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Home/ Questions/Q 3459670
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Editorial Team
  • 0
Editorial Team
Asked: May 18, 20262026-05-18T10:08:03+00:00 2026-05-18T10:08:03+00:00

I populate a list with search results appending li elements. I update DOM for

  • 0

I populate a list with search results appending li elements. I update DOM for each result.

for (var i = 0; i < topics.length; i++) {
    $("#searchResults").append(
        $("<li />")
            .append(result.Name)
            .addClass("example")
    );
};

I want to make a group of li elements first and update DOM-tree just once.

I try something like this:

var list = $([]);

for (var i = 0; i < topics.length; i++) {
    list.append(
        $("<li />")
            .append(result.Name)
            .addClass("example")
    );
};

$("#searchResults").append(list);

But div $("#searchResults") is empty.

Where is the problem?

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-18T10:08:03+00:00Added an answer on May 18, 2026 at 10:08 am

    Something like this should be much faster:

    var ul = $("<ul />");
    for (var i = 0, l=topics.length; i < l; i++) {
       $("<li />", { text: result.Name, "class": "example" }).appendTo(ul);
    };
    $("#searchResults").append(ul.contents());
    

    By using a document fragment ($("<ul />")) and appending to it, then appending at the end, we’re not messing with the entire DOM each append. Also we’re not repeatedly selecting #searchResults each loop…or checking .length would could also be expensive.

    Note: this method still uses the DOM to create elements (as opposed to a string), eliminating issues of result.Name having HTML that could screw things up, etc.

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