Insert PHP value as array member?

Question

I'm having a problem with inserting previously defined PHP value $video_url as one of the array members. Here the code:

$video_url= "[[+virtual_tour]]"; /* [[+virtual_tour]] is MODX placeholder which contains first video url */

echo $video_url; /* for test reasons. Outputs https://youtu.be/9bZkp7q19f0 */

$url = array(
'$video_url', /* URL of the first video */
'https://www.youtube.com/watch?v=e-ORhEE9VVg' /*URL of the second video */
);
// Extracts the YouTube ID from various URL structures
foreach ($url as $value) {
    if (preg_match('%(?:youtube(?:-nocookie)?\.com/(?:[^/]+/.+/|(?:v|e(?:mbed)?)/|.*[?&]v=)|youtu\.be/)([^"&?/ ]{11})%i', $value, $match)) {
        $id = $match[1];
echo $id; /* outputs only ID of second video, but not first */
    }
}

So I have something word by inserting $video_url into array. Thanks.

Mr. Llama · Accepted Answer · 2016-06-22 16:47:22Z

2

Try without quotes around $video_url:

$url = array(
   $video_url, /*URL of the first video**/
   'https://www.youtube.com/watch?v=e-ORhEE9VVg' /* URL of the second video */
);

Single quotes do not interpolate strings, meaning the value of $url[0] is literally $video_url.

Also, take care with comments, in PHP they are not #.

edited Jun 22, 2016 at 16:47

Mr. Llama

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answered Jun 22, 2016 at 16:43

Marcos Pérez Gude

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9 Comments

MyMadWeb Over a year ago

Tried without quotes, the same result

Marcos Pérez Gude Over a year ago

Are you fix your comments? Remove # and add //. Can you var_dump($url) to know what is inside it?

MyMadWeb Over a year ago

I've fixed comments (and in my original post). Here is var_dump($url) result: array(2) { [0]=> string(17) "youtu.be/9bZkp7q19f0" [1]=> string(43) "youtube.com/watch?v=e-ORhEE9VVg" }

Marcos Pérez Gude Over a year ago

It works like a charm: regex101.com/r/kW7jM2/1 . Why the second URL have 43 characters?

MyMadWeb Over a year ago

have no idea about 43 characters

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