jQuery: create any new object after creation of one object

Question

I want to insert one object like $('<p>Test</p>').prependTo('#TB_window'); just whenever #TB_window object is created.

How it can be made after the creation of a thickbox?

I want to use that in the wordpress site http://hogdal.dk/?p=60

Just see the pictures in the gallery.

I am using wordpress default thickbox.

just a note thickbox it is not supported now, you can use fancybox instead, check this for alternatives: http://jquery.com/demo/thickbox/ — mfadel, May 16 '12 at 08:39
@3nigma: I want to use that in the wordpress site http://hogdal.dk/?p=60 Just see the pictures in the gallery — itsazzad, May 16 '12 at 08:50

score 1 · Accepted Answer · answered May 16 '12 at 09:03

1

How about this:

http://jsfiddle.net/zerkms/YZwRu/

$(document).on('DOMNodeInserted', function(event) {
    var $target = $(event.target);

    if ($target.is('#TB_window')) {
        $target.append('<p>Test</p>');
    }
});

$('body').append('<p>a</p>');
$('body').append('<p id="TB_window">b</p>');
$('body').append('<p>c</p>');

In this example we use DOMNodeInserted event, which is triggered when another node is inserted into document. After that we check if that element has the required id, and if so - append new node to it

answered May 16 '12 at 09:03

zerkms

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Error: $(document).on is not a function in Error Console – itsazzad May 16 '12 at 09:13
@Sazzad Tushar Khan: for jquery < 1.7 use `.bind()` instead of `.on()` – zerkms May 16 '12 at 09:15
@Sazzad Tushar Khan: there is no issues with the code and jquery 1.7.1. http://jsfiddle.net/zerkms/YZwRu/1/ - it works perfectly. Please double check your version. You can check it in runtime with `alert($.fn.jquery);` – zerkms May 16 '12 at 09:19
jQuery(document).ready(function() { jQuery('body').append('
Test
'); }); jQuery(document).on('DOMNodeInserted', function(event) { var target = jQuery(event.target); if (target.is('#TB_window')) { target.append('
Test2
'); } }); – itsazzad May 16 '12 at 09:36
$target should be replaced by only target – itsazzad May 16 '12 at 09:38
@Sazzad Tushar Khan: "should"? Any reason for that? – zerkms May 16 '12 at 09:39
May be because its not PHP and its jQuery – itsazzad May 16 '12 at 09:41
@Sazzad Tushar Khan: it is javasctipt, and `$target` is perfectly valid name – zerkms May 16 '12 at 09:44
But its not working with me may be because i am using that in a PHP file in Wordpress – itsazzad May 16 '12 at 09:46

jQuery: create any new object after creation of one object

1 Answers1