jQuery: Best practice to populate drop down?

Andreas Grech was pretty close... it's actually this (note the reference to this instead of the item in the loop):

var $dropdown = $("#dropdown");$.each(result, function() {    $dropdown.append($("<option />").val(this.ImageFolderID).text(this.Name));});

$.getJSON("/Admin/GetFolderList/", function(result) {    var options = $("#options");    //don't forget error handling!    $.each(result, function(item) {        options.append($("<option />").val(item.ImageFolderID).text(item.Name));    });});

What I'm doing above is creating a new <option> element and adding it to the options list (assuming options is the ID of a drop down element.

PS My javascript is a bit rusty so the syntax may not be perfect

Sure - make options an array of strings and use .join('') rather than += every time through the loop. Slight performance bump when dealing with large numbers of options...

var options = [];$.getJSON("/Admin/GetFolderList/", function(result) {    for (var i = 0; i < result.length; i++) {        options.push('<option value="',          result[i].ImageFolderID, '">',          result[i].Name, '</option>');    }    $("#theSelect").html(options.join(''));});

Yes. I'm still working with strings the whole time. Believe it or not, that's the fastest way to build a DOM fragment... Now, if you have only a few options, it won't really matter - use the technique Dreas demonstrates if you like the style. But bear in mind, you're invoking the browser's internal HTML parser i*2 times, rather than just once, and modifying the DOM each time through the loop... with a sufficient number of options. you'll end up paying for it, especially on older browsers.

Note: As Justice points out, this will fall apart if ImageFolderID and Name are not encoded properly...