Android HttpClient会话后问题？如何知道会话是否已创建？

这是的后续问题，我已成功发布到服务器并收到200个代码，但当我尝试移动到另一个页面时，它无法识别我已登录。我想知道这是否是一个会话问题，或者我是否需要在发布后遵循重定向。我将如何执行重定向？再次感谢您的帮助。下面是一个简单的HttpClient/POST应用程序，我通过示例创建它，以帮助我快速测试任何更改

public class HttpClientTest extends Activity{

HttpClient client = new DefaultHttpClient();

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

    final Button btnFetch = (Button)findViewById(R.id.button);
    final TextView txtResult = (TextView)findViewById(R.id.content);
    final Button login = (Button)findViewById(R.id.button2);


    btnFetch.setOnClickListener(new Button.OnClickListener(){
        public void onClick(View v){
            getRequest(txtResult);
        }
    });

    login.setOnClickListener(new Button.OnClickListener(){
        public void onClick(View v){
            try {
                login(txtResult);
            } catch (UnsupportedEncodingException e) {
                // TODO Auto-generated catch block
                e.printStackTrace();
            } catch (IOException e) {
                // TODO Auto-generated catch block
                e.printStackTrace();
            }
        }
    });
}

public void getRequest(TextView txtResult){
    HttpGet request = new HttpGet("http://gc.gamestotal.com/i.cfm?f=com_empire&cm=3");
    try{
        HttpResponse response = client.execute(request);
        txtResult.setText(Parser.request(response));
    }catch(Exception ex){
        txtResult.setText("Failed!");
    }
}

public void login(TextView txtResult) throws UnsupportedEncodingException, IOException{
    String action = "i.cfm?&1028&p=login&se=4";
    String yourServer = "http://gc.gamestotal.com/";
    HttpPost post = new HttpPost(yourServer + action);
    List<NameValuePair> params = new ArrayList<NameValuePair>();
    params.add(new BasicNameValuePair("nic", "user"));
    params.add(new BasicNameValuePair("password", "password"));
    params.add(new BasicNameValuePair("server", "4"));
    UrlEncodedFormEntity entity = new UrlEncodedFormEntity(params, "UTF-8");
    post.setEntity(entity);

    try{
        HttpResponse response = client.execute(post);
        txtResult.setText(response.getEntity().toString());
    }catch(Exception ex){
        txtResult.setText("Failed!");
    }
}

}

---

我首先按下UI上的登录按钮，该按钮给出Http/1.1 200 OK响应代码，但当我按下btnFetch按钮，该按钮将我发送到一个页面，您必须登录才能访问该页面时，我会看到未登录页面。有什么想法吗？

经过大量的研究和示例之后，我终于登录到了我尝试的网站。显然，这是一个问题，我没有消费的实体从回应，即饼干。我创建了一个简单的活动，其主要目的是获取并发布到站点，然后将结果输出到LogCat。也许这可以帮助其他人

public class HttpClientTest extends Activity{

DefaultHttpClient client = new DefaultHttpClient();
HttpGet request;
HttpEntity entity;
List<Cookie> cookies;
HttpResponse response;
HttpPost post;

@Override
public void onCreate(Bundle savedInstanceState) {
    super.onCreate(savedInstanceState);
    try {
        getRequest();
    } catch (Exception e) {
        Log.d("My Activity", "Failed");
        e.printStackTrace();
    }
}

public void getRequest() throws Exception
{
    final String TAG = "MyActivity";
    request = new HttpGet("http://some.site.com/");
    response = client.execute(request);
    entity = response.getEntity();
    Log.d(TAG, "Login form get: " + response.getStatusLine());
    if(entity != null)
    {
        entity.consumeContent();
    }
    Log.d(TAG, "Initial set of cookies:");

    cookies = client.getCookieStore().getCookies();
    if (cookies.isEmpty())
    {
        Log.d(TAG, "None");
    }
    else
    {
        for(int i = 0; i<cookies.size(); i++)
        {
            Log.d(TAG, "- " + cookies.get(i));
        }
    }
    String action = "i.cfm?&1028&p=login&se=4";
    String yourServer = "http://some.site.com/";
    post = new HttpPost(yourServer + action);

    List<NameValuePair> params = new ArrayList<NameValuePair>();
    params.add(new BasicNameValuePair("nic", "username"));
    params.add(new BasicNameValuePair("password", "password"));
    params.add(new BasicNameValuePair("server", "4"));

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


    response = client.execute(post);
    entity = response.getEntity();

    Log.d(TAG, "Login form get: " + response.getStatusLine());
    if(entity != null){
        entity.consumeContent();
    }

    Log.d(TAG, "Post logon cookies:");
    cookies = client.getCookieStore().getCookies();
    if (cookies.isEmpty())
    {
        Log.d(TAG, "None");
    } 
    else
    {
        for (int i = 0; i < cookies.size(); i++) 
        {
            Log.d(TAG, "- " + cookies.get(i));
        }
    }


    request = new HttpGet("http://some.site.com/i.cfm?f=com_empire&cm=3");

    response = client.execute(request);
    Log.d(TAG, "Check for login: " + Parser.request(response));
    if(entity != null)
    {
        entity.consumeContent();
    }

}

}

---

最后一个日志log.dTAG，检查登录：+Parser.requestresponse；通过一个解析器类打印出该站点的html，我用它来验证它实际上是一个需要成功登录的页面。经过大量研究和许多示例之后，我终于成功登录到了我尝试的站点。显然，这是一个问题，我没有消费的实体从回应，即饼干。我创建了一个简单的活动，其主要目的是获取并发布到站点，然后将结果输出到LogCat。也许这可以帮助其他人

public class HttpClientTest extends Activity{

DefaultHttpClient client = new DefaultHttpClient();
HttpGet request;
HttpEntity entity;
List<Cookie> cookies;
HttpResponse response;
HttpPost post;

@Override
public void onCreate(Bundle savedInstanceState) {
    super.onCreate(savedInstanceState);
    try {
        getRequest();
    } catch (Exception e) {
        Log.d("My Activity", "Failed");
        e.printStackTrace();
    }
}

public void getRequest() throws Exception
{
    final String TAG = "MyActivity";
    request = new HttpGet("http://some.site.com/");
    response = client.execute(request);
    entity = response.getEntity();
    Log.d(TAG, "Login form get: " + response.getStatusLine());
    if(entity != null)
    {
        entity.consumeContent();
    }
    Log.d(TAG, "Initial set of cookies:");

    cookies = client.getCookieStore().getCookies();
    if (cookies.isEmpty())
    {
        Log.d(TAG, "None");
    }
    else
    {
        for(int i = 0; i<cookies.size(); i++)
        {
            Log.d(TAG, "- " + cookies.get(i));
        }
    }
    String action = "i.cfm?&1028&p=login&se=4";
    String yourServer = "http://some.site.com/";
    post = new HttpPost(yourServer + action);

    List<NameValuePair> params = new ArrayList<NameValuePair>();
    params.add(new BasicNameValuePair("nic", "username"));
    params.add(new BasicNameValuePair("password", "password"));
    params.add(new BasicNameValuePair("server", "4"));

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


    response = client.execute(post);
    entity = response.getEntity();

    Log.d(TAG, "Login form get: " + response.getStatusLine());
    if(entity != null){
        entity.consumeContent();
    }

    Log.d(TAG, "Post logon cookies:");
    cookies = client.getCookieStore().getCookies();
    if (cookies.isEmpty())
    {
        Log.d(TAG, "None");
    } 
    else
    {
        for (int i = 0; i < cookies.size(); i++) 
        {
            Log.d(TAG, "- " + cookies.get(i));
        }
    }


    request = new HttpGet("http://some.site.com/i.cfm?f=com_empire&cm=3");

    response = client.execute(request);
    Log.d(TAG, "Check for login: " + Parser.request(response));
    if(entity != null)
    {
        entity.consumeContent();
    }

}

}

---

最后一个日志log.dTAG，检查登录：+Parser.requestresponse；通过一个解析器类打印出该站点的html，我用它来验证它实际上是一个需要成功登录的页面

我知道，有点晚了，但我也遇到了这篇文章，我想在这里找到这个Ansare的链接：它说你应该重用处理会话管理的HttpClient，也就是说，将HttpClient设置为静态并只实例化一次。 但我不知道这是否是你要求的全部事实。对我来说是这样

---

干杯

我知道，有点晚了，但我也看到了这篇文章，我想在这里找到这个Ansare的链接：上面说你应该重用处理会话管理的HttpClient，也就是说，将HttpClient设置为静态并只实例化一次。 但我不知道这是否是你要求的全部事实。对我来说是这样

---

干杯

你需要问谁编写了你正在访问的Web应用程序。只有他们能告诉你他们想让你怎么做你想做的事。我害怕这样。感谢您的回复。可能是一些cookie问题。。请检查一下，我正在研究饼干，我开始认为这可能是一个参考问题。我将尝试编辑标题，以完全模仿Firefox。你需要问这个问题谁编写了你正在访问的Web应用程序。只有他们能告诉你他们想让你怎么做你想做的事。我害怕这样。感谢您的回复。可能是一些cookie问题。。请检查一下，我正在研究饼干，我开始认为这可能是一个参考问题。我将尝试编辑标题以完全模仿Firefox。