I want to test back() function in laravel but I am not sure how to do that .
For e.g
public function destroy($id)
{
$query = Query::findorfail($id);
$query->delete();
return back()->with('success','Query deleted successfully');
}
In the above method the url is redirected back . But Can I create a test so that I can assert back()
I have checked the TestResponse class. A function like assertRedirectBack is not yet provided.
I think assertRedirect is gonna work:
$response = $this->delete(...);
$response->assertRedirect($uri);
The previous url could be stored in session. I am not sure about that. So please try this:
$response->assertRedirect(session()->previousUrl());
Try with return redirect()->back();
Use TestCase's from (from MakesHttpRequests trait)
$response = $this
->from('{route}') // without this would redirect to '/'
->delete('/asset/{id}');
$response->assertRedirect('{route}');
Related
I'm sending an URL hashed and when i get it i have to show a view on Laravel, so i have those functions on the controller and also some routes:
This are my routes:
Route::post('/sendLink', 'Payment\PaymentController#getPaymentLink');
Route::get('/payment?hash={link}', 'Payment\PaymentController#show');
And this are the functions i have on my controller:
public function getPaymentLink (Request $request){
$budgetId = $request['url.com/payment/payment?hash'];
$link = Crypt::decryptString($budgetId);
Log::debug($link);
//here to the show view i wanna send the link with the id hashed, thats why i dont call show($link)
$view = $this->show($budgetId);
}
public function show($link) {
$config = [
'base_uri' => config('payment.base_uri'), ];
$client = new Client($config);
$banking_entity = $client->get('url')->getBody()->getContents();
$array = json_decode($banking_entity, true);
return view('payment.payment-data')->with('banking_entity', $array);
}
And this is getting a "Page not found" message error.
What i want to to is that when i the client clicks on the link i send him that has this format "url.com/payment/payment?hash=fjadshkfjahsdkfhasdkjha", trigger the getPaymentLink function so i can get de decrypt from that hash and also show him the view .
there is no need to ?hash={link} in get route
it's query params and it will received with $request
like:
$request->hash
// or
$request->get('hash')
You need to define route like this:
Route::get('/payment/{hash}', 'Payment\PaymentController#show');
You can now simply use it in your Controller method like below:
<?php
public function getPaymentLink (Request $request,$hash){
$budgetId = $hash;
// further code goes here
}
i'm trying to redirect this action method in my controller but i couldn't make it work. the action will goes to posts Resources #show
here is the controller.
public function readbyid(Request $request){
if(isset($request->data['id'])){
return redirect()->action(
'PostController#show', ['id' => $request->data['id']]
);
}else{
return redirect()->back();
}
}
Route:
Route::get('markread/{id}', 'NotifyController#readbyid' );
For optional parameters, use '?'
Route::get('markread/{id?}', 'NotifyController#readbyid');
Use $request->id instead $request->data['id']
you are not getting any value $request->data['id'] like this thats why it goes to the else part.
Just wondering if this is a good way to write ajax code to interact with Laravel routes?
Example my application's require to list all customer data and also list all country through ajax. I have 3 controller ApiController, CustomerController, CountryController.
So in my routes.php I have this routes
Route::get('api/v1/ajax/json/{class}/{function}', 'Api\v1\ApiController#ajaxreturnjson');
In the ApiController.php, I have below function to call other controller function to return the data I need.
class ApiController extends Controller
{
public function ajaxreturnjson(Request $request, $controller, $function){
$input = $request->input();
if($request->input('namespace') != ''){
$namespace = $request->input('namespace');
unset($input['namespace']);
}else{
$namespace = 'App\Http\Controllers';
}
$data = array();
try {
$app = app();
$controller = $app->make($namespace.'\\'.$controller);
$data = $controller->callAction($function, array($request)+$input);
} catch(\ReflectionException $e){
$data['error'] = $e->getMessage();
}
return response()->json($data);
}
}
So example to use the ajax, I just need to pass the class name, namespace and also the function name to the ajax url.
Example to retrieve all customer info.
$.ajax({
dataType:"json",
url:"api/v1/ajax/json/CustomerController/getList",
data:"namespace=\\App\\Http\\Controllers\\",
success:function(data){
}
})
So in this way, I don't have to create so many routes for different ajax request.
But I am not sure if this will cause any security issue or is this a bad design?
Personally, I would not do it this way. Sure, you could do it this way, but it's not very semantic and debugging it could be a pain.
Also, if someone else begins working on the project, when they look at your routes file, they won't have any idea how your app is structured or where to go to find things.
I think it's better to have a controller for each Thing.
I'm trying to have some after middleware so that if the request was ajax, only render part of the view. I'm having issues though.
My method in my controller:
$data = $this->repository->getUser($id);
$view = View::make('site.user')->withData($data);
return $view;
My middleware:
$response = $next($request);
if(Request::ajax()){
$response->renderSections()['content'];
}
return $response;
I get the error:
Call to undefined method Illuminate\Http\Response::renderSections()
To get the response you need to use a after Middleware which is you are doing now so, you have response as given below:
$response = $next($request);
Now you may try this:
$content = $response->getOriginalContent()->renderSections()['content']
Here is my controller code
function index() {
echo $this->session->flashdata('message');
$this->load->view('categoryView');
}
function delete() {
$products = $this->item_model->get_category_id($category_id);
if (count($products)) {
$message = 'Category Name is used by the product. Please change them to another category!';
}
else {
$category_id = $this->product_category_model->delete($category_id);
$message = ($category_id) ? 'Category Deleted Successfully' : 'Failed to Delete Category';
}
$this->session->set_flashdata('message', $message);
redirect('category', 'refresh');
}
After calling the delete function the flashdata has to be set and retrieve that value in index() function of the same controller but I can't.
Am also tried the $this->session->keep_flashdata('message'); before redirect to index function. But still am not get any value in index function.
Also changed $config['sess_expire_on_close'] = FALSE; to $config['sess_expire_on_close'] = TRUE; Still am not get the result.
I am wasted more time(nearly half day). Please anybody help to retrieve the flas data in codeigniter.
There are several probabilities:
1- Check your browser cookie. See if it allows cookies or if any other extension is intervening.
2- Refresh might not send the browser a new request (which thus mean a new URL). Try it for another controller see if it make any change
3- hard code the set_flashdata() with hard-coded value than a variable. set_flashdata("message", "Thank you");
In codeingitor you cannot echo anything in controller.
You have to set in $this->session->flashdata('message'); in view file not in index() function of controller
Try again by putting $this->session->flashdata('message'); in "categoryView" file where you want to display flash message.
You could try:
function set_flashdata_notification($notify_type, $msg, $error_code = null)
{
$ci =& get_instance();
$flashdata_data = set_notification_array($notify_type, $msg, $error_code);
$ci->session->set_flashdata($flashdata_data);
}
function delete() {
$products = $this->item_model->get_category_id($category_id);
if (count($products)) {
set_flashdata_notification('Category Name is used by the product.','Please change them to another category!');
redirect('path_to_view/view','refresh');
}