Java 登录后获取角色

当我发送登录后数据时，使用如下登录按钮：

loginbutton.setOnClickListener(new OnClickListener() {
        @Override
        public void onClick(View v) {
            // TODO Auto-generated method stub
            final String user = email.getText().toString().trim();
            final String pwd = password.getText().toString().trim();

            if (email.getText().toString().equals("")) {
                Utils.toast(context, "Username empty...");
            } else if (password.getText().toString().equals("")) {
                Utils.toast(context, "Password empty...");
            } else if (!isValidEmail(user)) {
                email.setError("Invalid Email");
            } else if (user.length() < 2) {
                Utils.toast(context, "Username to short...");
            } else if (pwd.length() < 2) {
                Utils.toast(context, "Password to short...");
            } else if (!isValidPassword(pwd)) {
                password.setError("Invalid Password");
            } else {
                progress.setVisibility(View.VISIBLE);
                SendfeedbackJob job = new SendfeedbackJob();
                job.execute(user, pwd);
            }
        }
    });

private class SendfeedbackJob extends AsyncTask<String, Void, String> {
    private static final String LOG_TAG = "UserLoginTask";
    @Override
    protected String doInBackground(String... params) {
        String user = params[0];
        String pwd = params[1];
        // do above Server call here
        ArrayList<NameValuePair> postParameters = new ArrayList<NameValuePair>();
        postParameters.add(new BasicNameValuePair("email", user ));
        postParameters.add(new BasicNameValuePair("password", pwd ));

        String responseString = null;

        try
        {
            HttpClient httpclient = new DefaultHttpClient();
            HttpPost httppost = new HttpPost("http://192.168.0.219:90/auth/login");

            // no idea what this does :)
            httppost.setEntity(new UrlEncodedFormEntity(postParameters));

            // This is the line that send the request
            HttpResponse response = httpclient.execute(httppost);

            HttpEntity entity = response.getEntity();
            String responseAsText = EntityUtils.toString(response.getEntity());
            Utils.log("daftar isi: " + responseAsText);
            JSONObject loginjson = new JSONObject(responseAsText);
            Utils.log("json object: " + loginjson);
            String roleString = loginjson.getString("role");
            Utils.log("role: " + roleString);
            if(roleString.equals("member")){
                Intent intent = new Intent(context, home.class);
                startActivity(intent);
            }else if(roleString.equals("studio")){
                Intent intent = new Intent(context, VendorDashboard.class);
                startActivity(intent);
            }
        }
        catch (Exception e)
        {
            Log.e(LOG_TAG, String.format("Error during login: %s", e.getMessage()));
        }
        return "processing";
    }

    @Override
    protected void onPostExecute(String message) {
        //process message
    }
}

{"token":"eyJ0eXAiOiJKV1QiLCJhbGciOiJIUzI1NiJ9.eyJzdWIiOiIxNCIsImlzcyI6Imh0dHA6XC9cLzE5Mi4xNjguMC4yMTk6OTBcL2F1dGhcL2xvZ2luIiwiaWF0IjoiMTQ0NTI0NzIyMSIsImV4cCI6IjE0NDY0NTY4MjEiLCJuYmYiOiIxNDQ1MjQ3MjIxIiwianRpIjoiNTY4MDEyZDUwNTg1NDFjM2UzNGVjMGViYzMzZDkzMGQifQ.zw5c5kLIlvPYIMhEzEnF_fCOu77XTq2prcDtSHJY7bk","role":"studio"}

如何获取角色（JSON响应反馈），以便在执行HTTP Post请求后使用if和elseif语句（在Utils.toast（context，responseEastText）；）之后）

在我使用asynctask后更新，它不会在单击时产生错误。 更新完成后。谢谢你，什里亚什先生

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问题结束。

您不能在UI线程上运行长时间运行的任务。您需要使用线程或异步任务…重复的…请描述正确的异步任务，我不理解。这样地？私有类SendfeedbackJob扩展了AsyncTask保护的String doInBackground（String user，String pwd），我这样调用onclick:SendfeedbackJob job=new SendfeedbackJob（）；作业执行（用户，pwd）；请参阅如何调用异步任务。因为有很多方法可以定义。我还更新了代码。像那样吗，什里亚什先生？但我仍然无法获得json对象角色。