﻿ SQL: A list of which salesmen work for which customers

SQL Exercise: A list of which salesmen work for which customers

SQL JOINS: Exercise-3 with Solution

From the following tables write a SQL query to find the salesperson(s) and the customer(s) he represents. Return Customer Name, city, Salesman, commission.

Sample table: customer

Sample table: salesman

Sample Solution:

``````SELECT a.cust_name AS "Customer Name",
a.city, b.name AS "Salesman", b.commission
FROM customer a
INNER JOIN salesman b
ON a.salesman_id=b.salesman_id;
``````

Output of the Query:

```Customer Name	city		Salesman	commission
Nick Rimando	New York	James Hoog	0.15
Brad Davis	New York	James Hoog	0.15
Graham Zusi	California	Nail Knite	0.13
Julian Green	London		Nail Knite	0.13
Fabian Johnson	Paris		Mc Lyon		0.14
Geoff Cameron	Berlin		Lauson Hen	0.12
Jozy Altidor	Moscow		Paul Adam	0.13
Brad Guzan	London		Pit Alex	0.11
```

Explanation:

The said SQL query that is used to select specific columns from two tables, customer and salesman, and join them using the 'salesman_id' column. The query selects the 'cust_name' column from the customer table as 'Customer Name', the 'city' column from the customer table, the 'name' column from the salesman table as 'Salesman', and the 'commission' column from the salesman table.
The query uses an inner join, which only returns rows where there is a match in both tables on the specified join column.

Visual Explanation:

Query Visualization:

Duration:

Rows:

Cost:

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Previous SQL Exercise: Customers and their cities for given range of orders.
Next SQL Exercise: A salesperson who gets a commission of at least 12%.

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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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