Joining 2 tables is not working correctly - php

Question

Idk what i am doing wrong.I am still new to php and mysqli and wanted to know what I am doing wrong.What I have on my 1st list are name of the food,description and what kind of food they are.On my 2nd list I have the ratings of the foods.
TableImg
So lets say for example I click on korean food all the foods with the cuisine_type = korean will be displayed on my page. When mysqli_fetch_array($query_type) it displays kimchi and ribs but when i replace $query_fetch it only displays the first one which is kimchi.My goal is to get all the food from cuisine_type display correctly and their ratings attached to them.
<?php
if(isset($_GET['type'])){
$type = (string)$_GET['type'];
}
(Displays only 1 of them)-> $query_fetch = mysqli_query($connect,"SELECT
cuisine.*,AVG(ratings.cuisine_rating) AS rt FROM cuisine
LEFT JOIN ratings ON cuisine.dish_name = ratings.cuisine_name WHERE
cuisine_type = '$type'");
(Displays everything)-> $query_type = mysqli_query($connect,"SELECT * FROM cuisine WHERE
cuisine_fetch ='$type'");
while($show = mysqli_fetch_array($query_type)){
echo "
<div class='box'>
<img class='foodimg' src='dish/".$show['dish_img'].".jpg'>
<figcaption class='title'>".$show['dish_name']."</figcaption>
<div class='text'>".$show['text']."</div>
</div>
</div>";
}
?>

Answer 1

What you want is to do your average using GROUPing.
SELECT c.*, avg(r.rating) AS rt FROM cuisine c
LEFT JOIN ratings r
ON c.dish_name = r.dish_name
WHERE c.cuisine_type = 'korean'
GROUP BY dish_name
The code can be tested via SQLFiddle.
You will need to replace where i hard coded korean with your php variable to make it work in your system.
The issue you had was you didn't specify how to group your rows, this caused the server to put them into all one group. Most DBMS would have errored when you tried to select columns not in the grouping (which was no columns) without an aggregate (like avg), though it appears yours does not, it allows you to group this way and simply return the first row for every non aggregate.
NOTE you still may need some kind of null handling for avg(r.rating), since Korean Ribs has no ratings, it will return an average rating of NULL. You can handle this by using IFNULL like this: IFNULL(avg(r.rating),2.5) AS rt. I would suggest using maybe 2.5 if rating is out of 5, so your no rating is the middle number, though you can use 0 or 5 or anything.

Answer 2

Assuming the TableImg is correct, you're not using the same names in the query as in the db.
SELECT
cuisine.*,
AVG(ratings.cuisine_rating) AS rt
FROM cuisine
LEFT JOIN ratings ON
cuisine.dish_name = ratings.cuisine_name
WHERE
cuisine_type = '$type'
Use this instead:
SELECT
cuisine.*,
AVG(ratings.Rating) AS rt
FROM cuisine
LEFT JOIN ratings ON
cuisine.dish_name = ratings.dish_name
WHERE
cuisine_type = '$type'