jQuery ajax basic authentication is not working

Question

I am trying to make jQuery AJAX request to some basic http authenticated url. here is my jQuery code:

jQuery.ajax({
  url: "https://abc.com/houses/1/appliance_hints",
  method: 'GET',
  cache: false,
  crossDomain: true,
  beforeSend: function(xhr){
    xhr.setRequestHeader('Authorization', 'Basic Z3NsYWI6cGFzczFnczFhYg==');
  },
  success: function(result){
    alert(result);
  }
});

I am setting request header 'Authorization' here. But here is what i see in request header in FF

Host    <host>
User-Agent  Mozilla/5.0 (Ubuntu; X11; Linux x86_64; rv:8.0) Gecko/20100101 Firefox/8.0
Accept  text/html,application/xhtml+xml,application/xml;q=0.9,*/*;q=0.8
Accept-Language en-us,en;q=0.5
Accept-Encoding gzip, deflate
Accept-Charset  ISO-8859-1,utf-8;q=0.7,*;q=0.7
Connection  keep-alive
Origin  <origin>
Access-Control-Request-Me...    GET
Access-Control-Request-He...    authorization,x-requested-with

Am i doing anything wrong ? I am getting status code 302 in the response. Is there any server side configuration error ?

Manuel van Rijn · Accepted Answer · 2011-11-15 08:14:23Z

3

I've noticed you've added crossDomain: true to you $.ajax call, so I think you are trying to fetch something from a remote domain?

Anyway to main problem you have is that you violate the Same origin policy.

If you have control over the remote_url (in your example case abc.com) you could set it to send some header to allow these calls.

example:

Access-Control-Allow-Origin: http://permitted_domain.com

php examples:

header('Access-Control-Allow-Origin: http://permitted_domain.com');
// or to allow all
header('Access-Control-Allow-Origin: *');

answered Nov 15, 2011 at 8:14

Manuel van Rijn

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Bruno Paula · Accepted Answer · 2017-12-02 18:54:21Z

0

var settings = {
  "async": true,
  "crossDomain": true,
  "url": "YOUGETURL",
  "method": "GET",
  "headers": {
    "authorization": "Basic Z3NsYWI6cGFzczFnczFhYg==",
    "cache-control": "no-cache"

  }
}

$.ajax(settings).done(function (response) {
  console.log(response);
});

answered Dec 2, 2017 at 18:54

Bruno Paula

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