I can't find any info about using $this->Cookie in view cells.
When i wrote code like this, error will arise:
namespace App\View\Cell;
use Cake\View\Cell;
class CityCell extends Cell {
public function display() {
$this->Cookie->config('encryption', false);
$cookie = $this->Cookie->read('city');
}
}
and the error is:
Error: Call to a member function read() on null
So can we use cookie in view cells ?
Thank You.
That of course won't work, view cells do not support the use of components, altough they can be thought of like "mini-controllers", they are not actually controllers in the terms of CakePHPs MVC system.
Depending on whether the cookies are encrypted, you can either use the request object to fetch them in your cell
$this->request->cookie('cookieName')
or
$this->request->cookies
see also API > \Cake\Network\Request::cookie()
or, in case they are encrypted, you have grab them via the Cookie component, and then for example pass them down from your controller to the view, and finally into the cell like
controller
public function controllerAction() {
// ...
$this->set('cookie', $this->Cookie->read('cookieName'));
}
cell
public function display($cookie) {
// ...
}
view
$this->cell('CellName', ['cookie' => $cookie]);
see also Cookbook > Views > View Cells > Passing Arguments to a Cell
Related
Hi i am working on a project in laravel 7.0, in back-end i have a table called Posts which contains 2 text language input one in french and the other is arabic added by the back-end application.
what i am trying to do is when the user uses the French Language i want the title_fr to be displayed on the view and same thing in Arabic language the title should be title_ar.
P.S data are stored in French and Arabic
I have tried the similar solution given in an other similar question but none of it worked in my case!
Any idea how i might get this to work ?
Thanks in advance.
You can do something similar to below. We have a model Post, this model has an attribute title. I also assume that you have an attribute that will return user's language from the User model.
class Post extends Model
{
public function getTitleAttribute(): string
{
return Auth::user()->language === 'fr' ? $this->title_fr : $this->title_ar;
}
}
FYI above is just a demo on what can be done. For a full blow solution I would recommend decorator pattern.
Also it might be worth considering using morph for things like that. You can have a service provider that will initiate the morph map for you post model relevant to the language that user has, I.e.
Class ModelProvider {
Protected $models = [
‘fr’ => [
‘post’ => App/Models/Fr/Post::class,
],
‘ar’ => [
‘post’ => App/Models/Ar/Post::class,
]
];
Public function boot() {
$language = Auth::user()->Settings->language;
Relation::morphMap($This->models[$language]);
}
}
Afterwards you just need to call to Relation::getMorphModel(‘post’) to grab Post class that will return correct language.
I.e. App/Models/Fr/Post can have a an attribute title:
Public function getTitleAttribute(): string {
Return $this->title_fr;
}
For example above you would also want to utilise interfaces to make sure that all models follow the same contract, something below would do the trick:
Interface I18nPostInterface {
Public function getTitleAttribute(): string
}
Also, depending on the database you use, to store titles (and other language data) in a JSON format in the database. MySQL 8 has an improve support for JSON data, but there are limitations with that.
So I was Able to fetch data from my database based on the Language selected by the user.
Like i said before I have a table called Posts and has columns id,title_fr and title_ar. I am using laravel Localization.
Inside my PostController in the index function i added this code:
public function index()
{
//
$post = Post::all();
$Frtitle = post::get()->pluck('title_fr');
$Artitle = post::get()->pluck('title_ar');
return view('post.index',compact('post','Frtitle','Artitle'));
}
if anyone has a better way then mine please let me know, i am sure
there is a better way.
I am trying to merge the contents of 2 arrays then use usort to get the posts with most views.
Trying to use usort to sort the contents of an array.
Am getting the follow error:
("Notice: Undefined property: Acme\DemoBundle\Entity\Article::$getViews in /.../PageController.php line 15")
Can someone point out what I am doing wrong?
Sort function inside of controller
private static function popularSort($articles, $posts, $articles2, $posts2)
{
return $articles->getViews() == $posts->getViews() ? 0 : ( $articles->getViews() < $posts->getViews()) ? 1: -1;
}
Sidebar action
$articles = $this->getDoctrine()->getRepository('AcmeDemoBundle:Article')
->getArticles();
$posts = $this->getDoctrine()->getRepository('AcmeDemoBundle:Post')
->getPosts();
$popular = array_merge($articles, $posts);
usort($popular, array($this, 'popularSort'));
getViews is a getter method for accessing the property views of the Entities Post and Article. So while accessing it u should access it as $articles->getViews().
But if you simply want to compare the property views of the two entities compare them using their property name instead of their getter
Assuming views as the name of the property, the call should be something like:
$posts->views and $articles->views.
You don't have a property called getViews in your Article class.
You probably have a property called views and a method getViews meaning you should call the actual method with the brackets like $article->getViews().
I've two controllers one is "Upload" which deals with images uploads and other is "Page" whid deals with the creation of pages of CMS now if in my "Upload" controller I load both the models i.e 'image_m' which deals with image upload and "page_m" which deals with the pages creation I've highlighted the relevant code my problem is if I access the variables in the view
$this->data['images'] = $this->image_m->get(); sent by this I can access in foreach loop as "$images->image_title, $images->image_path" etc
But the variable sent by this line ***$this->data['get_with_images'] = $this->page_m->get_no_parents();*** as $get_with_images->page_name, $get_with_images->page_id etc produces given error
A PHP Error was encountered
Severity: Notice
Message: Trying to get property of non-object
Filename: upload/index.php
Line Number: 20
what is the difference between these two access levels one for $image & other for $get_with_images because I can only access its values as $get_with_images
class Upload extends Admin_Controller {
public function __construct() {
parent::__construct();
***$this->load->model('image_m');
$this->load->model('page_m');***
}
public function index($id = NULL) {
//var_dump($this->data['images'] = $this->image_m->get_with_images());
//$this->data['images'] = $this->image_m->get_with_images();
***$this->data['images'] = $this->image_m->get();***
$this->data['subview'] = 'admin/upload/index';
if ($id) {
$this->data['image'] = $this->image_m->get($id);
count($this->data['image']) || $this->data['errors'][] = 'Page Could not be found';
}
$id == NULL || $this->data['image'] = $this->image_m->get($id);
/*this calls the page_m model function to load all the pages from pages table*/
***$this->data['get_with_images'] = $this->page_m->get_no_parents();***
You are not posting all your code so its hard to tell but is it because you used $this-> in the controller, but you haven't done the same thing in the view?
In this case i would recommend not using $this-> because its not necessary. Also its much better to check for errors etc when you call the model so do something like
if ( ! $data['images'] = $this->image_m->get($id) ) {
// Failure -- show an appropriate view for not getting any images
// am showing $data in case you have other values that are getting passed
$this->load->view( 'sadview', $data ); }
else {
// Success -- show a view to display images
$this->load->view( 'awesomeview', $data ); }
so we are saying if nothing came back - the ! is a negative - then show the failure view. Else $data['images'] came back, and it will be passed to the view. note i have not had to use $this-> for anything and it won't be needed in the view.
Would also suggest using separate methods - have one method to show all images and a separate method like returnimage($id) to show an image based on a specific validated $id.
====== Edit
You can access as many models as you want and pass that data to the View. You have a different issue - the problem is that you are waiting until the View to find out - and then it makes it more difficult to figure out what is wrong.
Look at this page and make sure you understand the differences between query results
http://ellislab.com/codeigniter/user-guide/database/results.html
When you have problems like this the first thing to do is make a simple view, and echo out directly from the model method that is giving you problems. Its probably something very simple but you are having to look through so much code that its difficult to discover.
The next thing is that for every method you write, you need to ask yourself 'what if it doesn't return anything?' and then deal with those conditions as part of your code. Always validate any input coming in to your methods (even links) and always have fallbacks for any method connecting to a database.
On your view do a var_dump($get_with_images) The error being given is that you are trying to use/access $get_with_images as an object but it is not an object.
or better yet on your controller do a
echo '<pre>';
var_dump($this->page_m->get_no_parents());
exit();
maybe your model is not returning anything or is returning something but the data is not an object , maybe an array of object that you still need to loop through in some cases.
in my grails app I need to get some data from database and show it in a gsp page.
I know that I need to get data from controller, for example
List<Event> todayEvents = Event.findAllByStartTime(today)
gets all Event with date today
Now, how can I render it in a gsp page?How can I pass that list of Event objects to gsp?
Thanks a lot
You can learn many of the basic concepts using Grails scaffolding. Create a new project with a domain and issue command generate-all com.sample.MyDomain it will generate you a controller and a view.
To answer your question create a action in a controller like this:
class EventController {
//Helpful when controller actions are exposed as REST service.
static allowedMethods = [save: "POST", update: "POST", delete: "POST"]
def showEvents() {
List<Event> todayEvents = Event.findAllByStartTime(today)
[eventsList:todayEvents]
}
}
On your GSP you can loop through the list and print them as you wish
<g:each in="${eventsList}" var="p">
<li>${p}</li>
</g:each>
Good luck
I am not sure if this is really what you meant, because in that case I suggest you to read some more on the grails :), but anyway, for your case you can use render, redirect as well but here I am taking simplest way:
In your controller you have:
def getAllElements(){
List<Event> todayEvents = Event.findAllByStartTime(today)
[todayEvents :todayEvents ]
}
and then in the GSP(I assume you know about grails conventions, as if you don't specify view name, it will by default render gsp page with the same name as the function in the controller, inside views/):
<g:each in="${todayEvents}" var="eventInstance">
${eventInstance.<propertyName>}
</g:each>
something like this.
My app has sales listing functionality that will allow the user to add 1 or more photos for the product that they want to sell.
I'm attempting to use the upload/filestore_image of ATK with a Join table to create the relationship - my models:
class Model_Listing extends Model_Table {
public $entity_code='listing';
function init(){
parent::init();
$this->addField('name');
$this->addField('body')->type('text');
$this->addField('status');
$this->addField('showStatus')->calculated(true);
}
function calculate_showStatus(){
return ($this->status == 1) ? "Sold" : "For Sale" ;
}
}
class Model_listingimages extends Model_Table {
public $entity_code='listing_images';
function init(){
parent::init();
$this->addField('listing_id')->refModel('Model_Listing');
$this->addField('filestore_image_id')->refModel('Model_Filestore_Image');
}
}
In my page manager class I have added the file upload to the crud:
class page_manager extends Page {
function init(){
parent::init();
$tabs=$this->add('Tabs');
$s = $tabs->addTab('Sales')->add('CRUD');
$s->setModel('Listing',array('name','body','status'),array('name','status'));
if ($s->form) {
$f = $s->form;
$f->addField('upload','Add Photos')->setModel('Filestore_Image');
$f->add('FileGrid')->setModel('Filestore_Image');
}
}
}
My questions:
I am getting a "Unable to include FileGrid.php" error - I want the user to be able to see the images that they have uploaded and hoped that this would be the best way to do so - by adding the file grid to bottom of the form. - EDIT - ignore this question, I created a FileGrid class based on the code in the example link below - that fixed the issue.
How do I make the association between the CRUD form so that a submit will save the uploaded files and create entries in the join table?
I have installed the latest release of ATK4, added the 4 filestore tables to the db and referenced the following page in the documentation http://codepad.agiletoolkit.org/image
TIA
PG
By creating model based on Filestore_File
You need to specify a proper model. By proper I mean:
It must be extending Model_Filestore_File
It must have MasterField set to link it with your entry
In this case, however you must know the referenced ID when the images are being uploaded, so it won't work if you upload image before creating record. Just to give you idea the code would look
$mymodel=$this->add('Model_listingimages');
$mymodel->setMasterField('listing_id',$listing_id);
$upload_field->setModel($mymodel);
$upload_field->allowMultiple();
This way all the images uploaded through the field will automatically be associated with your listing. You will need to inherit model from Model_Filestore_File. The Model_Filestore_Image is a really great example which you can use. You should add related entity (join) and define fields in that table.
There is other way too:
By doing some extra work in linking images
When form is submitted, you can retrieve list of file IDs by simply getting them.
$form->get('add_photos')
Inside form submission handler you can perform some manual insertion into listingimages.
$form->onSubmit(function($form) uses($listing_id){
$photos = explode(',',$form->get('add_photos'));
$m=$form->add('Model_listingimages');
foreach($photos as $photo_id){
$m->unloadDdata()->set('listing_id',$listing_id)
->set('filestore_image_id',$photo_id)->update();
}
}); // I'm not sure if this will be called by CRUD, which has
// it's own form submit handler, but give it a try.
You must be careful, through, if you use global model inside the upload field without restrictions, then user can access or delete images uploaded by other users. If you use file model with MVCGrid you should see what files they can theoretically get access to. That's normal and that's why I recommend using the first method described above.
NOTE: you should not use spaces in file name, 2nd argument to addField, it breaks javascript.