android - 在 Android 中维护用户会话

Question

我正在尝试为 Android 制作社交网络应用程序。我的问题是用户登录时如何维护用户会话？

请帮我找到上述问题的解决方案。

score 8 · Accepted Answer

尝试

public class Session {
private static String sessionId;
private static String userRole;

public static void setSessionId(String sessionId) {
    Session.sessionId = sessionId;
}

public static String getSessionId() {
    return sessionId;
}

}

使用此类并将其导入其他所有活动。您可以定义自己的函数来维护您的特定会话数据

score 0 · Accepted Answer

使用 SharedPreferences 创建会话。

public class Session {

    private SharedPreferences prefs;

    public Session(Context cntx) {
        // TODO Auto-generated constructor stub
        prefs = PreferenceManager.getDefaultSharedPreferences(cntx);
    }

    public void setusename(String usename) {
        prefs.edit().putString("usename", usename).commit();

    }

    public String getusename() {
        String usename = prefs.getString("usename","");
        return usename;
    }
}

现在，当你想像这样使用这个类时，在制作这个类之后，让这个类成为对象

private Session session;//global variable 
session = new Session(cntx); //in oncreate 
//and now we set sharedpreference then use this like

session.setusename("USERNAME");
now when ever u want to get username then same work for session object and call this

session.getusename();

祝你好运:) 密码相同

score 0 · Accepted Answer

http://www.devahead.com/blog/2011/06/extending-the-android-application-class-and-dealing-with-singleton/

请看上面的链接。它的细节相当好。

使用单例来维护用户会话。

score 0 · Accepted Answer

我DefaultHttpClient使用HttpRequestInterceptor和HttpResponseInterceptor。与此类似的东西：

public class HTTPClients {

    private static DefaultHttpClient _defaultClient;
    private static String session_id;
    private static HTTPClients _me;
    private HTTPClients() {

    }
    public static DefaultHttpClient getDefaultHttpClient(){
        if ( _defaultClient == null ) {
            _defaultClient = new DefaultHttpClient();
            _me = new HTTPClients();
            _defaultClient.addResponseInterceptor(_me.new SessionKeeper());
            _defaultClient.addRequestInterceptor(_me.new SessionAdder());
        }
        return _defaultClient;
    }

    private class SessionAdder implements HttpRequestInterceptor {

        @Override
        public void process(HttpRequest request, HttpContext context)
                throws HttpException, IOException {
            Log.d("SessionKeeper", "Adding session with the following string: " + session_id);
            if ( session_id != null ) {
                request.setHeader("Cookie", session_id);
            }
        }

    }

    private class SessionKeeper implements HttpResponseInterceptor {

        @Override
        public void process(HttpResponse response, HttpContext context)
                throws HttpException, IOException {
            Header[] headers = response.getHeaders("Set-Cookie");
            if ( headers != null && headers.length == 1 ){
                Log.d("SessionKeeper", "Keeping session with the following string: " + headers[0].getValue());
                session_id = headers[0].getValue();
            }
        }

    }
}

score 0 · Accepted Answer

当我尝试发送该会话 id 时，我在我的 android 客户端遇到了类似的问题，服务器端正在创建一个新会话......但是你在 android 客户端检查你没有创建 DefaulthttpClient 两次......只创建一次 httpclient 说主要活动并传递其他活动中的对象......不要创建第二个 HttpClient

android - 在 Android 中维护用户会话

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