localhost do not load due to php code - php

Question

my xampp localhost was working well till i add this code to my php file
<?php while ($notification) {?>
<li>
<?php
echo $notification['notification'];
?>
</li>
<?php
} ?>
now the page is not loading or partially loading
here $notification is
$notification_sql= "SELECT id FROM notifications WHERE user_id='{$_SESSION['user']}'";
$notification_query = mysqli_query($conn, $notification_sql);
$notification = mysqli_fetch_assoc($notification_query);

Answer 1

Before I begin I want to recommend you something: Avoid the use of the while statetements. Unless they are really needed - like in the exact way they are used in the PHP docs (and even then you can find alternatives) - they should be avoided all the time. I present the motive down under.
That said,... it's not $notification['notification'], but $notification['id'].
After you change it, you still remain with the issue: an infinite loop. Because you are using a while loop without changing the state of the loop condition. E.g_ you are validating the $notification array for beeing existent. Because it exists all the time - it's state never beeing changed in the loop - then the iterations will be infinite in number. In order to avoid this dangerous (!!!) situation, you can use the following codes.
Method 1:
Notice the difference: $notification is valid only for the period of a loop step. After each iteration $notification is newly created. And, when mysqli_fetch_assoc() returns FALSE on the (last + 1)-iteration step, then the $notification receives that value and, therefore, the loop ends.
<?php
$notification_sql = "SELECT id FROM notifications WHERE user_id='{$_SESSION['user']}'";
$notification_query = mysqli_query($conn, $notification_sql);
if ($notification_query) {
while ($notification = mysqli_fetch_assoc($notification_query)) {
?>
<li>
<?php
echo $notification['id'];
?>
</li>
<?php
}
mysqli_free_result($notification_query);
}
?>
Method 2:
Or, if you want to fetch the results in an array and to output its items later, then you can do it like this:
<?php
$notification_sql = "SELECT id FROM notifications WHERE user_id='{$_SESSION['user']}'";
$notification_query = mysqli_query($conn, $notification_sql);
$notifications = array();
if ($notification_query) {
while ($row = mysqli_fetch_assoc($notification_query)) {
$notifications[] = $row['id'];
}
mysqli_free_result($notification_query);
}
// OTHER STUFF AFTER FETCHING...
?>
<?php
// LOOPING AT A LATER TIME.
foreach ($notifications as $notificationId) {
?>
<li>
<?php
echo $notificationId;
?>
</li>
<?php
}
?>
Other recommendations:
Use prepared statements in order to avoid MySQL injection.
Use exception handling in order to catch all errors and handle them correspondingly. Especially when you run database operations.
Use PDO instead of mysqli.
Here I have provided full code examples of prepared statements combined with exception handling (using mysqli library):
Can't insert info in a server, but i can as localhost (See EDIT 2)
Login using MySqli Prepared Statement (See solution 1 & 2)
Good luck.