SQL Exercise: Customers who holds a grade less than 300

Last update on April 22 2025 10:56:45 (UTC/GMT +8 hours)

9. Customers with Grade Less Than 300

From the following tables write a SQL query to find those customers with a grade less than 300. Return cust_name, customer city, grade, Salesman, salesmancity. The result should be ordered by ascending customer_id.

Sample table: customer

 customer_id |   cust_name    |    city    | grade | salesman_id 
-------------+----------------+------------+-------+-------------
        3002 | Nick Rimando   | New York   |   100 |        5001
        3007 | Brad Davis     | New York   |   200 |        5001
        3005 | Graham Zusi    | California |   200 |        5002
        3008 | Julian Green   | London     |   300 |        5002
        3004 | Fabian Johnson | Paris      |   300 |        5006
        3009 | Geoff Cameron  | Berlin     |   100 |        5003
        3003 | Jozy Altidor   | Moscow     |   200 |        5007
        3001 | Brad Guzan     | London     |       |        5005

Sample table: salesman

 salesman_id |    name    |   city   | commission 
-------------+------------+----------+------------
        5001 | James Hoog | New York |       0.15
        5002 | Nail Knite | Paris    |       0.13
        5005 | Pit Alex   | London   |       0.11
        5006 | Mc Lyon    | Paris    |       0.14
        5007 | Paul Adam  | Rome     |       0.13
        5003 | Lauson Hen | San Jose |       0.12

Sample Solution:

-- Selecting specific columns from the 'customer' and 'salesman' tables
SELECT a.cust_name, a.city, a.grade, 
       b.name AS "Salesman", b.city 
-- Specifying the tables to retrieve data from ('customer' as 'a' and 'salesman' as 'b')
FROM customer a 
-- Performing a left outer join based on the salesman_id, including unmatched rows from 'customer'
LEFT OUTER JOIN salesman b 
ON a.salesman_id = b.salesman_id 
-- Filtering the results based on the condition that 'grade' is less than 300
WHERE a.grade < 300 
-- Sorting the result set by customer_id in ascending order
ORDER BY a.customer_id;

Output of the Query:

cust_name	city		grade	Salesman	city
Nick Rimando	New York	100	James Hoog	New York
Jozy Altidor	Moscow		200	Paul Adam	Rome
Graham Zusi	California	200	Nail Knite	Paris
Brad Davis	New York	200	James Hoog	New York
Geoff Cameron	Berlin		100	Lauson Hen	San Jose

Explanation:

The said SQL query is selecting the customer name, city, and grade from the customer table aliased as a and the salesman name and city from the salesman table aliased as b.
It is joining these tables on the 'salesman_id' column and only selecting the customers whose grade is less than 300.
The results are ordered by the 'customer_id' column.
Additionally, it is using a LEFT OUTER JOIN which will retrieve all records from the left table and the matching records from the right table. If no match is found on the right table, it will return NULL for the right table's fields.

Visual Explanation:

Practice Online

Query Visualization:

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For more Practice: Solve these Related Problems:

• Write a SQL query to find customers with a grade less than 300 and display customer name, city, grade along with their salesperson’s name and city, sorted by customer_id.
• Write a SQL query to list customers whose grade is under 300, joining with the salesman table to include the salesperson’s details, and ordering by customer_id.
• Write a SQL query to retrieve details of customers with grade below 300 and their associated salesperson information, filtering out records with NULL grade.
• Write a SQL query to display customer name, city, grade, and corresponding salesperson details for customers with grade less than 300, ordered by ascending customer_id.

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