PHP Notice: Undefined offset: 2

Question

I'm using this code to embed Gists in WordPress. It works fine, Gists show up, but I keep getting an error in my debug.log file: PHP Notice: Undefined offset: 2 in /functions.php on line 333

(line 333 is esc_attr($matches[2]))

    wp_embed_register_handler( 'gist', '/https?:\/\/gist\.github\.com\/([a-z0-9]+)(\?file=.*)?/i', 'embed_handler_gist' );

function embed_handler_gist( $matches, $attr, $url, $rawattr ) {

    $embed = sprintf(
            '<script src="https://gist.github.com/%1$s.js%2$s"></script>',
            esc_attr($matches[1]),
            esc_attr($matches[2])
            );

    return apply_filters( 'embed_gist', $embed, $matches, $attr, $url, $rawattr );

}

The error is specific. Pretty sure that if you check if the `$matches[2]` has any value whatsoever, you'll find out that it doesn't. — André Silva, Oct 28 '15 at 00:06
Can someone help on how I can check if the ` $matches[2] ` has any value? How do I program an error check? @Isuckatprogramming — Irina, Oct 28 '15 at 01:29
Always append an empty entry to the array that way there will always be at least two entries? i.e. before the `$embed = sprintf ..` add: `$matches[] = "";` — Ryan Vincent, Oct 28 '15 at 02:48
@RyanVincent Thank you! I appended an empty entry to the array and it made the error disappear — Irina, Oct 28 '15 at 13:37

score 3 · Answer 1 · answered Oct 28 '15 at 01:59

3

That error simply means that $matches[2] doesn't exist. You can avoid this error by checking for its existence before attempting to access the variable.

if( isset($matches[2]) )
    esc_attr($matches[2]);

or if you are looking to assign a default value:

$value = isset($matches[2]) ? $matches[2] : false;

isset()

answered Oct 28 '15 at 01:59

Chris

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What a solution! The `isset()` is the key success! Thanks for saving my day! – Jodyshop Jul 22 '19 at 00:46

score 1 · Answer 2 · answered Oct 28 '15 at 00:29

I ran into an offset issue when I was trying to pull a media list from a mysql database. When I had only one entry in the database it gave me an offset error. Check to see if you have both variables returned. If you only get one, maybe you can program an error check with an sizeof($array) or count($array). Just a thought :)

PHP Notice: Undefined offset: 2

2 Answers2