Question

我正在制作基本的登录应用程序，下面的代码显示单击登录按钮时的条件

btn_login.setOnClickListener(new View.OnClickListener() {
            @Override
            public void onClick(View v) {
                // TODO Auto-generated method stub
                final String username = txt_username.getText().toString();
                final String password = txt_password.getText().toString();
                // mengecek kolom yang kosong
                if (username.trim().length() > 0 && password.trim().length() > 0) {
                    if (conMgr.getActiveNetworkInfo() != null
                            && conMgr.getActiveNetworkInfo().isAvailable()
                            && conMgr.getActiveNetworkInfo().isConnected()) {
                        checkLogin(username, password);
                    } else {
                        Toast.makeText(getApplicationContext() ,"No Internet Connection", Toast.LENGTH_LONG).show();
                    }
                } else {
                    // Prompt user to enter credentials
                    Toast.makeText(getApplicationContext() ,"Kolom tidak boleh kosong", Toast.LENGTH_LONG).show();
                }
            }
        });`

然后进行checklogin程序

 private void checkLogin(final String username, final String password) {
    pDialog = new ProgressDialog(this);
    pDialog.setCancelable(false);
    pDialog.setMessage("Logging in ...");
    showDialog();
    txt_username.setText("");
    txt_password.setText("");

    StringRequest strReq = new StringRequest(Request.Method.POST, url, new Response.Listener<String>() {

        @Override
        public void onResponse(String response) {
            Log.e(TAG, "Login Response: " + response.toString());
            hideDialog();

            try {
                JSONObject jObj = new JSONObject(response);
                success = jObj.getInt(TAG_SUCCESS);

                // Check for error node in json
                if (success == 1) {
                    String username = jObj.getString(TAG_USERNAME);
                    String id = jObj.getString(TAG_ID);

                    Log.e("Successfully Login!", jObj.toString());

                    Toast.makeText(getApplicationContext(), jObj.getString(TAG_MESSAGE), Toast.LENGTH_LONG).show();

                    // menyimpan login ke session
                    SharedPreferences.Editor editor = sharedpreferences.edit();
                    editor.putBoolean(session_status, true);
                    editor.putString(TAG_ID, id);
                    editor.putString(TAG_USERNAME, username);
                    editor.commit();

                    // Memanggil main activity
                    Intent intent = new Intent(Login.this, MainActivity.class);
                    intent.putExtra(TAG_ID, id);
                    intent.putExtra(TAG_USERNAME, username);
                    finish();
                    startActivity(intent);
                } else {
                    Toast.makeText(getApplicationContext(),
                            jObj.getString(TAG_MESSAGE), Toast.LENGTH_LONG).show();

                }
            } catch (JSONException e) {
                // JSON error
                e.printStackTrace();
            }

        }
    }, new Response.ErrorListener() {

        @Override
        public void onErrorResponse(VolleyError error) {
            Log.e(TAG, "Login Error: " + error.getMessage());
            Toast.makeText(getApplicationContext(),
                    error.getMessage(), Toast.LENGTH_LONG).show();

            hideDialog();

        }
    }) {

        @Override
        protected Map<String, String> getParams() {
            // Posting parameters to login url
            Map<String, String> params = new HashMap<String, String>();
            params.put("username", username);
            params.put("password", password);

            return params;
        }
    };

    // Adding request to request queue
    AppController.getInstance().addToRequestQueue(strReq, tag_json_obj);
}

当我第一次点击登录时，我没有问题，它可以读取我服务器中的PHP文件，然后执行请求，但是当我第二次点击时，它无法执行请求并且没有t显示登录按摩，我的应用程序只显示先前请求的结果。

Answer 1

首先，我必须提到默认情况下，Volley会自动缓存您的请求。

一个解决方案是：

// Create your request
StringRequest strReq = new StringRequest(Request.Method.POST.......

// Disable the cache option before you add it to the queue.
strReq.setShouldCache(false);

// Adding request to request queue
AppController.getInstance().addToRequestQueue(strReq, tag_json_obj);

不能做两次StringRequest

1 个答案: