Show selected data as array value instead of string - php

Question

I'm trying to put data from my database into seperate arrays within another array. This works but when I'm trying to fetch the 'user_id' information, it only shows one number so it works like a string. How can I get it to work like an array and get the entire user_id?
$fetch = mysqli_query($con, "SELECT * FROM spotify_userdata");
$return_arr = [];
while ($row = mysqli_fetch_array($fetch, MYSQL_ASSOC)) {
$return_arr[] = array(
$row_array['user_id'] = $row['user_id'],
$row_array['name'] = $row['name'],
$row_array['artists'] = $row['artists'],
);
}
$user = json_encode($return_arr[0]);
echo $user[2];
This code returns 1 so it show the third number of the user_id. How can I get it to show the entire user_id like this: 111434343

Answer 1

You have many things in your code that's wrong:
Remove the last array item's comma
Change
$return_arr = [];
To
$return_arr = array();
3.Add:
$row_array = array()
at the begginning of all that code
At the end your code must be like this:
$fetch = mysqli_query($con, "SELECT * FROM spotify_userdata");
$row_array = array();
while ($row = mysqli_fetch_array($fetch, MYSQL_ASSOC)) {
$return_arr = array(
$row_array['user_id'] = $row['user_id'],
$row_array['name'] = $row['name'],
$row_array['artists'] = $row['artists'],
);
}
$user = json_encode($return_arr[0]);

Answer 2

According to your code you should use:
echo $user['user_id'];
But the real problem is - where is $row_array initialized?!?
And even bigger problem - why use "=" inside array creation in the while ... it seems to me that "=>" would fit better, don't you think?