Question

I’m sending a POST request with a string in the .send() part which is "id=jeff&command=test".

How do I parse and use these values in Java via the HttpServletRequest object?

I’ve tried changing the content type to text/html and application/x-www-form-urlencoded.
My java server is embedded jetty btw.

var text = form.command.value;
var xmlhttp = new XMLHttpRequest();

    xmlhttp.onreadystatechange=function()
    {
          if (xmlhttp.readyState==4 && xmlhttp.status==200)
          {
            xmlhttp.responseText
          }
        }

    xmlhttp.open("POST", 'go', true);
    xmlhttp.setRequestHeader("Content-type", "application/x-www-form-urlencoded");
    xmlhttp.send("id=jeff&command=" + text);

here’s my Java code

public class GoHandler extends HttpServlet 
{
Commands c = Commands.getInstance();
@Override
protected void doPost(HttpServletRequest request, HttpServletResponse response) throws IOException
{
    BufferedReader reader = request.getReader();
    String str = "";
    while ((str = reader.readLine()) != null)
    {
        System.out.println(str);
    }

    String p = request.getParameter("id");
    String input = request.getParameter("command"); 
    System.out.print(p);

here is my output when I make the request from a browser

id=jeff&command=test (this is from the buffered reader)

null (this is from my String p which should be the id)

here’s Chrome’s toolkit thing..

Request URL:http://localhost:8080/go
Request Method:POST
Status Code:200 OK
Request Headersview source
Accept:*/*
Accept-Charset:ISO-8859-1,utf-8;q=0.7,*;q=0.3
Accept-Encoding:gzip,deflate,sdch
Accept-Language:en-US,en;q=0.8
Connection:keep-alive
Content-Length:20
Content-Type:application/xml
Host:localhost:8080
Origin:http://localhost:8080
Referer:http://localhost:8080/
User-Agent:Mozilla/5.0 (Windows NT 6.1; WOW64) AppleWebKit/537.4 (KHTML, like Gecko)     Chrome/22.0.1229.79 Safari/537.4
Request Payload
id=jeff&command=test
Response Headersview source
Content-Length:66
Content-Type:application/json;charset=utf-8
Server:Jetty(8.1.7.v20120910)

my response from the server

{"flavor":"You are in the goHandler"}

Answer 1

You can get it as buffered reader. This will give full string. The you can parse it manually. If your separator is “\n” then,

BufferedReader reader = request.getReader();
while ((str = reader.readLine()) != null)
  jb.append(str);

Edit:

Can you try one/both of these http headers?! Here is your updated code.

var text = form.command.value;
var xmlhttp = new XMLHttpRequest();

xmlhttp.onreadystatechange=function()
{
      if (xmlhttp.readyState==4 && xmlhttp.status==200)
      {
        alert(xmlhttp.responseText);
      }
    }
var params = "id=jeff&command=" + encodeURIComponent(text);
xmlhttp.open("POST", "go", true);
xmlhttp.send(params);

Using encodeURIComponent() to convert the URI into valid ASCII.

Updated the JavsScript since the default content-type is “application/x-www-form-urlencoded” so removed it. Now try to access all your parameters in the servlet.