jQuery - Can threads/asynchronous be done?

Question

I\'m currently using AJAX with Django Framework.

I can pass asynchronous POST/GET to Django, and let it return a

Answer 1

Try processing retrieved data incrementally.

At piece below , elements generated in blocks of 250 , primarily utilizing jQuery deferred.notify() and deferred.progress().

When all 10,000 items processed , the deferred object is resolved with the collection of 10,000 elements. The elements are then added to document at single call to .html() within deferred.then()'s .done() callback ; .fail() callback cast as null .

Alternatively , could append elements to the document in blocks of 250 , within deferred.progress callback ; instead of at the single call within deferred.done , which occurs upon completion of the entire task.

setTimeout is utilized to prevent "freeze the web browser" condition .

$(function() {
// 10k items 
var arr = $.map(new Array(10000), function(v, k) {
  return v === undefined ? k : null
});
  
var len = arr.length;
var dfd = new $.Deferred();
// collection of items processed at `for` loop in blocks of 250
var fragment = [];
var redraw = function() {
  for (i = 0 ; i < 250; i++)
    {
        // configurationVariableChart.row.add(
        // $(
        fragment.push('' +
                '' + arr[i] + '' +
            '')
        // )[0]);
    };
    arr.splice(0, 250);
    console.log(fragment, arr, arr.length);
    return dfd.notify([arr, fragment])
};

$.when(redraw())
// `done` callbacks
.then(function(data) {
  $("#results").html(data.join(","));
  delete fragment;
}
  // `fail` callbacks      
 , null
  // `progress` callbacks
 , function(data) {
  // log ,  display `progress` of tasks
     console.log(data);
     $("progress").val(data[1].length);
     $("output:first").text(Math.floor(data[1].length / 100) + "%");
     $("output:last").text(data[1].length +" of "+ len + " items processed");
     $("#results").html("processing data...");
     if (data[0].length) {
         var s = setTimeout(function() {
             redraw()
         }, 100)
     } else {
       clearTimeout(s);
       dfd.resolve(data[1]);
     }
})
})

jsfiddle http://jsfiddle.net/guest271314/ess28zLh/