CODEWITHSUNDEEP
jqueryajaxjquery-deferredpromise
Bruno Teixeira
Bruno Teixeira

Reputation: 575

Jquery Ajax Promise does not work

I'm making a app and I'm unable to fire a promise, I breaked the code to the a simple example that still does not work, I'm using EasyPHP...

ajaxDialog = function( destiny ) {
    // Promise to let me know when complete
    return $.ajax({
      url: destiny,
      dataType: 'json',
    }).promise();

  };

  teste = ajaxDialog('data.json');

  teste.done( function() { alert("sadasassa"); })

Upvotes: 0

Views: 2227

Answers (4)

Sebastien
Sebastien

Reputation: 1328

Check if this would work:

return $.ajax({
  url: destiny,
  dataType: 'json',
}).promise(function(){alert("This is a test");});

I don't know what would that change but everything is worth trying!

Upvotes: 0

Bergi
Bergi

Reputation: 664297

a simple example that still does not work

The JavaScript works well. Probably your server does not work and does not return a JSON document from the data.json url.

Use

teste.done( function() {
   alert("sadasassa");
}).fail( function(x, s, e) {
   alert(s+": "+e);
});

and debug with the Networks tab in your browser devtools what is happening.

Upvotes: 0

Ruan Mendes
Ruan Mendes

Reputation: 92274

Your ajax is failing and going to the fail handler. If you do the following you will see an alert. 

  // This does get called
  teste2.fail( function() { alert("sadasassa"); });

Here's a jfiddle that proves that if you do reach your server, it gets called http://jsfiddle.net/L6bJ2/367/

$(function() {

    ajaxDialog = function( destiny ) {
        // Promise to let me know when complete
        return $.ajax({
          url: '/echo/json/',
          dataType: 'json',
        }).promise();    
    };

    teste2 = ajaxDialog('data.json');    
    teste2.done( function() { alert("sadasassa"); });

});

Upvotes: 2

Hoffmann
Hoffmann

Reputation: 14719

try this:

when(ajaxDialog('data.json')).then(function() { alert("aaaa"); });

Upvotes: 0

Related Questions