使用Symfony API在Android上保存Http会话

使用Symfony API在Android上保存Http会话,android,symfony,http,session,cookies,Android,Symfony,Http,Session,Cookies,我正在尝试向在Symfony中开发的API发送一些请求 我使用以下方式通过它进行连接: if (param.getRequestType().equals("post")) { HttpPost httppost = new HttpPost(url); try { httppost.setEntity(new UrlEncodedFormEntity(param.getParam()));

我正在尝试向在Symfony中开发的API发送一些请求

我使用以下方式通过它进行连接:

   if (param.getRequestType().equals("post")) {
            HttpPost httppost = new HttpPost(url);
            try {
                httppost.setEntity(new UrlEncodedFormEntity(param.getParam()));
                HttpResponse res = httpclient.execute
                        (httppost);
                return EntityUtils.toString(res.getEntity());
            } catch (Exception e) {
                e.printStackTrace();
            }
        }
        return null;
    }
它运行良好,API返回我已连接的状态

然后我想使用另一个API请求(get),但我的会话似乎没有保存,而且永远不会工作

get请求如下所示:

@Override
    protected String doInBackground(String... uri) {
        CookieStore cookieStore = new BasicCookieStore();
        HttpContext localContext = new BasicHttpContext();
        localContext.setAttribute(ClientContext.COOKIE_STORE, cookieStore);
        HttpClient httpclient = new DefaultHttpClient();
        HttpGet httpget = new HttpGet("http://***.**/***/faq");
        String responseString = null;
        try {
            HttpResponse response = httpclient.execute(httpget, localContext);
            StatusLine statusLine = response.getStatusLine();
            ByteArrayOutputStream out = new ByteArrayOutputStream();
            response.getEntity().writeTo(out);
            responseString = out.toString();
            System.out.println("HERE --> " + responseString);
            out.close();

        } catch (IOException e) {
            e.printStackTrace();
        }
        return "error";
    }
我认为这是因为Android不能让我的会话保持活跃。我说得对吗?我怎么能保留它


Symfony正在使用Fosuserbundle作为登录部分。

我编辑了我的问题:我修改了get请求以使用CookieStore。但我仍然有完全相同的问题我编辑了我的问题:我修改了get请求以使用CookieStore。但我仍然有完全相同的问题