jQuery / .when - package up arguments into function?

Question

The jQuery doco shows an example where two $.ajax calls are made under the control of a $.when (page down the 'Examples' heading) .

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

That's fine but what I would like to do is to define those $.ajax calls within a function (where I might do other stuff to prepare for the call).

So something like this :

$.when( myAjaxCall1(), myAjaxCall2() )
  .then( goodFetches, badFetches );

I have tried that with a functions that look like this :

var myAjaxCall1 = function()
{
   return $.ajax('https://httpbin.org/get?Z=Y,X,W');
};
var myAjaxCall2 = function()
{
   return $.ajax('https://httpbin.org/get?A=B,C,D');
};

I do get a three element array as an argument to goodFetches for each of myAjaxCall1 and myAjaxCall2 but it's not the same information as I receive if I just put the $.ajax calls straight into the $.when as shown in the above example from the documentation.

How should I restructure myAjaxCall1 and myAjaxCall2 so that the behaviour is equivalent to :

$.when( $.ajax( "https://httpbin.org/get?Z=Y,X,W" ), 
        $.ajax( "https://httpbin.org/get?A=B,C,D" ) )
  .then( goodFetches, badFetches );

I think you can't do that, since the $.ajax call waiting the response from the server so you have no return value while you call the return before the server give the response. — Fadhly Permata
– Fadhly Permata, Commented Jul 13, 2016 at 4:09
@FadhlyPermata - You can just return the promise from the function which is the same thing that $.ajax() returns. — jfriend00
– jfriend00, Commented Jul 13, 2016 at 4:38
Your naming of the callbacks goodFetches, badFetches suggests you don't exactly understand what $.when does. It's more like .then(allGood, anyBad) (and just like with any promise, only one of the two will get called). — Bergi
– Bergi, Commented Jul 13, 2016 at 5:50
The behaviour of $.when when being passed ajax promises is indeed abysimal. Switch to a proper promise library if you can. — Bergi
– Bergi, Commented Jul 13, 2016 at 5:54
Your snippets with and without the ajax calls wrapped in functions are exactly equivalent. $.when does not behave differently. I'd suggest you show us what those goodFetches and badFetches callbacks are and what goes wrong in them (and what you would have expected instead) — Bergi
– Bergi, Commented Jul 13, 2016 at 5:56

jfriend00 · Accepted Answer · 2016-07-13 04:46:12Z

It will work just fine in the functions. You are probably just not accessing the arguments properly from the result. You get a series of arguments that are each a sub array that contains three elements: [data, statusText, jqXHR]. Those arguments will be the same whether you have your ajax calls in a function or not as long as the function returns the promise.

Here's an example from the jQuery doc:

$.when( $.ajax( "/page1.php" ), $.ajax( "/page2.php" ) ).done(function( a1, a2 ) {
  // a1 and a2 are arguments resolved for the page1 and page2 ajax requests, respectively.
  // Each argument is an array with the following structure: [ data, statusText, jqXHR ]
  var data = a1[ 0 ] + a2[ 0 ]; // a1[ 0 ] = "Whip", a2[ 0 ] = " It"
  if ( /Whip It/.test( data ) ) {
    alert( "We got what we came for!" );
  }
});

It makes no difference if the $.ajax() calls are directly in the $.when() arguments or if you return the promise from a function. Same result.

So, you can do this and you will get the same data back in the .then() handler:

var myAjaxCall1 = function() {
   return $.ajax('https://httpbin.org/get?Z=Y,X,W');
};

var myAjaxCall2 = function() {
   return $.ajax('https://httpbin.org/get?A=B,C,D');
};

$.when( myAjaxCall1(), myAjaxCall2() ).then(function(a1, a2) {
      // a1[0] is data from first ajax call
      // a2[0] is data from second ajax call
});

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