Undefined PHP Variable : Undefined variable: pagination in

Question

I'm using the following code to paginate mysql results using this tutorial https://www.youtube.com/watch?v=6SuCSIJ5wNU.

Eventhough the code is working perfectly, i get an undefined variable. ": Undefined variable: pagination in ".

<?php

$get_data = $connect -> query("SELECT NULL FROM movies"); // get the table in the database
$row_count = $get_data -> rowCount(); // count rows

// pagination starts
if( isset($_GET['page']) ) {
    $page = preg_replace("#[^0-9]#", "", $_GET['page']);
} else {
    $page = 1;
}

$perPage = 2;
$lastPage = ceil($row_count / $perPage);

if($page < 1) {
    $page = 1;
} else if($page > $lastPage) {
    $page = $lastPage;
}

$limit = "LIMIT ".($page - 1) * $perPage . ", $perPage";

if($lastPage != 1) {

    if($page != 1) {
        $prev = $page - 1;
        $pagination.='<a href="index.php?page='.$prev.'">Previous</a>';
    }

    if($page != $lastPage) {
        $next = $page + 1;
        $pagination.='<a href="index.php?page='.$next.'">Next</a>';
    }

}

?>

Answer 1

You never defined $pagination before you started appending to it (.=).

These two are equivalent:

$foo .= 'bar';
$foo = $foo . 'bar';

To do the concatentation, PHP has to retrieve the value of $foo, which hasn't been defined, so you get your warning. Then the concatenation occurs and is stored in $foo, so the error doesn't occur again the next time you try it.