javascript - jquery : wait until all ajax calls finish then continue - Stack Overflow

I do not like any answer at all, the best way (since Jquery 1.5+) is to use Deferred objects, those are objects to manipulate async calls, you can solve :

$.when($.ajax("/page1.php"), $.ajax("/page2.php"))    .then(myFunc, myFailure);

This way myFunc executes after the 2 ajax calls are made, and myFailure if either one has an error.
You can read more about it in the jquery official documentation:JQuery Deferred Object

First off, you might want to consider doing the startup code in a single call.
Second: Instead of waiting just call another function call. for the above code it should look something like:

for (j=1; j <= 7; j++){
  (function(index) {
    $.getJSON('my.php', {id:index}, 
      function(data) {
         $.each(data, function(index2, array){
         ........
         }); 

         if (j === 7) {
            initDoneDoMoreStuff()
         }
      });
    })(j)
} 

or trigger:

for (j=1; j <= 7; j++){
  (function(index) {
    $.getJSON('my.php', {id:index}, 
      function(data) {
         $.each(data, function(index2, array){
         ........
         }); 

         if (j === 7) {
            $(document).trigger("initdone");
         }
      });
    })(j)
}

$(document).bind("initdone", function() {....});

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