Question

我正在尝试以编程方式登录mobile.twitter.com。以下是手动执行的步骤： 1）转到https://mobile.twitter.com/session/new

This has a web form of structure:


<?xml version="1.0" encoding="utf-8"?>
<!DOCTYPE html PUBLIC "-//WAPFORUM//DTD XHTML Mobile 1.1//EN" "http://www.openmobilealliance.org/tech/DTD/xhtml-mobile11.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
  <head>
    <meta name="HandheldFriendly" content="True"/>
<meta name="viewport" content="width=device-width, initial-scale=1.0, maximum-scale=1.0, user-scalable=0"/>
<meta name="twitter-redirect-srcs" content="{&quot;pwreset-iphone&quot;:true,&quot;android&quot;:true,&quot;email&quot;:true}">
<meta name="twitter-redirect-url" content="twitter://timeline">
    <link href="https://ma-0.twimg.com/twitter-mobile/9635f415e4f7c5dce70ba32d2e0b334893f16c7f/images/ic_favicon.png" rel="icon" type="image/png" />
    <title>Sign in to Twitter!</title>
    <link href="https://ma-0.twimg.com/twitter-mobile/9635f415e4f7c5dce70ba32d2e0b334893f16c7f/assets/m2s.css" media="screen" rel="stylesheet" type="text/css" />
  </head>
  <body class="images nojs sessions-page sessions-new-page">
    <div id="container">
      <div id="brand_bar">
        <table id="top">
          <tr>
            <td class="modal-left">
                <a href="https://mobile.twitter.com/" class="brandmark"><img alt="Twitter" height="28" src="https://ma-0.twimg.com/twitter-mobile/9635f415e4f7c5dce70ba32d2e0b334893f16c7f/images/sprites/larry_28px.gif" /></a>
              <span class="title">Sign in to Twitter!</span>
            </td>
            <td class="modal-right">
                <a href="/signup" class="w-button-common w-button-bright">Sign up</a>
            </td>
          </tr>
        </table>
      </div>
      <div id="main_content">
        <div class="header">
  <h2>Sign in</h2>
</div>
<div class="body">
  <form action="https://mobile.twitter.com/session" method="post">
    <span class="m2-auth-token"><input name="authenticity_token" type="hidden" value="2a59e713a2fba05f4ac9" /></span>
    <fieldset class="inputs">
      <label for="username">Username</label>
      <div class="input-wrapper">
        <input autocapitalize="off" autocorrect="off" class="text-input" name="username" id="username" placeholder="" type="text" value="">
      </div>
      <div class="hint">You can also use your email address</div>
      <label for="password">Password</label>
      <div class="input-wrapper">
        <input autocomplete="off" class="text-input" name="password" id="password" placeholder="" type="password" value="">
      </div>
    </fieldset>
    <span class="w-button-common w-button-bright"><input name="commit" type="submit" value="Sign in" /></span>
  </form>
</div>


      </div>
<div id="footer">
    <div class="signup">
      <a href="/account/resend_password">Forgot password?</a>
    </div>
    <div class="signup">
      <a href="/signup/sms">Already using Twitter via SMS?</a>
    </div>
</div>
    </div>
  </body>
</html>

所以有4个输入参数：用户名，密码，access_token和提交。表单的操作网址为https://mobile.twitter.com/session

以下是我尝试登录的代码：

String url = "https://mobile.twitter.com/session/new";
    String url2 = "https://mobile.twitter.com/session";

DefaultHttpClient httpclient = new DefaultHttpClient();
CookieStore cookieStore = new BasicCookieStore();
HttpContext httpContext = new BasicHttpContext();
httpContext.setAttribute(ClientContext.COOKIE_STORE, cookieStore);
HttpGet httpget = new HttpGet(url);
HttpResponse response = httpclient.execute(httpget);
HttpEntity entity = response.getEntity();

String formCode = EntityUtils.toString(entity);

Map<String,String> formParams =getFormParams(formCode, "myusername", "mypwd");




// make sure cookies is turn on
CookieHandler.setDefault(new CookieManager());

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


HttpPost httpost = new HttpPost(url2);

List <NameValuePair> nvps = new ArrayList <NameValuePair>();
for(String s:formParams.keySet()){
    nvps.add(new BasicNameValuePair(s, formParams.get(s)));
}

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

response = httpclient.execute(httpost,httpContext);
entity = response.getEntity();
formCode = EntityUtils.toString(entity);
Log.d("FormCode",formCode);

（请注意，在上面的代码中，getFormParams返回正确的映射，即具有4个名称值对的映射）。我正在从android执行此代码。当我在Log.debug中看到FormCode时，我总是得到相同的主页。两种情况下的response.getStatusLine（）。getStatusCode（）返回200.那么为什么我总是登陆登录页面？我已经花了一个多星期来解决这个问题。我迫切需要一些帮助......

Answer 1

我不能在中国使用Twitter。但我可以告诉你我登录WebChat的代码。我认为他们将是一样的。关键是你每次都使用相同的客户端和上下文。标题应该设置合适。我将clint和context设置为静态，所以我可以随时随地使用它们，这只是为了简单。

public static HttpClient client;
    public static HttpContext context;
    //登陆
    static{
        String username = "ewallet2014@qq.com";
        String password = "xxxxxxxx";
        try {
            dologin(username, password);
        } catch (Exception e) {
            e.printStackTrace();
        }
    }

    public  static void dologin(String name, String passwd) throws Exception {
        // TODO Auto-generated method stub
        client = new DefaultHttpClient();
        context = new BasicHttpContext();
        HttpPost post = new HttpPost(
                "https://mp.weixin.qq.com/cgi-bin/login?lang=zh_CN");
        List<NameValuePair> params = new ArrayList<NameValuePair>();
        params.add(new BasicNameValuePair("username", name));
        params.add(new BasicNameValuePair("pwd", passwd));
        params.add(new BasicNameValuePair("f", "json"));

        post
                .setHeader("Accept-Language",
                        "zh-CN,zh;q=0.8,en-US;q=0.6,en;q=0.4");
        post
                .setHeader("accept",
                        "application/json, text/javascript, */*; q=0.01");
        post
                .setHeader(
                        "user-agent",
                        "Mozilla/5.0 (Windows NT 6.1; WOW64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/30.0.1599.101 Safari/537.36");
        post.setHeader("Origin","https://mp.weixin.qq.com");
        post.setHeader("Referer","https://mp.weixin.qq.com/");
        post.setHeader("Origin","https://mp.weixin.qq.com");

        post.setEntity(new UrlEncodedFormEntity(params, HTTP.UTF_8));

        HttpResponse response = client.execute(post, context);
        int stat = response.getStatusLine().getStatusCode();
        //System.out.println(stat);

        if (stat == 200) {
            String str = EntityUtils.toString(response.getEntity());
            System.out.println(str);
            //token = StringUtils.substringBetween(str, "token=", "\"}");
            System.out.println("login success! ");
        }
        EntityUtils.consume(response.getEntity());
    }

无法使用Apache HttpClient库登录

1 个答案: