision
Structure:
Field | Type | Null | Key | Default | Extra |
---|---|---|---|---|---|
salesman_id | int(11) | YES | |||
commision_amt | int(11) | YES |
Data:
salesman_id | commision_amt |
---|---|
101 | 10000 |
103 | 4000 |
104 | 8000 |
102 | 6000 |
105 | 11000 |
Sample Solution:
SQL Code(MySQL):
CREATE TABLE salemast(salesman_id int, salesman_name varchar(255), yearly_sale int);
INSERT INTO salemast VALUES (101, 'Adam', 250000);
INSERT INTO salemast VALUES (103, 'Mark', 100000);
INSERT INTO salemast VALUES (104, 'Liam', 200000);
INSERT INTO salemast VALUES (102, 'Evan', 150000);
INSERT INTO salemast VALUES (105, 'Blake', 275000);
INSERT INTO salemast VALUES (106, 'Noah', 50000);
SELECT * FROM salemast;
CREATE TABLE commision (salesman_id int, commision_amt int);
INSERT INTO commision VALUES (101, 10000);
INSERT INTO commision VALUES (103, 4000);
INSERT INTO commision VALUES (104, 8000);
INSERT INTO commision VALUES (102, 6000);
INSERT INTO commision VALUES (105, 11000);
SELECT * FROM commision;
SELECT s.salesman_name,c.commision_amt
FROM salemast s LEFT JOIN
commision c
ON s.salesman_id=c.salesman_id
WHERE c.commision_amt<10000;
Sample Output:
salesman_name|commision_amt| -------------|-------------| Mark | 4000| Liam | 8000| Evan | 6000|
SQL Code Editor:
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