Resend the same ajax call by it's variable

Question

Here's my code :

var request = $.ajax({
    url: 'some.url'    
});
request.done(function(result){
    console.log(result);
});
request.fail(function(result){
    console.log(result);
});

My question is: Is there any way to initiate the same ajax call with just the variable name (In this case it's request) ???

Like: request.SomeMethodToCallSameAjax();

Try looking at http://stackoverflow.com/questions/8881614/how-do-i-resend-a-failed-ajax-request — William, May 02 '14 at 15:21

score 1 · Accepted Answer · answered May 02 '14 at 15:20

1

You can create a function:

var myAjax = function() {
  return request = $.ajax({
     url: 'some.url'    
  });
}

myAjax.done(function(result) { console.log(result); }
myAjax.fail(function(result) { console.log(result); }

When you need to:

//Call my function
myAjax();

answered May 02 '14 at 15:20

alexmngn

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Thanks :) That's what I wanted. But I think `myAjax.done()` wont work. Instead there should be `myAjax().done()` – Jalay May 05 '14 at 06:20

Resend the same ajax call by it's variable

1 Answers1