Question

我经历过不同的教程和本网站，但找不到合适的解决方案。另一方面，我已经看到应用程序登录到网站并请求更多信息，所以我确信有一种方法可以实现这一点，但也许我的方法都是错误的。

以下是我要做的事情：我想登录需要用户身份验证的网站，然后阅读并解析只有在用户登录后才能访问的网站。问题：在将凭据发布到网站后，我收到一个cookie，这个cookie在我的HttpClient中似乎没有保留，即使文档建议确实应该这样做。

以下是我的一些代码：

DefaultHttpClient httpclient = new DefaultHttpClient();
HttpPost httpost = new HttpPost(LOGIN_URL);

List<NameValuePair> nvps = new ArrayList<NameValuePair>();
nvps.add(new BasicNameValuePair(USER_FIELD, login));
nvps.add(new BasicNameValuePair(PASS_FIELD, pw));
nvps.add(new BasicNameValuePair(REMEMBERME, "on"));

httpost.setEntity(new UrlEncodedFormEntity(nvps, HTTP.UTF_8));

HttpResponse response = httpclient.execute(httpost);
HttpEntity entity = response.getEntity();

if (entity != null) {
  entity.consumeContent();
}

List<Cookie> cookies = httpclient.getCookieStore().getCookies();

当我输出“cookies”的内容时，一切似乎都很好（我收到一个会话）：

- [version: 0][name: ASP.NET_SessionId][value: xxx][domain: xxx][path: /][expiry: null]

据我了解，只要我不关闭它，cookie /会话就会被保存并在我的HttpClient中使用。

阅读下一页（受限制）时，请使用以下代码：

HttpGet httpget2 = new HttpGet(RESTRICTED_URL);
response = httpclient.execute(httpget2);
entity = response.getEntity();
InputStream data = entity.getContent();
// data will be parsed here
if (entity != null) {
    entity.consumeContent();
}
// connection will be closed afterwards

如果我输出GET请求的响应（使用response.getStatusLine()），我收到“200 OK”消息，但解析返回的站点显示登录丢失（我只检索登录）形式）。

感谢任何帮助。

Answer 1

在我必须登录的应用程序中。首先，我必须运行GET，然后运行POST，然后再运行GET。 First get将为我的连接实例化Jsession Id。 POST将验证我的ID，然后原始获取GET将返回真实内容。

以下代码适用于在JBoss中运行的应用

public boolean login() {
    HttpGet  httpGet = new HttpGet(  "http://localhost:8080/gwt-console-server/rs/identity/secure/sid/");
    HttpPost httpPost = new HttpPost("http://localhost:8080/gwt-console-server/rs/identity/secure/j_security_check");
    HttpResponse response = null;

    List<NameValuePair> nvps = new ArrayList<NameValuePair>();
    nvps.add(new BasicNameValuePair(USER_FIELD, userName));
    nvps.add(new BasicNameValuePair(PASS_FIELD, password));

    try {
        httpPost.setEntity(new UrlEncodedFormEntity(nvps, HTTP.UTF_8));

        response = httpClient.execute(httpGet);
        EntityUtils.consume(response.getEntity());

        response = httpClient.execute(httpPost);
        EntityUtils.consume(response.getEntity());

        response = httpClient.execute(httpGet);
        String sessionId =EntityUtils.toString(response.getEntity());

        String cookieId =""; 
        List<Cookie> cookies = ((AbstractHttpClient) httpClient).getCookieStore().getCookies();
        for (Cookie cookie: cookies){
            if (cookie.getName().equals("JSESSIONID")){
                cookieId = cookie.getValue();
            }
        }

        if(sessionId!= null && sessionId.equals(cookieId) ){
            return true;
        }
    } catch (ClientProtocolException e) {
        // TODO Auto-generated catch block
        e.printStackTrace();
    } catch (IOException e) {
        // TODO Auto-generated catch block
        e.printStackTrace();
    }
    return false;   
}

Answer 2

假设您的httpclient对象在两种情况下都是相同的，并且假设RESTRICTED_URL与LOGIN_URL位于同一个域中，那么我认为您应该有所作为。{ / p>

您可能希望使用Wireshark或代理或其他东西来检查您正在进行的HTTP请求，以查看cookie是否实际附加到请求。可能是cookie被附加，在这种情况下，还有其他错误导致您的第二个请求失败。

Answer 3

你必须使用单身模式DefaultHttpClient httpclient使sessioncookie仍然保持登录会话。

这是Mainactivity类：

public static DefaultHttpClient httpClient;

@Override
public void onCreate(Bundle savedInstanceState) {
    super.onCreate(savedInstanceState);
    RequestPage request = new RequestPage();
    request.post("http://www.example.com/login.php");

    RequestPage requestProfile =new RequestPage();
    requestProfile.post("http://www.example.com/profile.php");
}

这是RequestPage类：

private InputStream post(String url){
    String paramUsername = "username";
    String paramPassword = "pass";

    if(MainActivity.httpClient==null){
        MainActivity.httpClient = new DefaultHttpClient();
    }
    DefaultHttpClient httpClient = MainActivity.httpClient;

    // In a POST request, we don't pass the values in the URL.
    //Therefore we use only the web page URL as the parameter of the HttpPost argument
    HttpPost httpPost = new HttpPost(url);

            // Because we are not passing values over the URL, we should have a mechanism to pass the values that can be
    //uniquely separate by the other end.
    //To achieve that we use BasicNameValuePair             
    //Things we need to pass with the POST request
    BasicNameValuePair usernameBasicNameValuePair = new BasicNameValuePair("username", paramUsername);
    BasicNameValuePair passwordBasicNameValuePAir = new BasicNameValuePair("password", paramPassword);

    // We add the content that we want to pass with the POST request to as name-value pairs
    //Now we put those sending details to an ArrayList with type safe of NameValuePair
    List<NameValuePair> nameValuePairList = new ArrayList<NameValuePair>();
    nameValuePairList.add(usernameBasicNameValuePair);
    nameValuePairList.add(passwordBasicNameValuePAir);

    try {
        // UrlEncodedFormEntity is an entity composed of a list of url-encoded pairs. 
        //This is typically useful while sending an HTTP POST request. 
        UrlEncodedFormEntity urlEncodedFormEntity = new UrlEncodedFormEntity(nameValuePairList);

        // setEntity() hands the entity (here it is urlEncodedFormEntity) to the request.
        httpPost.setEntity(urlEncodedFormEntity);

        try {
            // HttpResponse is an interface just like HttpPost.
            //Therefore we can't initialize them
            HttpResponse httpResponse = httpClient.execute(httpPost);

            // According to the JAVA API, InputStream constructor do nothing. 
            //So we can't initialize InputStream although it is not an interface


            return httpResponse.getEntity().getContent();

        } catch (ClientProtocolException cpe) {
            System.out.println("First Exception caz of HttpResponese :" + cpe);
            cpe.printStackTrace();
        } catch (IOException ioe) {
            System.out.println("Second Exception caz of HttpResponse :" + ioe);
            ioe.printStackTrace();
        }

    } catch (UnsupportedEncodingException uee) {
        System.out.println("An Exception given because of UrlEncodedFormEntity argument :" + uee);
        uee.printStackTrace();
    }

    return null;
}

Answer 4

你可以这样做，虽然它是一种解决方法。

@Override
public void onCreate(Bundle savedInstanceState) {
    super.onCreate(savedInstanceState);
    setContentView(R.layout.activity_main);

    WebView webv = (WebView)findViewById(R.id.MainActivity_webview);         
    webv.setWebViewClient(new WebViewClient(){
            @Override
            public boolean shouldOverrideUrlLoading(WebView view, String url) {
                view.loadUrl(url);
                return true;
            }
    });

    String postData = FIELD_NAME_LOGIN + "=" + LOGIN +
            "&" + FIELD_NAME_PASSWD + "=" + PASSWD;

    // this line logs you in and you stay logged in
    // I suppose it works this way because in this case WebView handles cookies itself
    webv.postUrl(URL, EncodingUtils.getBytes(postData, "utf-8"));
}

Android：使用DefaultHttpClient登录网站并保留会话/ cookie

4 个答案: