﻿ SQL - Salesmen, customer involved in inventory management

# SQL Exercises: Salesmen, customer involved in inventory management

## SQL UNION: Exercise-3 with Solution

3. From the following tables, write a SQL query to find all those salespeople and customers who are involved in the inventory management system. Return salesperson ID, customer ID.

Sample table: orders

Sample table: customer

Sample Solution:

``````SELECT salesman_id, customer_id
FROM customer
UNION
(SELECT salesman_id, customer_id
FROM orders)
``````

Sample Output:

```salesman_id	customer_id
5005		3001
5007		3003
5001		3007
5002		3008
5006		3004
5003		3009
5001		3002
5002		3005
```

Code Explanation:

The said query in SQL which selects the distinct set of salesman IDs and customer IDs from two tables - customer and orders - and combines them using the UNION operator.
The UNION operator combines the results of two SELECT statements into a single table. It automatically removes any duplicate rows, so the resulting table will have a distinct set of salesman_id and customer_id pair.

Relational Algebra Expression:

Relational Algebra Tree:

## Query Visualization:

Duration:

Rows:

Cost:

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Previous SQL Exercise: Display distinct salesman and their cities.
Next SQL Exercise: Largest and smallest orders are produced on each date.

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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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