How to fix invisible value for php variable inside input field - php

Question

I'm trying to display the users nickname into a html form input text field. It was working, but now quit showing up randomly. There is data in the database for the user. The fields are just blank and show nothing. I have plenty other fields working, but this does not and idk why. Feel free to edit the title I did not know how to word it.
When I try to implode the $nickname
$string_nickname = implode(',', $nickname);
I get the error
Warning: implode(): Invalid arguments passed in
C:\xampp\htdocs\Client-Projects\Crossfire\CoinSubmission.php on line
26
So, I tried parsing it to just a string since it's not an array, like..
$string_nickname = $nickname['nickname'];
But when I do this the field in the form is just completely blank and it gives no error.
Here is the input field where I insert nickname.
<label>NICKNAME:</label>
<input type="text" name="nickname" id="nickname" value="<?php echo $string_nickname; ?>" readonly>
and where I select and set nickname.
$result = mysqli_query($con,"SELECT adminlogin.nickname FROM adminlogin INNER JOIN coinsub ON coinsub.profileid=adminlogin.profileid") or die(mysqli_error($con));
$nickname = mysqli_fetch_assoc($result);
$string_nickname = implode(',', $nickname);
Please let me know if you need anymore info.

Answer 1

$string_nickname = '';
if(!empty($nickname)){
$string_nickname = implode(',', $nickname);
}

Answer 2

Do it with object orientated mysqli
$arr = array();
while ($row = $result->fetch_assoc()) {
$arr[] = $row['nickname'];
}
$string_nickname = implode(',', $arr);
You are fetching one value, I think you need more than one. Just create a new array, push to it and you are done. Even if you have one value in the database, this works The above array contains all the nicknames we get from the database. Once you implode them, they are like
nickname1, nickname2, nickname3, nicknamen
So your whole code becomes something like this
$result = $con->query("SELECT adminlogin.nickname FROM adminlogin INNER JOIN coinsub ON coinsub.profileid=adminlogin.profileid") or die(mysqli_error($con));
$arr = array();
while ($row = $result->fetch_assoc()) {
$arr[] = $row['nickname'];
}
$string_nickname = implode(',', $arr);

Answer 3

If you want to implode try this :
$string_nickname = implode(', ', (array)$nickname);
For select mysqli try this :
$selectquery = $con->prepare("SELECT adminlogin.nickname FROM adminlogin
LEFT JOIN coinsub ON coinsub.profileid=adminlogin.profileid");
$selectquery->execute();
$result = $selectquery->get_result();
$user = $result->fetch_assoc();
$nickname = $user['nickname'];
You can also use while statement as follow :
while($user=$result->fetch_assoc()){
echo $user['nickname'];
}