Question

I’m new to Java and I’m getting the HttpServletRequest, but I have no clue how to respond to the request using the HttpServletResponse.

Here is my example code:

public void handle(String target, HttpServletRequest request,
                   HttpServletResponse response, int dispatch)
       throws IOException {
  // Scan request into a string
  Scanner scanner = new Scanner(request.getInputStream());
  StringBuilder sb = new StringBuilder();
  while (scanner.hasNextLine()) {
    sb.append(scanner.nextLine());
  }

This is the sample request I’m getting:

GET / HTTP/1.1
Host: 10.10.10.100:8800
User-Agent: Mozilla/5.0 (Windows NT 6.1; WOW64; rv:10.0) Gecko/20100101 Firefox/10.0
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
Connection: keep-alive

By default the response is

HTTP/1.1 200 

But I want the rosponse something like

POST something back to the GET Request

how can I do it. and where am I suppose to add the code...??
I'm actually pretty lost in this whole thing and rather uncomfortable with Java still, so I have no idea what I'm missing. Any pointers are greatly appreciated. Thanks!

Answer 1

Add the response post at the end of handle method like this

public void handle(String target, HttpServletRequest request,
                   HttpServletResponse response, int dispatch)
       throws IOException {
  // Scan request into a string
  Scanner scanner = new Scanner(request.getInputStream());
  StringBuilder sb = new StringBuilder();
  while (scanner.hasNextLine()) {
    sb.append(scanner.nextLine());
  }
  response.getOutputStream().println("This is servlet response");
}