发送XML到服务器

转自：传智播客教程

1.客户端

public static boolean sendXML(String path, String xml)throws Exception{
        byte[] data = xml.getBytes();
        URL url = new URL(path);
        HttpURLConnection conn = (HttpURLConnection)url.openConnection();
        conn.setRequestMethod("POST");
        conn.setConnectTimeout(5 * 1000);
        conn.setDoOutput(true);//如果通过post提交数据，必须设置允许对外输出数据
        conn.setRequestProperty("Content-Type", "text/xml; charset=UTF-8");
        conn.setRequestProperty("Content-Length", String.valueOf(data.length));
        OutputStream outStream = conn.getOutputStream();
        outStream.write(data);
        outStream.flush();
        outStream.close();
        if(conn.getResponseCode()==200){
            return true;
        }
        return false;
    }

 2.服务端

public ActionForward getXML(ActionMapping mapping, ActionForm form,
            HttpServletRequest request, HttpServletResponse response)
            throws Exception {
        InputStream inStream = request.getInputStream();
        byte[] data = StreamTool.readInputStream(inStream);
        String xml = new String(data, "UTF-8");
        System.out.println(xml);
        return mapping.findForward("result");
    }

 3.测试示例：

public void testSendXMLRequest() throws Throwable{
        String xml = "<?xml version=\"1.0\" encoding=\"UTF-8\"?><persons><person id=\"23\"><name>liming</name><age>30</age></person></persons>";
        HttpRequest.sendXML("http://10.10.97.51:8180/videoweb/video/manage.do?method=getXML", xml);
    }