SQL Exercises: Customer lives in a city other than the salesman's
SQL Query on Multiple Tables: Exercise-3 with Solution
From the following tables, write a SQL query to find those salespeople who generated orders for their customers but are not located in the same city. Return ord_no, cust_name, customer_id (orders table), salesman_id (orders table).
Sample table: salesman
Sample table: customer
Sample table: orders
Sample Solution:
SELECT ord_no, cust_name, orders.customer_id, orders.salesman_id
FROM salesman, customer, orders
WHERE customer.city <> salesman.city
AND orders.customer_id = customer.customer_id
AND orders.salesman_id = salesman.salesman_id;
Output of the query:
ord_no cust_name customer_id salesman_id 70004 Geoff Cameron 3009 5003 70003 Geoff Cameron 3009 5003 70011 Jozy Altidor 3003 5007 70001 Graham Zusi 3005 5002 70007 Graham Zusi 3005 5002 70012 Julian Green 3008 5002
Code Explanation:
The said query in SQL that joins the 'salesman', 'customer', and 'orders' tables. The result set includes the order number (ord_no), customer name (cust_name), customer ID (customer_id), and salesman ID (salesman_id). The WHERE clause specifies multiple conditions for the join.
There is a first condition that must be fulfilled in order for the city column in the customer table to not be the same as the city column in the salesman table.
The second and third conditions specify the join conditions between the orders table and the customer and salesman tables, respectively, which is that the customer_id and salesman_id columns of the orders table must be equal to the corresponding columns in the customer and salesman tables.
Relational Algebra Expression:

Relational Algebra Tree:

Explanation:

Visual presentation :

Practice Online
Query Visualization:
Duration:

Rows:

Cost:

Note: The pictorial represetation above is based on hypothetical table for the purpose of explanation only. Your answer may not match.
Have another way to solve this solution? Contribute your code (and comments) through Disqus.
Previous SQL Exercise: Customers along with the salesmen who works for them.
Next SQL Exercise: Find out customers who made the order.
What is the difficulty level of this exercise?
Test your Programming skills with w3resource's quiz.
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
- Weekly Trends
- Python Interview Questions and Answers: Comprehensive Guide
- Scala Exercises, Practice, Solution
- Kotlin Exercises practice with solution
- MongoDB Exercises, Practice, Solution
- SQL Exercises, Practice, Solution - JOINS
- Java Basic Programming Exercises
- SQL Subqueries
- Adventureworks Database Exercises
- C# Sharp Basic Exercises
- SQL COUNT() with distinct
- JavaScript String Exercises
- JavaScript HTML Form Validation
- Java Collection Exercises
- SQL COUNT() function
- SQL Inner Join
We are closing our Disqus commenting system for some maintenanace issues. You may write to us at reach[at]yahoo[dot]com or visit us at Facebook