servicenow xml java basic auth失败

Question

我现在尝试从服务中下载xml文件以查找事件列表，并使用以下代码：

public static void main(String[] args) {
    try {
        String webPage = "https:/ServicnowURL//incident_list.do?XML&sysparm_query=u_im_record_type...";
        String name = "uname";
        String password = "Pwd";

        String authString = name + ":" + password;
        System.out.println("auth string: " + authString);

        byte[] authEncBytes = Base64.encodeBase64(authString.getBytes("UTF-8"));
        String authStringEnc = new String(authEncBytes);
        System.out.println("Base64 encoded auth string: " + authStringEnc);

        URL url = new URL(webPage);
        HttpURLConnection urlConnection = (HttpURLConnection) url.openConnection();
        urlConnection.setRequestProperty("Authorization", "Basic %s" + authStringEnc);
        urlConnection.setRequestMethod("POST");
        urlConnection.setRequestProperty( "Content-type", "text/xml");
        urlConnection.setRequestProperty( "Accept", "*/*" );
        urlConnection.setRequestProperty( "Accept-Encoding", "gzip" );
        System.out.println(urlConnection.getResponseCode());

        InputStream is;
        if(urlConnection.getResponseCode()==HttpURLConnection.HTTP_OK){
            is = urlConnection.getInputStream();
            System.out.println(is);
        }
        else{
            is=urlConnection.getErrorStream();
            System.out.println(is);
        }
    } catch (MalformedURLException e) {
        e.printStackTrace();
    } catch (IOException e) {
        e.printStackTrace();
    }
}

上述代码获得401未经授权的错误

如果我通过删除网址中的“XML”将网址设为“https：/ServicnowURL//incident_list.do？sysparm_query = u_im_record_type ...”，那么我的回复就会成功。