Question

There is only a one line difference and was wondering if it was at all possible!

function sortHot() {
    var order = [];

    $("#container").children(".submission-preview").each(
        function() {
            var key = $(this).attr('id');

            var value = $("div#" + key.replace("sub", "votes")).html() * key.replace("sub", "");

            order.push([key, value]);
        }
    );

    return order;
}

function sortNew() {
    var order = [];

    $("#container").children(".submission-preview").each(
        function() {
            var key = $(this).attr('id');

            var value = key.replace("sub", "");

            order.push([key, value]);
        }
    );

    return order;
}

function sortTop() {
    var order = [];

    $("#container").children(".submission-preview").each(
        function() {
            var key = $(this).attr('id');

            var value = $("div#" + key.replace("sub", "votes")).html();

            order.push([key, value]);
        }
    );

    return order;
}

Answer 1

You could use a switch with an argument passed to the function:

function sortBy(sortOption) {
    var order = [];

    $("#container").children(".submission-preview").each(
        function() {
            var key = $(this).attr('id');
            var value;

            switch (sortOption) {
                case "hot":
                    value = $("div#" + key.replace("sub", "votes")).html() * key.replace("sub", "");
                    break;

                case "new":
                    value = key.replace("sub", "");
                    break;

                case "top":
                    value = $("div#" + key.replace("sub", "votes")).html();
                    break;

                // You may want to provide a default case here, or
                // have one of the above options be the default
            }

            order.push([key, value]);
        }
    );

    return order;
}

Then call it like this, for the three respective options:

sortBy("hot");
sortBy("new");
sortBy("top");