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SQL Exercise: Salesmen who works either for one or more customer

SQL JOINS: Exercise-13 with Solution

From the following tables write a SQL query to list all salespersons along with customer name, city, grade, order number, date, and amount. Condition for selecting list of salesmen : 1. Salesmen who works for one or more customer or, 2. Salesmen who not yet join under any customer, Condition for selecting list of customer : 3. placed one or more orders, or 4. no order placed to their salesman.

Sample table: customer


Sample table: salesman


Sample table: orders


Sample Solution:

SELECT a.cust_name,a.city,a.grade, 
b.name AS "Salesman", 
c.ord_no, c.ord_date, c.purch_amt 
FROM customer a 
RIGHT OUTER JOIN salesman b 
ON b.salesman_id=a.salesman_id 
RIGHT OUTER JOIN orders c 
ON c.customer_id=a.customer_id;

Output of the Query:

cust_name	city		grade	Salesman	ord_no	ord_date	purch_amt
Brad Guzan	London			Pit Alex	70009	2012-09-10	270.65
Nick Rimando	New York	100	James Hoog	70002	2012-10-05	65.26
Geoff Cameron	Berlin		100	Lauson Hen	70004	2012-08-17	110.50
Brad Davis	New York	200	James Hoog	70005	2012-07-27	2400.60
Nick Rimando	New York	100	James Hoog	70008	2012-09-10	5760.00
Fabian Johnson	Paris		300	Mc Lyon		70010	2012-10-10	1983.43
Geoff Cameron	Berlin		100	Lauson Hen	70003	2012-10-10	2480.40
Jozy Altidor	Moscow		200	Paul Adam	70011	2012-08-17	75.29
Nick Rimando	New York	100	James Hoog	70013	2012-04-25	3045.60
Graham Zusi	California	200	Nail Knite	70001	2012-10-05	150.50
Graham Zusi	California	200	Nail Knite	70007	2012-09-10	948.50
Julian Green	London		300	Nail Knite	70012	2012-06-27	250.45

Explanation:

The said SQL query is performing a right outer join on three tables: the customer table alias a, the salesman table alias b, and the orders table alias c.
First, it performs a right outer join on the customer table and the salesman table using the 'salesman_id' column.
Then it performs another right outer join on the result of the first join and the orders table using the 'customer_id' column.
It is then selecting the 'cust_name', 'city', 'grade', 'name' as 'Salesman', 'ord_no', 'ord_date', and 'purch_amt' columns from the three tables.
This query will select all the rows from orders table and any matching rows from customer table and salesman table. If there is no match, it will return NULL for the non-matching columns of customer table and salesman table.

Visual Explanation:

Result of a list for the salesmen who works either for one or more customer or not yet join under any of the customer who placed either one or more orders or no order to their supplier
Result of a list for the salesmen who works either for one or more customer or not yet join under any of the customer who placed either one or more orders or no order to their supplier

Practice Online


Query Visualization:

Duration:

Query visualization of Display the list for the salesmen who works either for one or more customer or not yet join under any of the customers who placed either one or more orders or no order to their supplier - Duration

Rows:

Query visualization of Display the list for the salesmen who works either for one or more customer or not yet join under any of the customers who placed either one or more orders or no order to their supplier - Rows

Cost:

Query visualization of Display the list for the salesmen who works either for one or more customer or not yet join under any of the customers who placed either one or more orders or no order to their supplier - Cost

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Previous SQL Exercise: Salesmen works either for one or more customer or not.
Next SQL Exercise: Salesmen work for one or more customers or yet to join.

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SQL: Tips of the Day

Grouped LIMIT in PostgreSQL: Show the first N rows for each group?

db=# SELECT * FROM xxx;
 id | section_id | name
----+------------+------
  1 |          1 | A
  2 |          1 | B
  3 |          1 | C
  4 |          1 | D
  5 |          2 | E
  6 |          2 | F
  7 |          3 | G
  8 |          2 | H
(8 rows)

I need the first 2 rows (ordered by name) for each section_id, i.e. a result similar to:

id | section_id | name
----+------------+------
  1 |          1 | A
  2 |          1 | B
  5 |          2 | E
  6 |          2 | F
  7 |          3 | G
(5 rows)

PostgreSQL v9.3 you can do a lateral join

select distinct t_outer.section_id, t_top.id, t_top.name from t t_outer
join lateral (
    select * from t t_inner
    where t_inner.section_id = t_outer.section_id
    order by t_inner.name
    limit 2
) t_top on true
order by t_outer.section_id;

Database: PostgreSQL

Ref: https://bit.ly/3AfYwZI

 





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