AJAX - Undefined PHP variable

I have the following problem, the following script sends a keyword a PHP file hosted in another domain (I already added the CROS headers), this PHP returns me some "echos of different variables" (title, thumbnail, url, etc.) And it works but randomly returns me "Undefined variables".

The first thing was to add an if (isset ()) to my variables in PHP and the error does not appear anymore but the results returned by my searches are much smaller (Before adding it averaged 10 to 20 results, Now I get 5 results).

Can this be a problem with my script?

My form.php

<form method="POST" action="" id="form-busqueda">
<input type="text" name="keyword">
<button id="search" name="search">Search</search>
<div id="results"></div>
            <script>
            jQuery(function($){
                var pluginUrl = '<?php echo plugin_dir_url( __FILE__ ); ?>' ;
                $('[id^="form-busqueda"]').on('submit', function(e) {
                    e.preventDefault();
                    $.ajax({
                        type : 'POST',
                        url  : 'http://localhost/ladoserver/script.php',
                        data : $(this).serialize(),
                        beforeSend: function(){
                            $('#results').html('<img src="'+pluginUrl+'../../assets/img/loading.gif" />');
                        }
                    }).done(function(data) {
                        $('#results').html(data);
                    });
                });
            });
        </script>
</form>

My script.php (dlPage is a function that create cURL connection):

<?php
if (isset($_POST['keyword'])) {
    $search = $_POST['keyword'];
    $html = dlPage("http://example.com/" . $search);
    //where I search and get with simple_html_dom example:
    $video = $videos->find('div.example2>a', 0);
    $title = $video->innertext;
    $url = $video->attr['href'];
    $id = $video->attr['id'];
    $thumbnail = $video->find('div.thumb', 0)->innertext;
    echo $title;
    echo $url;
    echo $id;
    echo $thumbnail[0];
}
?>

I've updated my code, I didn't put all the code because I thought that it isn't relevant, my script.php works fine with pure PHP. The problem appear when I use AJAX.

I'm getting the following error:

Notice: Undefined variable: title in C:\xampp\htdocs\webs\ladoserver\script.php on line 13

Notice: Undefined variable: title in C:\xampp\htdocs\webs\ladoserver\script.php on line 13

Notice: Undefined variable: url in C:\xampp\htdocs\webs\ladoserver\script.php on line 14

The undefined variable is coming from your PHP file (/ladoserver/script.php).

What generates the variables being returned? The most common "cause" of this, is by only setting the variables within a block of code that might not be executed (eg within an if block, or in a loop that iterates 0 times)

You could get around the error (assuming you're okay with blank values) by defining each of the variables at the top of your script.

<?php
$title = "";
$thumbnail = "";
$url = "";
$id = "";
?>

---

Edit: @snip1377 reminded me that you can also just use isset at the end of your script before the output as well.

Here's some sample code for your $thumbnail variable, which you could apply to all your variables being returned

<?php
if (isset($thumbnail))
{
    echo $thumbnail;
}
else
{
    echo "";
}
?>

Alternativaely, you can use a ternary operator:

<?php
    echo (isset($thumbnail)) ? $thumbnail : '';
?>

---

Edit again: just to illustrate what I mean about how the variables might not get defined within a script, here is an example that could cause that undefined error:

<?php
if ($_POST['value'] == 1)
{
    // This will never be reached unless $_POST['value'] is exactly 1
    $return_val = 1;
}

echo $return_val;
?>

This will give the undefined warning, if $_POST['value'] is anything other than 1.

Similarly, if $_POST['value'] were 0 in the following code, it would have that undefined warning as well:

<?php
for ($i=0; $i<$_POST['value']; $i++)
{
    // This will never be reached if $_POST['value'] is less than 1
    $return_val = $i;
}
echo $return_val;
?>

In the examples above, you can simply define $return_val at the top of the script, and you won't get the error anymore.