Question

Consider the following Jquery external script snippet for creating youtube iframes in modal windows:

$(function() {

    $(document).ready(function() {  

    var myPhpVa = '<?php echo embeddable($ytid);?>';

    $('a[name="modal"]').click(function(e) {

        $('<iframe width="560" height="315" src=myPhpVa frameborder="0" allowfullscreen>').appendTo('#dcontent');

    }

});

For some reason I get object not found. The PHP function is declared above this javascript, the declaration is:

function embeddable($ytid){
        return $embeddable = 'http://www.youtube.com/embed/'.$ytid;
    }

Any help for passing this PHP variable is greatly appreciated.

Answer 1

First of all – assigning returned value is useless. Remove it:

function embeddable($ytid){
    return 'http://www.youtube.com/embed/'.$ytid;
}

The same thing about $(document).ready(function() { and $(function() { – they are equal and doubled – remove one of them.

After you assign myPhpVa variable – write

alert(myPhpVa);

and check what it shows you.