如何向客户端发送XML文件？具体步骤如下：

import java.io.BufferedInputStream;
import java.io.IOException;
import java.io.PrintWriter;
import java.net.MalformedURLException;
import java.net.URL;
import java.net.URLConnection;

import javax.servlet.ServletException;
import javax.servlet.http.HttpServlet;
import javax.servlet.http.HttpServletRequest;
import javax.servlet.http.HttpServletResponse;

public class ResourceServlet extends HttpServlet {

  public void doGet(HttpServletRequest request, HttpServletResponse response)
      throws ServletException, IOException {

    //get web.xml for display by a servlet
    String file = “/WEB-INF/web.xml”;

    URL url = null;
    URLConnection urlConn = null;
    PrintWriter out = null;
    BufferedInputStream buf = null;
    try {
      out = response.getWriter();
      url = getServletContext().getResource(file);
      //set response header
      response.setContentType(“text/xml”);

      urlConn = url.openConnection();
      //establish connection with URL presenting web.xml
      urlConn.connect();
      buf = new BufferedInputStream(urlConn.getInputStream());
      int readBytes = 0;
      while ((readBytes = buf.read()) != -1)
        out.write(readBytes);
    } catch (MalformedURLException mue) {
      throw new ServletException(mue.getMessage());
    } catch (IOException ioe) {
      throw new ServletException(ioe.getMessage());
    } finally {
      if (out != null)
        out.close();
      if (buf != null)
        buf.close();
    }
  }
  public void doPost(HttpServletRequest request, HttpServletResponse response)
      throws ServletException, IOException {

    doGet(request, response);
  }
}