How to pass arguments to the chained promises callbacks

Question

How can i pass arguments to Sequence promises using q.promises

var argumentsArray = ["PARAM1", "PARAM2", "PARAM3"]
var arr = [];
function aJaxCall(argumentsArray) {
    var deferred = Q.defer();
    $.ajax({
        type: 'POST',
        /*
            Here i want to get the passed parameter and use it
            in my post request
        */
        url: 'someurl.com' + PARAMS,
        // data: data,
        dataType: 'text',
        success: function (data) {
            arr.push(data);
            deferred.resolve(data);
        }
    });
    return deferred.promise;
}

var result = Q();   
for (var i = 0, len = argumentsArray.length; i < len; i++) {
    /* Pass the argumentsArray[i]
        for each then call, i want to be able to pass
        the current argumentsArray value
    */
    result = result.then(aJaxCall);
}

So far i tried with properties to the aJaxCall function itself and then handle it with arguments.calee.prop

for (var i = 0, len = argumentsArray.length; i < len; i++) {
    aJaxCall.prop = argumentsArray[i];
    result = result.then(aJaxCall);
}

and then within the aJaxCall function i tried to get the parameters like this:

arguments.callee.prop

but it is always binding the last element in the argumentsArray, cause the for loop finish and the .then is fired later in the cycle. Any suggestions are highly appreciated.

Don't create a custom `deferred` (where you have forgotten error handling)! Just assimilate the jqXHR promise that jQuery gives you: `return Q($.ajax(…));` — Bergi, Aug 09 '14 at 10:36
i know that callee is depricated, but i couldn't come up with something that works, so i pretty much tried all i can think of — sla55er, Aug 09 '14 at 10:37

joeltine · Accepted Answer · 2014-08-09T10:41:46.877

1

You can create a method that returns an anonymous function that calls ajaxCall with the passed parameters:

var callWithArgs = function(arg) {
  return function () {
     return ajaxCall(arg);
  }
}

for (var i = 0, len = argumentsArray.length; i < len; i++) {
    result = result.then(callWithArgs(argumentsArray[i]));
}

edited Aug 09 '14 at 10:41

answered Aug 09 '14 at 10:04

joeltine

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this is not working, since .then expects a function that return promise – sla55er Aug 09 '14 at 10:38
1

Easy fix. Just added a return ajaxCall. – joeltine Aug 09 '14 at 10:42
tried this already - the ajax are not called at all – sla55er Aug 09 '14 at 10:45
my bad, this is actaully working... – sla55er Aug 09 '14 at 10:55
OK, sounds good. Also, looking at the Q documentation, I don't see a use case for "var q = Q(); q = q.then(...);". Are you sure it's doing what you want? – joeltine Aug 09 '14 at 10:57
yes, that is what im after. you can see the var q = Q() in the here http://documentup.com/kriskowal/q/#tutorial/sequences – sla55er Aug 09 '14 at 11:00

How to pass arguments to the chained promises callbacks

1 Answers1