Playframework WSRequest POST登录

Question

我使用Java和Playframework 1.2.3创建了一个Web爬虫.现在,我想抓取一些受经典登录/密码表单保护的网页.

事实上,这就像做这个游戏测试:

@Test
public void someTestOfASecuredAction() {
    Map<String, String> loginUserParams = new HashMap<String, String>();
    loginUserParams.put("username", "admin");
    loginUserParams.put("password", "admin");
    Response loginResponse = POST("/login", loginUserParams);

    Request request = newRequest();
    request.cookies = loginResponse.cookies; // this makes the request authenticated
    request.url = "/some/secured/action";
    request.method = "POST";
    request.params.put("someparam", "somevalue");
    Response response = makeRequest(request);
    assertIsOk(response); // Passes!
}

但不是通过播放生成的网站,而是与外部网站.

所以,我设法使用播放网络服务器来做到这一点:

Map<String,String> params = new HashMap<String, String>();
params.put( "telecom_username", 
            Play.configuration.getProperty("telecom.access.user") );

params.put( "telecom_password", 
            Play.configuration.getProperty("telecom.access.pass") );

HttpResponse response = WS.url(loginUrl)
                        .setParameters(params)
                        .followRedirects(true)
                        .post();

当我这样做时,如果我查看response.getString(),我发现在继续之前设置了cookie的重定向页面,但是,如果我得到一个受保护的页面,我仍然没有登录.它就像是cookie从未设置过,并且HttpResponse对象没有任何与cookie相关的功能,就像之前测试代码中的响应一样.

我也尝试过ws.url()上的authenticate()方法,但它也不起作用.

我真的不知道我正在尝试做什么是通过使用播放网络服务器,但我可以使用这个^^的帮助

非常感谢 !

Answer 1

好的,我找到了一种方法,但我做得很难,这里是:

首先,我们存储会话cookie的GET,请考虑我正在使用的字符集,并且我知道我正在寻找的cookie的名称,您可以将它们全部存储起来.此外,您可能想要加密它们.

HttpResponse wsResponse = WS.url(comercialYComunUrl).get();
String responseString = wsResponse.getString("ISO-8859-1");

if (wsResponse.getStatus() == 200) {
  List<Header> headers = wsResponse.getHeaders();
  // get all the cookies
  List<String> cookies = new ArrayList<String>();
  for (Header header: headers) {
    if (header.name.equals("Set-Cookie")) {
      cookies = header.values;
    }
  }
  // look for the session cookies
  String sessionCookie = "";
  for (String cookie : cookies) {
    if (cookie.toUpperCase().contains("ASPSESSIONID")) {
      sessionCookie = cookie.split(";", 2)[0];
    }
  }
  // store it on the session
  session.put("COOKIE", sessionCookie);
}

现在邮报:

String url = "http://www.url.com/";
String charset = "ISO-8859-1";
String param1 = "value1";
String param2 = "value2";
String param3 = "value3";

String query = String.format("param1=%s&param2=%s&param2=%s", 
     URLEncoder.encode(param1, charset),
     URLEncoder.encode(param2, charset),
     URLEncoder.encode(param3, charset));

URLConnection connection = new URL(url).openConnection();
connection.setDoOutput(true); // Triggers POST.
connection.setRequestProperty("Accept-Charset", charset);
connection.setRequestProperty("Content-Type",
    "application/x-www-form-urlencoded;charset=" + charset);
connection.addRequestProperty("Cookie", session.get("COOKIE"));
OutputStream output = null;
try {
     output = connection.getOutputStream();
     output.write(query.getBytes(charset));
} finally {
     if (output != null) try { output.close(); } catch (IOException logOrIgnore) {}
}
InputStream responseStream = connection.getInputStream();

StringWriter writer = new StringWriter();
IOUtils.copy(responseStream, writer);
String response = writer.toString();

这对我有用,这是我的来源,这是一个很棒的帖子: 如何使用java.net.URLConnection来触发和处理HTTP请求？

- - - - - - - - - - - - - - 编辑 - - - - - - - - - - - ----------------

好吧,我对答案都不满意,所以我找到了一个更好的方法:

String url = "http://www.url.com/";
String charset = "ISO-8859-1";

String param1 = "value1";
String param2 = "value2";
String param3 = "value3";

WSRequest wsRequest = WS.url(url);
wsRequest.parameters.put("param1", param1);
wsRequest.parameters.put("param2", param2);
wsRequest.parameters.put("param3", param3);
wsRequest.headers.put("Cookie", session.get("COOKIE"));
HttpResponse wsResponse = wsRequest.post();
String responseString = wsResponse.getString(charset);

它的工作原理^.^