How to print out data within function beforeAction? I want to make some verification before each action in a controller, therefore if some condition occurs in beforeAction I should print out data and prevent further execution, for example, JSON:
[
status: "error",
msg: "access denied"
]
I try to even inner redirect to another controller, but it doesn't work.
public function beforeAction($action)
{
$request = Yii::$app->request;
if ( ! checkByToken($request->get('token')) && $this->getRoute() != 'web/abonent/token_error') {
\Yii::$app->runAction('web/abonent/token_error');
return true;
}
return parent::beforeAction($action); // TODO: Change the autogenerated stub
}
But maybe there an another concept of doing so. I just need to check the condition before any actions and print our result or let the action execute.
To prevent further execution:
public function beforeAction($action) {
return false; // key point
}
To print out data within beforeAction:
public function beforeAction($action) {
// set response format = json:
Yii::$app->response->format = Response::FORMAT_JSON;
// then, set the response data:
Yii::$app->response->data = [
'status' => 'error',
'msg' => 'access denied'
];
return false;
}
I think will be better
public function beforeAction($action)
{
$request = Yii::$app->request;
if ( ! checkByToken($request->get('token')) && $this->getRoute() != 'web/abonent/token_error') {
$action = 'error';
}
return parent::beforeAction($action); // TODO: Change the autogenerated stub
}
Action name must be 'actionError'
Related
I am trying to setup google indexing api in codeigniter, I have done all steps on google cloud and search console part.
It works, but returning success message on all options event when url is not submited, that is why I want to get exact response from google instead of a created success message.
How can I display exact response from google return $stringBody;? or check for the correct response ?
Here is my controller :
namespace App\Controllers;
use App\Models\LanguageModel;
use App\Models\IndexingModel;
class IndexingController extends BaseController
{
public function initController(\CodeIgniter\HTTP\RequestInterface $request, \CodeIgniter\HTTP\ResponseInterface $response, \Psr\Log\LoggerInterface $logger)
{
parent::initController($request, $response, $logger);
$this->indexingModel = new IndexingModel();
}
public function GoogleUrl()
{
checkPermission('indexing_api');
$data['title'] = trans("indexing_api");
$data["selectedLangId"] = inputGet('lang');
if (empty($data["selectedLangId"])) {
$data["selectedLangId"] = $this->activeLang->id;
}
echo view('admin/includes/_header', $data);
echo view('admin/indexing_api', $data);
echo view('admin/includes/_footer');
}
/**
* indexing Tools Post
*/
public function indexingToolsPost()
{
checkPermission('indexing_api');
$slug = inputPost('slug');
$urltype = inputPost('urltype');
$val = \Config\Services::validation();
$val->setRule('slug', trans("slug"), 'required|max_length[500]');
if (!$this->validate(getValRules($val))) {
$this->session->setFlashdata('errors', $val->getErrors());
return redirect()->to(adminUrl('indexing_api?slug=' . cleanStr($slug)))->withInput();
} else {
$this->indexingModel->AddUrlToGoogle($slug, $urltype);
$this->session->setFlashdata('success', trans("msg_added"));
resetCacheDataOnChange();
return redirect()->to(adminUrl('indexing_api?slug=' . cleanStr($slug)));
}
$this->session->setFlashdata('error', trans("msg_error"));
return redirect()->to(adminUrl('indexing_api?slug=' . cleanStr($slug)))->withInput();
}
}
And This is my model :
namespace App\Models;
use CodeIgniter\Model;
use Google_Client;
class IndexingModel extends BaseModel {
public function AddUrlToGoogle($google_url, $Urltype){
require_once APPPATH . 'ThirdParty/google-api-php-client/vendor/autoload.php';
$client = new Google_Client();
$client->setAuthConfig(APPPATH . 'ThirdParty/google-api-php-client/xxxxxxxxx.json');
$client->addScope('https://www.googleapis.com/auth/indexing');
$httpClient = $client->authorize();
$endpoint = 'https://indexing.googleapis.com/v3/urlNotifications:publish';
$array = ['url' => $google_url, 'type' => $Urltype];
$content = json_encode($array);
$response = $httpClient->post($endpoint,['body' => $content]);
$body = $response->getBody();
$stringBody = (string)$body;
return $stringBody;
}
public function AddUrlToBing($google_url, $Urltype){
}
public function AddUrlToYandex($google_url, $Urltype){
}
}
This is a success response when I try it out of codeigniter and print_r($stringBody);
{ "urlNotificationMetadata": { "url": "https://example.com/some-text", "latestUpdate": { "url": "https://example.com/some-text", "type": "URL_UPDATED", "notifyTime": "2023-01-29T01:51:13.140372319Z" } } }
And this is an error response :
{ "error": { "code": 400, "message": "Unknown notification type. 'type' attribute is required.", "status": "INVALID_ARGUMENT" } }
But In codeigniter I get a text message "url submited" even if url not submited.
Currently you are not handling the actual response of IndexingModel->AddUrlToGoogle(). It seems your code has a validation before, so it claims, if no validation error occurs, its always a success.
So the first question to ask is, why your validation is not working here - or is it?
Secondly you could handle the actual response in any case:
IndexingController
class IndexingController extends BaseController
public function indexingToolsPost()
{
if (!$this->validate(getValRules($val))) {
// validation error
$this->session->setFlashdata('errors', $val->getErrors());
return redirect()->to(adminUrl('indexing_api?slug=' . cleanStr($slug)))->withInput();
} else {
// no validation error
$apiResponseBody = $this->indexingModel->AddUrlToGoogle($slug, $urltype);
if(array_key_exists('error', $apiResponseBody)) {
// its an error!
// either set the actual messsage
$this->session->setFlashdata('error', $apiResponseBody['error']['message']);
// OR translate it
$this->session->setFlashdata('error', trans($apiResponseBody['error']['message']));
} else {
// Its a success!
$this->session->setFlashdata('success', trans("msg_added"));
}
// ...
}
return redirect()->to(adminUrl('indexing_api?slug=' . cleanStr($slug)))->withInput();
}
And in the model, return the response as an array:
IndexingModel
public function AddUrlToGoogle($google_url, $Urltype) {
// ...
$response = $httpClient->post($endpoint,['body' => $content]);
return json_decode($response->getBody() ?? '', true); // return an array
}
I want to ask how to to check the data if we input data if there is the same data that cannot be inserted.
My code:
$obj = new Pengajuan();
// $obj->id_pengajuan = $req->input('kode');
$obj->id_nasabah = $req->input('id_nasabah');
$obj->tgl_pengajuan = $req->input('tgl_pengajuan');
$obj->besar_pinjaman = $req->input('besar_pinjaman');
$obj->status = $req->input('status');
$simpan = $obj->save();
if ($simpan == 1) {
$status = "Tersmpan";
} else {
$status = "Gagal";
}
echo json_encode(array("status" => $status));
Above of the code add a validation like below:
$this->validate([
'id_nasabah' =>'unique:pengajuans'
]) ;
And then rest of your controller code.
try this:
public function store(Request $request)
{
$this->validate($request,[
'id_nasabah'=>'required|exists:your_table_name,id',
'tgl_pengajuan'=>'more_validations',
'besar_pinjaman'=>'more_validations',
]);
//if $this->validate() fails, it will return a response 422 code error
//else it will continue with the creation of your object
//is a good idea to use a try-catch in case of something goes wrong
try
{
$pengajuan=Pengajuan::create($request->only('id_nasabah','tgl_pengajuan','besar_pinjaman'));
return response->json([
'pengajuan'=>$pengajuan,
'status'=>'Tersmpan',
],200);//return a http code 200 ok
}
catch(Exception $e)
{
//'message'=>'this will return what is the error and line'
return response()->json([
'message'=>$e->getMessage().'/'.$e->getLine(),
'status'=>'Gagal',
],422);
}
}
I will try to insert & also update data using session in Codeigniter, but data not inserted into the database even its print save successfully.
Here is my controller:
public function save($user_id)
{
$this->load->model('Users');
$code=$this->input->post('code');
$name=$this->input->post('name');
$address=$this->input->post('address');
$user_data= array(
'code' =>$code,
'name'=>$name,
'address'=>$address,
'active'=>1
);
if($this->Users->save($user_data,$user_id))
{
$this->session->set_flashdata('msg',"save sucesss");
}else {
$this->session->set_flashdata('msg',"not save");
}
redirect('home');
}
& this is my model:
public function save($data,$id)
{
if (id=='') {
// code...
$this->db->insert('user',$data);
return true;
}else
{
$this->db->where('id',$id)
->update('user',$data);
return true;
}
return false;
}
Data insert if I removed if in model!
You have the model always returning true no matter the outcome of the database operation. You should use the return value from insert() or update() so the "message" reports what actually happens.
Note that the argument to save has a default value. Now you can call the save URL without an argument and it will automatically do an insert.
public function save($user_id = NULL)
{
$this->load->model('users');
$user_data = array(
'code' => $this->input->post('code'),
'name' => $this->input->post('name'),
'address' => $this->input->post('address'),
'active' => 1
);
if($this->Users->save($user_data, $user_id))
{
$msg = "save sucesss";
}
else
{
$msg = "not save";
}
$this->session->set_flashdata('msg', $msg);
redirect('home');
}
public function save($data, $id)
{
if(empty($id))
{
// code...
// insert returns TRUE on success, FALSE on failure
return $this->db->insert('user', $data);
}
// update() accepts a third argument, a "where" array
// and returns TRUE on success, FALSE on failure
return $this->db->update('user', $data, array('id' => $id));
}
Now have an accurate report on the database operations.
the first check is data is coming in save controller or not if it's not getting the data then fix it. If coming then pass it in a model in the correct format and it will definitely be inserted in the database.
use following printing data
echo $data;
var_dump($data);
print($data);
print_r($data);
First thing is to rename your model calling eg:
$this->load->model('users');
and use this to call your method:
$this->users->save($user_data,$user_id)
your model should look like this then:
public function save($data, $id) {
if ($id) {
$this->db->where('id', $id)
->update('user', $data);
return true;
}
$this->db->insert('user', $data);
return true;
}
if you want to use your flashdata on the next request, use this:
$this->session->keep_flashdata('item');
$this->session->keep_flashdata(array('item1', 'item2', 'item3'));
because flashdata is only for the next request:
CodeIgniter supports “flashdata”, or session data that will only be available for the next request, and is then automatically cleared.
I'm completely lost as to why this is happening, and it happens about 50% of the time.
I have a check to see if a user exists by email and last name, and if they do, run some code. If the user doesn't exist, then create the user, and then run some code.
I've done various testing with dummy data, and even if a user doesn't exist, it first creates them, but then runs the code in the "if" block.
Here's what I have.
if (User::existsByEmailAndLastName($params->email, $params->lastName)) {
var_dump('user already exists');
} else {
User::createNew($params);
var_dump("Creating a new user...");
}
And here are the respective methods:
public static function existsByEmailAndLastName($email, $lastName) {
return User::find()->where([
'email' => $email,
])->andWhere([
'last_name' => $lastName
])->one();
}
public static function createNew($params) {
$user = new User;
$user->first_name = $params->firstName;
$user->last_name = $params->lastName;
$user->email = $params->email;
$user->address = $params->address;
$user->address_2 = $params->address_2;
$user->city = $params->city;
$user->province = $params->province;
$user->country = $params->country;
$user->phone = $params->phone;
$user->postal_code = $params->postal_code;
return $user->insert();
}
I've tried flushing the cache. I've tried it with raw SQL queries using Yii::$app->db->createCommand(), but nothing seems to be working. I'm totally stumped.
Does anyone know why it would first create the user, and then do the check in the if statement?
Editing with controller code:
public function actionComplete()
{
if (Yii::$app->basket->isEmpty()) {
return $this->redirect('basket', 302);
}
$guest = Yii::$app->request->get('guest');
$params = new CompletePaymentForm;
$post = Yii::$app->request->post();
if ($this->userInfo || $guest) {
if ($params->load($post) && $params->validate()) {
if (!User::isEmailValid($params->email)) {
throw new UserException('Please provide a valid email.');
}
if (!User::existsByEmailAndLastName($params->email, $params->lastName)) {
User::createNew($params);
echo "creating new user";
} else {
echo "user already exists";
}
}
return $this->render('complete', [
'model' => $completeDonationForm
]);
}
return $this->render('complete-login-or-guest');
}
Here's the answer after multiple tries:
Passing an 'ajaxParam' parameters with the ActiveForm widget to define the name of the GET parameter that will be sent if the request is an ajax request. I named my parameter "ajax".
Here's what the beginning of the ActiveForm looks like:
$form = ActiveForm::begin([
'id' => 'complete-form',
'ajaxParam' => 'ajax'
])
And then I added this check in my controller:
if (Yii::$app->request->get('ajax') || Yii::$app->request->isAjax) {
return false;
}
It was an ajax issue, so thanks a bunch to Yupik for pointing me towards it (accepting his answer since it lead me here).
You can put validation like below in your model:
public function rules() { return [ [['email'], 'functionName'], [['lastname'], 'functionforlastName'], ];}
public function functionName($attribute, $params) {
$usercheck=User::find()->where(['email' => $email])->one();
if($usercheck)
{
$this->addError($attribute, 'Email already exists!');
}
}
and create/apply same function for lastname.
put in form fields email and lastname => ['enableAjaxValidation' => true]
In Create function in controller
use yii\web\Response;
if (Yii::$app->request->isAjax && $model->load(Yii::$app->request->post())) {
Yii::$app->response->format = Response::FORMAT_JSON;
return ActiveForm::validate($model);
}
else if ($model->load(Yii::$app->request->post()))
{
//place your code here
}
Add 'enableAjaxValidation' => false to your ActiveForm params in view. It happens because yii sends request to your action to validate this model, but it's not handled before your if statement.
I use Laravel5.2 to build a E-commerce platform and get some troubles.When I loginout user state and I use Session::forget('user') or $request->session()->forget('user'), it works in current page,and I redirect to the login page and dd(Session::all()) find the session 'user' still exist!!! so I am confused to know what is wrong with it? plz tell me the reason,thanks all.
ps. loginout code
public function logout(Request $request)
{
if ($request->session()->has('user')) {
$is_forgotten = $request->session()->forget('user');
if ($is_forgotten === null)
echo json_encode(['result' => 1]);
exit;
}
echo json_encode(['result'=>0,'msg'=>'loginout error']);
exit;
}
The session doesn't get saved until the response is sent and events/middleware are triggered.
Try doing:
public function logout(Request $request)
{
if ($request->session()->has('user')) {
$is_forgotten = $request->session()->forget('user');
if ($is_forgotten === null)
return Response::json(['result' => 1]);
return '';
}
return Response::json(['result'=>0,'msg'=>'loginout error']);
}
Or if your using the helpers:
public function logout(Request $request)
{
if ($request->session()->has('user')) {
$is_forgotten = $request->session()->forget('user');
if ($is_forgotten === null)
return response()->json(['result' => 1]);
return '';
}
return response()->json(['result'=>0,'msg'=>'loginout error']);
}