How can I reorder/sort a NodeList in JavaScript?

Question

I have what I think should be a straightforward question; let me quickly explain:

In my JavaScript, food.xml is read in with:

getMenuXml.open("GET","food.xml",false);
getMenuXml.send();
xmlDoc=getMenuXml.responseXML;
xmlFoodList = xmlDoc.getElementsByTagName("food");

so that now I have a NodeList xmlFoodList with all the food elements. Great so far. The problem is I want to sort the nodes based on an element <category> inside. I can read that with:

xmlFoodList[i].getElementsByTagName("category")[0].childNodes[0].nodeValue

Later in my code, the food items are displayed in a list, and as you'd expect, I want food of the same category to be listed together. So, my question is: How can I reorder the nodes in xmlFoodList based on their category?

Notes: I can't change food.xml coming in, and I don't want to edit my later code to do the sorting as the list is populated. I don't want to convert the NodeList to an array, since I'd have to rewrite a lot of later code. Performance isn't actually much of a concern, so feel free to clone/nested loop all you want. Thanks for your time.

Answer 1

You can order the elements of the NodeList, if you convert them to an array first:

var foods = xmlDoc.getElementsByTagName("food");
var foodsArray = Array.prototype.slice.call(foods, 0);

Then you can use the sort method:

foodsArray.sort(function(a,b) {
    var aCat = a.getElementsByTagName("category")[0].childNodes[0].nodeValue;
    var bCat = b.getElementsByTagName("category")[0].childNodes[0].nodeValue;
    if (aCat > bCat) return 1;
    if (aCat < bCat) return -1;
    return 0;
});

This is highly dependent on your XML schema though - if, for example, you had foods which were in more than one category they would only be sorted by the first category in the code above.

Answer 2

使用Javascript库来获取节点列表操作的现成函数是个好主意，例如重新排序，排序，foreach等。我建议使用YUI-3，请参考YUI-3 NodeList 。

Answer 3

Take a look at this: Xml, xsl Javascript sorting . Worst case scenario, you transform the data into precisely the same xml, but sorted. As long as you're paying the transformation penalty, you might consider transforming it into a form more useful for whatever the next step is.