SQL Exercises: Customer with more than one current order

Last update on January 12 2026 12:47:47 (UTC/GMT +8 hours)

9. From the following table, write a SQL query to find those salespersons and customers who have placed more than one order. Return ID, name.

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 table: orders

ord_no      purch_amt   ord_date    customer_id  salesman_id
----------  ----------  ----------  -----------  -----------
70001       150.5       2012-10-05  3005         5002
70009       270.65      2012-09-10  3001         5005
70002       65.26       2012-10-05  3002         5001
70004       110.5       2012-08-17  3009         5003
70007       948.5       2012-09-10  3005         5002
70005       2400.6      2012-07-27  3007         5001
70008       5760        2012-09-10  3002         5001
70010       1983.43     2012-10-10  3004         5006
70003       2480.4      2012-10-10  3009         5003
70012       250.45      2012-06-27  3008         5002
70011       75.29       2012-08-17  3003         5007
70013       3045.6      2012-04-25  3002         5001

Sample Solution:

-- Selecting specific columns (customer_id as "ID", cust_name as "NAME") from the 'customer' table (aliased as 'a')
SELECT customer_id as "ID", cust_name as "NAME"

-- Filtering rows in the 'customer' table (aliased as 'a') where the count of associated orders is greater than 1
FROM customer a
WHERE 1 <
    (SELECT COUNT(*)
     FROM orders b
     WHERE a.customer_id = b.customer_id)

-- Performing a UNION operation with the result set of a subquery
UNION

-- Selecting specific columns (salesman_id as "ID", name as "NAME") from the 'salesman' table (aliased as 'a')
SELECT salesman_id as "ID", name as "NAME"

-- Filtering rows in the 'salesman' table (aliased as 'a') where the count of associated orders is greater than 1
FROM salesman a
WHERE 1 <
    (SELECT COUNT(*)
     FROM orders b
     WHERE a.salesman_id = b.salesman_id)

-- Ordering the result set based on the second column ("NAME")
ORDER BY 2

Sample Output:

ID	        NAME
3009		Geoff Cameron
3005		Graham Zusi
5001		James Hoog
5003		Lauson Hen
5002		Nail Knite
3002		Nick Rimando

Code Explanation:

The said query in SQL which selects customer_id and cust_name columns from the customer table and renames them as "ID" and "NAME" respectively.
The subquery checks whether at least one order placed by each customer or not. If the count of orders for a customer is greater than 1, then that customer is included in the results.
The UNION operator is used to combine the results of this query with another query that selects salesman_id and name columns from the salesman table, renames them as "ID" and "NAME" respectively, and checks if there is at least one order placed by each salesman. The final result is then ordered by the "NAME" column.

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