I find a problem with sending receiving parameter

Posted by kawtousse on Stack Overflow See other posts from Stack Overflow or by kawtousse
Published on 2010-03-22T15:41:05Z Indexed on 2010/03/22 15:51 UTC
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how to get the xml translation to html dropdownlist with ajax.

I send the parameter with GET method but the JSP FILE THAT GENERATES THE XML DONT RECEIVE IT.

var url="responsexml.jsp";
url=url+"?projectCode="+prj.options[prj.selectedIndex].value;
xmlhttp.onreadystatechange=stateChanged;
xmlhttp.open("GET",url,true);
xmlhttp.send(null);
and then in responsexml.jsp I do like that:
<% 
String projectcode= (String)request.getParameter("projectCode");
System.out.println("++++projectCode:=" +projectcode);
Session s = null;
Transaction tx;     

try {
                         s = HibernateUtil.currentSession();    
                         tx = s.beginTransaction();




Query query = s.createQuery("SELECT from Wa wa where wa.ProjectCode='"+projectcode+"'");
response.setContentType("text/xml");
PrintWriter output = response.getWriter();
output.write( "<?xml version=\"1.0\" encoding=\"utf-8\"?>");

//response.setHeader("Cache-Control", "no-cache");

  //constriure le xml
if(projectcode!=null)
{
for(Iterator it=query.iterate();it.hasNext();)
{   
    if(it.hasNext())
    {                                                                                                                               

        Wa object=(Wa)it.next();
//out.print( "<item id=\"" +object.getIdWA() + "\" name=\"" + object.getWAName() + "\" />");

  output.write("<wa>");
  output.write( "<item id=\"" + object.getIdWA() + "\" name=\""   + object.getWAName() + "\" />");


    output.write("</wa>");
    }   }
}
} catch (HibernateException e) {
    e.printStackTrace();
}

%>
</body>
</html>

With this code I dont have my xml file. I got this error: The server did not understand the request, or the request was invalid. Erreur de traitement de la ressource http://www.w3.o... PLEASE HELP.

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