关于android给服务器传值问题

Question

程序中通过activity的EditText输入的值传入服务器的action中，判断登录，可是一直提示失败，请各位大侠看看。
loginBtn.setOnClickListener(new OnClickListener() {

        @Override
        public void onClick(View arg0) { 
            String path="http://10.0.2.2:8080/KOA_Product/execute";
            new UsersAsyncTask().execute(path);
        }      
    });
    private class UsersAsyncTask extends AsyncTask<String, Void, String>{

    @Override
    //子线程中写方法
    protected String doInBackground(String... arg0) {
        String path=arg0[0];
        HashMap<String, String> params=new HashMap<String, String>();
        params.put("name",nameEt.getText().toString() );
        params.put("password",passwordEt.getText().toString());
        params.put("type","mobile");
        String jsonInfo=HttpUtils.submitPostData(path, params, "GBK");
    return  jsonInfo;
    }
    @Override
    protected void onPostExecute(String result) {
        super.onPostExecute(result);
        UserInfoDto user=null;
        try {
            user = getUserByJsonData(result);
        } catch (JSONException e1) {
            // TODO Auto-generated catch block
            e1.printStackTrace();
        }
        if (user!=null) {
            Toast.makeText(LoginActivity.this, "登录成功", Toast.LENGTH_SHORT).show();
            if (remberPasswordCb.isChecked()) {
                //记住用户名和密码
                Editor editor=sp.edit();
                editor.putString("name",nameEt.getText().toString() );
                editor.putString("password", passwordEt.getText().toString());
                editor.commit();
            }
            //跳转界面
            Intent intent=new Intent(LoginActivity.this, MainActivity.class);
            startActivity(intent);
        }else{
            Toast.makeText(LoginActivity.this, "用户名或者密码不正确！",Toast.LENGTH_SHORT).show();
        }
    }
  }

Answer 1

看一下这个地方

user = getUserByJsonData(result);

user的返回值是多少，怀疑是这个地方出现异常了