Echo all While outputs into a single variable

Question

This is my code:

$sql = "SELECT * FROM notifications";
if($result = mysqli_query($link, $sql)) {
    if(mysqli_num_rows($result) > 0) {  
        while($row = mysqli_fetch_array($result)) {
            $notification = "
<a class='d-flex align-items-center dropdown-item' href=".$row["link"].">
<div class='dropdown-list-image mr-3'><i class='fas fa-bell'></i>
<div class='bg-success status-indicator'></div>
</div>
<div class='font-weight-bold'>
<div class='text-truncate'><span>".$row["omschrijving"].".</span></div>
<p class='small text-gray-500 mb-0'>".$row["klant"]." - ".$row["createdat"]."</p>
</div>
</a>    
";
        }
    }
}

and further in my page is:

echo "<h6 class='dropdown-header'>Notificaties</h6>
$notification"

I really want to have the output as a single variable for different reasons.

Now it's only outputting the first notification from the database.

Is it possible to do like: echo $notification for each row there is in the table?

Use concatenation with `$notification .=` – Barmar Aug 17 '20 at 21:22 — Barmar, Aug 17 '20 at 21:22

Alberto Sinigaglia · Accepted Answer · 2020-08-17T21:57:50.760

0

Concatenating the notifications probably will do it

$notification = '';
while($row = mysqli_fetch_array($result))
    $notification .="
        <a class='d-flex align-items-center dropdown-item' href=".$row["link"].">
        <div class='dropdown-list-image mr-3'><i class='fas fa-bell'></i>
        <div class='bg-success status-indicator'></div>
        </div>
        <div class='font-weight-bold'>
        <div class='text-truncate'><span>".$row["omschrijving"].".</span></div>
        <p class='small text-gray-500 mb-0'>".$row["klant"]." - ".$row["createdat"]."</p>
        </div>
        </a>   ";

edited Aug 17 '20 at 21:57

answered Aug 17 '20 at 21:22

Alberto Sinigaglia

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It's showing up more now, I got 6 notifications. the weird thing is that i have 2 notifications in my Database. and also got : Notice: Undefined variable: notification in C:\xampp\htdocs\parameters.php on line 83 – Tim Nijland Aug 17 '20 at 21:52
Before the `while` you need to define the variable `$notification = '';` – AbraCadaver Aug 17 '20 at 21:56
@TimNijland you need to initialize the variable before using it (added the line necessary in the snipped) – Alberto Sinigaglia Aug 17 '20 at 21:58
"How to concatenate a string in a loop" is probably a mega-duplicates on Stack Overflow @Berto Please close instead of answering. – mickmackusa Aug 17 '20 at 22:01
Thank you so much @Berto99! – Tim Nijland Aug 17 '20 at 22:12

score -1 · Answer 2 · answered Aug 17 '20 at 21:33

Looks like you're writing over the $notification variable every instance in your while loop. You should look at concatenating that variable. Additionally you want to be aware that HTML inside a PHP variable may not always play nice. I would also recommend using heredoc syntax. It would look something like this:

$notification .= <<<EOD
HTML goes here
EOD;

Echo all While outputs into a single variable

2 Answers2