﻿ SQL: Customers who holds a grade less than 300

# SQL Exercise: Customers who holds a grade less than 300

## SQL JOINS: Exercise-9 with Solution

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

Sample table: salesman

Sample Solution:

``````SELECT a.cust_name,a.city,a.grade,
b.name AS "Salesman", b.city
FROM customer a
LEFT OUTER JOIN salesman b
ON a.salesman_id=b.salesman_id
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:

## Query Visualization:

Duration:

Rows:

Cost:

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Previous SQL Exercise: Customer who works either through a salesman or by own.
Next SQL Exercise: Customer have placed no order or one or more orders.

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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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