查询多个MySQL表并将数据编码为JSON

Question

I'll admit, I don't know tons about MySQL queries or JSON for that matter but I would like to. Currently when I get data from my database I query one table which brings values into another query and so on. I don't think this is the best practice to go about receiving all the data I need to display. Below is an example of how my MySQL queries currently work in a PHP foreach function:

So, is there a better way of completing the query(s) I want and how would I be able to encode it as JSON?

Any help would be appreciated, thank you.

Answer 1

From what I understand, your code is doing something like this:

<?php
$friends = query("SELECT Friends");
while($row = fetch_object($friends)){
    $friend_dets = query("SELECT Friend_dets WHERE Friend_ID = $row->Friend_ID");
    $output[] = fetch_assoc($friend_dets);
}
echo json_encode($output);
?>

With this, you're making the process more complex than it needs to be. You can get all of the information you need with one query with a JOIN like this:

SELECT Name, Status, WhateverElseYouWant
FROM Friends
    JOIN Profiles ON (Friends.friend-profile-id = Profiles.profile-id)
WHERE Friends.profile-id = MyCurrentProfileID

That will give you the name, status and whatever else of everyone who is friends with MyCurrentProfileID . Then, you just need to put the result in an array and json_encode it:

<?php
$friends = query($QueryFromAbove);
while($row = fetch_object($friends)){
    $output[] = fetch_assoc($friend_dets);
}
echo json_encode($output);
?>

Answer 2

You can make 1 SQL query for this list, can you tell me whats your table Friends and Profiles info.

In Friends table you must have id to profiles to indicate is the friens of what profile.

The query may be like : select profile.* from profile, friends, where friends.id_profile = profile.id and profile.id = current-profile-id

In your graph you use current profile id in friends table and not profile this may be error