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Home/ Questions/Q 783651
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Editorial Team
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Editorial Team
Asked: May 14, 20262026-05-14T20:34:56+00:00 2026-05-14T20:34:56+00:00

I want to use document.createDocumentFragment() to create an optimized collection of HTML elements that

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I want to use document.createDocumentFragment() to create an optimized collection of HTML elements that contain “.data” coming from jQuery (v 1.4.2), but I’m kind of stuck on how to get the data to surface from the HTML elements.

Here’s my code:


var genres_html = document.createDocumentFragment();
$(xmlData).find('genres').each(function(i, node) {
    var genre = document.createElement('a');
    $(genre).addClass('button')
        .attr('href', 'javascript:void(0)')
        .html( $(node).find('genreName:first').text() )
        .data('genreData', { id: $(node).find('genreID:first').text() });
    genres_html.appendChild( genre.cloneNode(true) );
});

$('#list').html(genres_html);

// error: $('#list a:first').data('genreData') is null
alert($('#list a:first').data('genreData').id);

What am I doing wrong here? I suspect it’s probably something with .cloneNode() not carrying over the data when the element is appended to the documentFragment. Sometimes there are tons of rows so I want to keep things pretty optimized, speed-wise.

Thanks!

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-14T20:34:56+00:00Added an answer on May 14, 2026 at 8:34 pm

    You’re running cloneNode on a jQuery object. You start off with native API, then convert it to a jQuery object, then switch back.

    I suppose you could do:

    genres_html.appendChild( genre.get(0).cloneNode(true) );
    

    But then I suspect you would lose your data.


    EDIT:

    If you want jQuery, instead of creating a fragment, try creating an empty jQuery object, then pushing each genre into it:

    var genres_html = $();
    ...
    genres_html.push( genre );
    

    EDIT:

    Give this a try. I’m no DOM expert, but it may work for you.

    var genres_html = document.createDocumentFragment();
    $(xmlData).find('genres').each(function(i, node) {
        var genre = document.createElement('a');
        genre.setAttribute('class','button');
        genre.setAttribute('href', 'javascript:void(0)');
        var $node = $(node);
        genre.setAttribute('genreData', $node.find('genreID:first').text() );
        genre.innerHTML = $node.find('genreName:first').text();
        genres_html.appendChild( genre.cloneNode(true) );   // Not sure why you would need to make a clone??
    });
    
    var list = document.getElementById('list');
    list.appendChild(genres_html);
    
    // error: $('#list a:first').data('genreData') is null
    alert($('#list a:first').attr('genreData'));
    

    Let me know if it works.

    EDIT: Changed my error with innerHTML

    EDIT2: Using native innerHTML to append to #list

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