我是Android的新手(也是Java),如果我的问题是一个基本命题,那就很抱歉！
我必须编写一个Android应用程序,在后台登录到aspx网页,从中获取一些数据,然后从网页中注销. (以编程方式完成所有操作)

基本上,该过程喜欢从Gmail获取电子邮件列表：
1.转到“https://mail.google.com”,然后登录
2.点击“联系人”(==转到“https://mail.google.com/mail/?shva=1\u0026amp;zx=dzi4xmuko5nz#contacts”)
3.使用HttpsURLConnection(或类似的东西)获取页面,并在(例如Map或String)对象中获取电子邮件
4.单击“注销”链接

我希望,这是可以理解的.看互联网,我找到了解决方案只有“取出部分”,所以这不是问题.但我对“点击部分”一无所知.

  ......
    // Get the connection
    URL myurl = new URL("https://mail.google.com");
    HttpsURLConnection con = (HttpsURLConnection) myurl.openConnection();

    // complete the fields
    con.setRequestProperty("Email","myacc");
    con.setRequestProperty("Passwd","mypass");

    /* 
     * in this part, should make sign in, and go directly to contacts... 
     * I don't have any idea how to do it...
     */

    // for the present, just write out the data
    InputStream ins = con.getInputStream();
    BufferedReader in = new BufferedReader(new InputStreamReader(ins));

    String inputLine;
    while ((inputLine = in.readLine()) != null) {
        Log.d("Page:"," "+inputLine);
    }

    in.close();

    /*
     * And here should be the "Sign out" part
     */
  ......

任何帮助都会很棒,谢谢你！
(对不起,如果我的英语不太好……)

编辑：问题解决了.谢谢！

 .......    
    String GMAIL_CONTACTS = "https://mail.google.com/mail/?shva=1#contacts";
    String GMAIL_LOGIN = "https://mail.google.com";

        DefaultHttpClient httpClient = new DefaultHttpClient();
        HttpPost httpPost = new HttpPost(GMAIL_LOGIN);

        List<NameValuePair> nameValuePairs = new ArrayList<NameValuePair>(3);
        nameValuePairs.add(new BasicNameValuePair("Email", MY_ACC));
        nameValuePairs.add(new BasicNameValuePair("Passwd", MY_PASS));
        nameValuePairs.add(new BasicNameValuePair("signIn", "Sign In"));

        httpPost.setEntity(new UrlEncodedFormEntity(nameValuePairs));

        // Execute HTTP Post Request 
        HttpResponse response = httpClient.execute(httpPost);
        Log.d(TAG, "response stat code " + response.getStatusLine().getStatusCode());

        if (response.getStatusLine().getStatusCode() < 400) {

            String cookie = response.getFirstHeader("Set-Cookie")
                    .getValue();
            Log.d(TAG, "cookie: " + cookie);

            // get the contacts page 
            HttpGet getContacts = new HttpGet(GMAIL_CONTACTS);
            getContacts.setHeader("Cookie", cookie);
            response = httpClient.execute(getContacts);

            InputStream ins = response.getEntity().getContent();
            BufferedReader in = new BufferedReader(new InputStreamReader(
                    ins));

            String inputLine;
            while ((inputLine = in.readLine()) != null) {
                Log.d(TAG, " " + inputLine);
            }

            in.close();
        } else {
            Log.d(TAG, "Response error: "
                    + response.getStatusLine().getStatusCode());
        }
 .......

解决方法:

“点击”基本上是向服务器发送请求并显示返回信息.

1 /找出要为该请求调用的URL(如果是网页,请参阅firebug)

2 /找出参数是什么,找出方法是GET还是POST

3 /以编程方式重现.

4 /“登录”阶段可能意味着使用cookie,服务器为您提供,并且您必须在每次请求后发回

但是,你的方法是错误的.您不应该尝试通过网址连接直接登录谷歌. (你也应该使用HttpClient).而且,请求属性不是参数.它们是标题.

我强烈建议你从简单的东西开始,以便在java,GET,POST,参数,标题,响应,cookie中熟悉HTTP …

编辑

收到回复后,您需要检查一下

response.getStatusLine().getStatusCode() < 400

它会告诉您登录成功. (2xx成功,3xx被移动等等.4xx是请求中的错误,5xx是服务器端错误; Gmail响应302登录以建议重定向到收件箱).然后,您会注意到响应“Set-Cookie”中有一个特定的标题,其中包含您想要进一步连接的cookie,因此：

String cookie = response.getFistHeader("Set-Cookie");

然后,您应该能够调用请求来获取联系人：

HttpGet getContacts = new HttpGet(GMAIL_CONTACTS);
getContacts.setHeader("Cookie", cookie);
response = httpClient.execute(getContacts);
InputStream ins = response.getEntity().getContent();

它应该是那样的.