Average and comparative left join MySQL - php

Question

I'm trying to fetch information from 2 different MySQL tables.
The primary table is this one:
From this one I use the information to get the average rate from this:
How can I write an SQL query that will get me the average rating by counting all the rows with rating_house = house_id, and sort it by highest rating and if equal ratings, the one with the most rates.
This is what I have come up with myself:
$sql = "SELECT l.location_address, "
. "r.rating_structure+r.rating_inventory+r.rating_service/3 AS average "
. "FROM houses h "
. "LEFT JOIN rating r ON h.house_id = r.rating_house "
. "LEFT JOIN location l ON h.house_address = l.location_id "
. "WHERE h.house_deleted IS NULL SORT BY average DESC LIMIT 10";
$result = $db_connect->prepare($sql);
if($result->execute()){
while($user_data = $result->fetch(PDO::FETCH_ASSOC)){
$user_data['location_address']."<br>";
}
}
However I get no output?

Answer 1

For performance use INNER JOINs instead of LEFT JOINs if you know that the related rows must exist.
SELECT l.location_address,
(SUM(r.rating_structure)+SUM(r.rating_inventory)+SUM(r.rating_service))/3 AS average
FROM houses h
INNER JOIN rating r ON h.house_id = r.rating_house
INNER JOIN location l ON h.house_address = l.location_id
WHERE h.house_deleted IS NULL
GROUP BY l.location_address
ORDER BY 2,
(SUM(r.rating_structure)+SUM(r.rating_inventory)+SUM(r.rating_service)) DESC
LIMIT 10

Answer 2

Below query should return houses sorted by average ratings:
select h.house_id, (sum(hr.rating_structure) + sum(rating_inventory) + sum(rating_service))/3 as "average"
from house h left outer join house_rating hr on h.house_id = hr.rating_house
group by h.house_id
order by average