?
Solved

convert ajax xmlhttprequest to jquery ajax

Posted on 2013-05-17
4
Medium Priority
?
1,103 Views
Last Modified: 2013-05-18
Hi, I am trying to convert the following code to jquery, but I am having issues.  This uses cross site scripting, but I changed the setting on my browser.  I can successfully use the non jquery code.

      Ajax=new XMLHttpRequest();
        Ajax.open('POST', url, true);
      Ajax.setRequestHeader("Content-type", "application/x-www-form-urlencoded");
      Ajax.send(data);

Here is what I currently have:
jQuery.ajax({
      url: loginUrl,
      contentType: "text/html; charset=Windows-1256",
      dataType: "JSONP",
      async: true,
      success: function (responseText, status) {
            alert("success" + responseText);
      },
      error: function (xhr, ajaxOptions, thrownError) {
            responseText("error=" + xhr.status + "=" + thrownError);
      }
0
Comment
Question by:jackjohnson44
[X]
Welcome to Experts Exchange

Add your voice to the tech community where 5M+ people just like you are talking about what matters.

  • Help others & share knowledge
  • Earn cash & points
  • Learn & ask questions
  • 2
  • 2
4 Comments
 
LVL 82

Accepted Solution

by:
leakim971 earned 2000 total points
ID: 39176412
Try this :
$.post(loginUrl, function(responseText) { alert("success" + responseText); });
0
 

Author Comment

by:jackjohnson44
ID: 39176423
Actually it might be working better than I thought.  It is not hitting my success area, but if I look in fiddler I can see a 200 response in the header, and the xml response in the text view.  For some reason, everything is working, but my error area is still being hit.

This is what I have:
var postData = {};
postData["source"] = source;
var loginUrl = url + "LogIn?source=" + source + "&version=" + version;
jQuery.ajax({
      url: loginUrl,
      //dataType: "xml",
      data: postData,
      async: true,
      type: "POST",
      success: function (responseText, status) {
            alert("ASDF");
      },
      error: function (xhr, ajaxOptions, thrownError) {
            responseText("xhr.status=" + xhr.status + "\najaxOptions=" + JSON.stringify(ajaxOptions) + "\nthrownError=" + thrownError);
      }
});
0
 
LVL 82

Expert Comment

by:leakim971
ID: 39176479
please provide a link to a simple page which reside on your site
0
 

Author Closing Comment

by:jackjohnson44
ID: 39177702
Your answer will work.  I had an issue due to cross site scripting.  Thanks!
0

Featured Post

Is Your Team Achieving Their Full Potential?

74% of employees feel they are not achieving their full potential. With Linux Academy, not only will you strengthen your team's core competencies but also their knowledge of of the newest IT topics.

With new material every week, we'll make sure that you stay ahead of the game.

Question has a verified solution.

If you are experiencing a similar issue, please ask a related question

International Data Corporation (IDC) prognosticates that before the current the year gets over disbursing on IT framework products to be sent in cloud environs will be $37.1B.
Today, the web development industry is booming, and many people consider it to be their vocation. The question you may be asking yourself is – how do I become a web developer?
The viewer will learn how to dynamically set the form action using jQuery.
The viewer will learn the basics of jQuery, including how to invoke it on a web page. Reference your jQuery libraries: (CODE) Include your new external js/jQuery file: (CODE) Write your first lines of code to setup your site for jQuery.: (CODE)
Suggested Courses

770 members asked questions and received personalized solutions in the past 7 days.

Join the community of 500,000 technology professionals and ask your questions.

Join & Ask a Question