Select locations from two tables - mysql

Question

I am new to SQL and I'm having difficulties writing the following query.
Scenario
A user has two addresses, home address (App\User) and listing address (App\Listing). When a visitor searches for listings for a Suburb or postcode or state, if the user's listing address does not match - but if home address does match - they will be in the search result too.
For example: if a visitor searches for Melbourne, I want to include listings from Melbourne and also the listings for the users who have an address in Melbourne.
Expected output:
user_id first_name email suburb postcode state
1 Mathew mathew.afsd#gmail.com Melbourne 3000 VIC
2 Zammy Zamm#xyz.com Melbourne 3000 VIC
Tables
users:
id first_name email
1 Mathew mathew.afsd#gmail.com
2 Zammy Zamm#xyz.com
3 Tammy tammy#unknown.com
4 Foo foo#hotmail.com
5 Bar bar#jhondoe.com.au
listings:
id user_id hourly_rate description
1 1 30 ABC
2 2 40 CBD
3 3 50 XYZ
4 4 49 EFG
5 5 10 Efd
addresses:
id addressable_id addressable_type post_code suburb state latitude longitude
3584 1 App\\User 2155 Rouse Hill NSW -33.6918372 150.9007221
3585 2 App\\User 3000 Melbourne VIC -33.6918372 150.9007221
3586 3 App\\User 2000 Sydney NSW -33.883123 151.245969
3587 4 App\\User 2008 Chippendale NSW -33.8876392 151.2011224
3588 5 App\\User 2205 Wolli Creek NSW -33.935259 151.156301
3591 1 App\\Listing 3000 Melbourne VIC -37.773923 145.12385
3592 2 App\\Listing 2030 Vaucluse NSW -33.858935 151.2784079
3597 3 App\\Listing 4000 Brisbane QLD -27.4709331 153.0235024
3599 4 App\\Listing 2000 Sydney NSW -33.91741 151.231307
3608 5 App\\Listing 2155 Rouse Hill NSW -33.863464 151.271504

Answer 1

Try this. You can check it here.
SELECT l.*
FROM listings l
LEFT JOIN addresses a_l ON a_l.addressable_id = l.id
AND a_l.addressable_type = "App\\Listing"
AND a_l.suburb = "Melbourne"
LEFT JOIN addresses a_u ON a_u.addressable_id = l.user_id
AND a_u.addressable_type = "App\\User"
AND a_u.suburb = "Melbourne"
WHERE a_l.id IS NOT NULL OR a_u.id IS NOT NULL

Answer 2

As per my understanding of your question, for any suburb - supplied by a visitor, you want to include all the listings where either User's address is the same as the suburb supplied or the Listing's address is the same as the suburb supplied.
Assuming addressable_id column is related to Id of Users table and Listings table, based on value in addressable_type column, you can use the following query to join and get the desired result:
Select l.*
From Listings l
inner join Addresses a on ((a.addressable_id = l.user_Id and a.addressable_type = 'App\\User') or (a.addressable_id = l.Id and a.addressable_type = 'App\\Listings'))
inner join Addresses a1 On a1.addressable_id = a.addressable_id and a1.Suburb = 'Melbourne'

Answer 3

try this,
SELECT
a.addressable_id AS `userid`,
b.first_name AS `username`
FROM
addresses AS a JOIN users AS b ON a.addressable_id=b.id
WHERE
a.suburb = 'Melbourne';
if < addressable_id > has relation with < id > in listing table,
SELECT
a.addressable_id AS `userid`,
b.first_name AS `username`
FROM
addresses AS a JOIN users AS b ON a.addressable_id=b.id AND addressable_type='App\\User'
WHERE
a.suburb = 'Melbourne'
UNION
SELECT
b.user_id AS `userid`,
c.first_name AS `username`
FROM
addresses AS a JOIN listings AS b ON a.addressable_id=b.id AND addressable_type='App\\Listing'
JOIN users AS c ON b.user_id=c.id
WHERE
a.suburb = 'Melbourne';