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Sorting XML doesn't return a value

What I did was parse the the var a, sort and then display it. It doesn't return value or error.

var a = '<root><ap by="A" on="2021/10/01 14:03:24" lvl="4"></ap><ap by="A" on="2021/10/01 14:03:59" lvl="2"></ap><ap by="A" on="2021/10/01 14:04:01" lvl="5"></ap><ap by="A" on="2021/10/01 14:04:05" lvl="1"></ap></root>';


        $($.parseXML(a), function (log) {
        var ap = $(log).find($(this)).clone();

        ap.sort(function (a, b) {
            return (parseInt($(a).attr("lvl")) - parseInt($(b).attr("lvl")));
        });

        ap.find("ap").each(function () {
        var ap = $(this);
        var by = ap.attr("by");
        var on = kendo.parseDate(ap.attr("on"));
        var lvl = parseInt(ap.attr("lvl"));

                 
        if (apdate.html() != "") $("<br>").appendTo(apdate);
        $("<span" + props + ">").text(by + " (" + kendo.toString(on, User.datetimeFormat) + ") (" + lvl + ")").appendTo(apdate);

        });

I was able to display the data unsorted by using this. I assumed that the problem is in my sorting code.

        $($.parseXML(a).find("ap").each(function () {
        var ap = $(this);
        var by = ap.attr("by");
        var on = kendo.parseDate(ap.attr("on"));
        var lvl = parseInt(ap.attr("lvl"));
        var props = "";
          
        if (apdate.html() != "") $("<br>").appendTo(apdate);
        $("<span" + props + ">").text(by + " (" + kendo.toString(on, User.datetimeFormat) + ") (" + lvl + ")").appendTo(apdate);

7 months ago · Juan Pablo Isaza

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Accepted Answer · 2022-07-04T10:49:09.907Z

I was referencing to the answer here link

From that link: $.get is an asynchronous call to load an xml file, with a callback for when it has completed. You do not have a file, you have a string - the string does not need to be "loaded" asynchronously (as it's already loaded).

So there you have:

$.get("remote_file_url.xml", callback_when_loaded(xml) { ...

but here you want:

var a = "<root>...";
var xml = $.parseXML(a)
var log = $(xml);
var ap = log.find("ap")...

then you can use log as a jquery object. $.parseXML(a) will give you an XML document object, but to use jquery methods, it needs to be a jquery object, so wrap in $() - this is the equivalent of using this in an event handler: this = a DOM node and $(this) = jquery object.

Later on you have ap.find("ap") as ap already = $(xml).find("ap") this is the equivalent of $(xml).find("ap").find("ap")

Updated code (with some minor adjustments to remove irrelevant "kendo" code)

var a = '<root><ap by="A" on="2021/10/01 14:03:24" lvl="4"></ap><ap by="A" on="2021/10/01 14:03:59" lvl="2"></ap><ap by="A" on="2021/10/01 14:04:01" lvl="5"></ap><ap by="A" on="2021/10/01 14:04:05" lvl="1"></ap></root>';

var log = $($.parseXML(a))
console.log(log.find("ap").length)

var ap = log.find("ap").clone();
//console.log(ap)

ap.sort(function(a, b) {
  return (parseInt($(a).attr("lvl")) - parseInt($(b).attr("lvl")));
});
//console.log(ap)

ap.each(function() {
  var ap = $(this);
  var by = ap.attr("by");
  //var on = kendo.parseDate(ap.attr("on"));
  var lvl = parseInt(ap.attr("lvl"));

  apdate = $("#output")
  if (apdate.html() != "") 
    $("<br>").appendTo(apdate);
  $("<span>").text(by + "(" + lvl + ")").appendTo(apdate);

});

<script src="https://cdnjs.cloudflare.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script>
<div id='output'></div>