Hi I am trying to get the content within a div element that also happens to be within a form into my controller. I dont want to use ajax. How may I get that done ?
<div id="editorcontents" name="editorcontents">
</div>
Then in controller
Use Input;
$content = Input::get('editorcontents');
In your controller, do something like this. Look up the correct way in the docs (https://laravel.com/docs/5.4/eloquent#retrieving-models). For example, if you want ALL input, you would do Input::all();, instead of Input::where('editorcontents')->get();
public function index() {
$content = Input::where('editorcontents')->get();
return view('your_view.blade.file', compact('content'));
}
Then in your view your would now have $content, that you passed from your controller.
start of by looking at it, add this at top of your view: {{ dd($content) }}. This will die dump $content.
Go ahead and remove that line and do something like (docs here https://laravel.com/docs/5.4/blade#loops):
<div id="editorcontents" name="editorcontents">
#forelse ($content as $value)
<li>{{ $value->body }}</li>
#empty
<p>No content</p>
#endforelse
</div>
Related
I have a simple controller function that fetch all records from db. but when i am trying to show all these records it show nothing. In fact it shows me hard coded foreach loop like this.
#foreach ($compactData as $value) {{ $value->Name }} #endforeach
this is my contoller function.
public function showallProducts()
{
$productstock = Product::all()->stocks;
$productoldprice = Product::all()->OldPrices;
$productcurrentprice = Product::all()->CurrentPrice;
$compactData=array('productstock', 'productoldprice', 'productcurrentprice');
return view('welcome', compact($compactData));
}
this is my view
<!doctype html>
<html lang="{{ app()->getLocale() }}">
<head>
</head>
<body>
<div class="flex-center position-ref full-height">
<div class="content">
<div class="title m-b-md">
Laravel
</div>
<div class="title m-b-md">
All products
</div>
<table>
<tbody>
#foreach ($compactData as $value)
{{ $value->Name }}
#endforeach
</tbody>
</table>
</div>
</div>
</body>
why it is behaving like this. any solution?? I am using phpstorm version 17. Is their any setting issue to run project because what ever project I ran it gives me the only page which i ran with only html?
My route is.
Route::get('/', function () {
$action = 'showallProducts';
return App::make('ProductController')->$action();
});
Have you checked your $compactData variable? Please dd($compactData) to see what it contains.
Problem 1
You are accessing a relational property as a property of Eloquent collection, like this:
Product::all()->stocks
which is not correct. Because the Collection object doesn't have the property stocks but yes the Product object might have a stocks property. Please read the Laravel documentation about Collection.
Problem 2
$compactData = array('productstock', 'productoldprice', 'productcurrentprice');
This line creating an array of 4 string, plain string not variable. So, your $compactData is containing an array of 4 string. If you want to have a variable with associative array then you need to do the following:
$compactData = compact('productstock', 'productoldprice', 'productcurrentprice');
Problem 3
return view('welcome', compact($compactData));
Here you are trying to pass the $compactDate to the welcome view but unfortunately compact() function doesn't accept variable but the string name of that variable as I have written in Problem 2. So, it should be:
return view('welcome', compact('compactData'));
Problem 4
Finally, in the blade you are accessing each element of the $compactData data variable and print them as string which might be an object.
You most likely have a problem with your web server.
Try to use Laravel Valet as development environnement.
Edit : I found this : Valet for Windows
I think you didn't mention the blade in the name of the view file by which it is saved. So change the name of the file by which it is save to something like:
filename.blade.php
and try again.
Explanation:
#foreach ($compactData as $value) this is the syntax of blade template engine, and to parse and excute it, you have to mention the blade extension in the name.
I have a page system in Laravel - where I pass data from controller to view.
$this->data['title'] = $row->title;
$this->data['breadcrumb'] = $row->bc;
Now I passed it as follows:
return View::make('Themes.Page', $this->data);
In the view file, I access the data as follows:
{{$breadcrumb}}
What I am trying to do now is to pass this data in nested views:
$this->layout->nest('content',$page, $this->data);
(Content is the {{content}} in the view which will be replaced with $page contents. I want to pass the $this->data just as before but now I get an error:
Variable breadcrumb not defined.
Note: Laravel Version 4.2 $this->layout is set in constructor to a
template file (Themes.Page)
Actually you don't need to pass any separate data to your partial page(breadcrumb)
controller page
$this->data['title'] = $row->title;
$this->data['breadcrumb'] = $row->bc;
return View::make('idea.show',array("data"=>$this->data));
main view page
<div>
<h1>here you can print data passed from controller {{$data['title']}}</h1>
#include('partials.breadcrumb')
</div>
your partial file
<div>
<h1>here also you can print data passed from controller {{$data['title']}}</h1>
<ul>
<li>....<li>
<li>....<li>
</ul>
</div>
for more information on this you can check following links http://laravel-recipes.com/recipes/90/including-a-blade-template-within-another-template or watch this video https://laracasts.com/series/laravel-5-fundamentals/episodes/13
You should pass data as follows
return View::make('Themes.Page')->with(array(
'data'=>$this->data));
or (since you're passing only 1 variable)
return View::make('Themes.Page')->with('data', $this->data);
and further you can pass it on to nested views as by referencing $data
$dataForNestedView = ['breadcrumb' => $row->bc];
return View::make('Themes.Page', $this->data)->nest('content', 'page.content', $dataForNestedView);
In the Themes.Page view render nested view:
<div>
{{ $content }} <!-- There will be nested view -->
</div>
And in the nested page.content view you can call:
<div>
{{ $breadcrumb }}
</div>
*div tag is only for better understanding.
Okay, after intense search, I figured out that there was a bug in the Laravel version 4.2 I was using.
Laravel 5 works.
For laravel 4.2, a better option would be to pass array of data objects using View::share('data',$objectarray) while passing the data from controller.
Thanks everyone for help
I followed a tutorial on Tutsplus about creating an ecommerce website using Laravel. The problem I'm having right now is when trying to route to a subfolder. In the tutorial, the instructor included a feature where you can view products by ID. And this is how he did it:
// StoreController.php
public function getView($id) {
return View::make('store.view')->with('store', Store::find($id));
}
This piece of code seems to be passing an id from the stores table. I think when a product is clicked, that's when the id is passed
// Routes.php
Route::controller('store', 'StoreController');
Also some of the templates:
// store\index.blade.php
<h2>Stores</h2>
<hr>
<div id="stores row">
#foreach($stores as $store)
<div class="stores col-md-3">
<a href="/store/products/view/{{ $store->id }}">
{{ HTML::image($store->image, $store->title, array('class' => 'feature', 'width'=>'240', 'height' => '127')) }}
</a>
<h3>{{ $store->title }}</h3>
<p>{{ $store->description }}</p>
</div>
#endforeach
</div><!-- end product -->
So.. How it goes is when I click on a product, it leads me to domain:8000/store/view/6 where 6 is the id.
This works fine but what I want to know is how do I route through a subfolder? Let's say I want it to be like this: store/view/products/6 considering that I have a folder called products and my view.blade.php is inside that like this: store/products/view.
In my StoreController class, I tried changing this
public function getView($id) {
return View::make('store.view')->with('store', Store::find($id));
}
to this
public function getView($id) {
return View::make('store.product.view')->with('store', Store::find($id));
}
but it does not seem to work giving me nothing but a Controller Method Not Found Error.
First, the view name View::make('store.product.view') has nothing to do with the URL.
You have to change the route:
Route::controller('store/view', 'StoreController');
And then adjust the name of your method in the controller because it should be the same as the segment of the URL after store/view
public function getProducts($id) {
return View::make('store.product.view')->with('store', Store::find($id));
}
I strongly recommend you read the Laravel docs on the topic
Halo here is my need;
i want to include different view as apart of different view in laravel php frame work.
class DashboardController extends BaseController {
public function comments( $level_1=''){
// process data according to $lavel_1
return View::make('dashboard.comments', $array_of_all_comments);
}
public function replys( $level_2=''){
// process data according to $lavel_1
return View::make('dashboard.replys', $array_of_all_replys);
}
these both data can now accessed from
www.abc.com/dashboard/comments
www.abc.com/dashboard/replys
And in my view what i need is to generate replys according to the comments id ($lavel_2)
// dashboard/comments.blade.php
#extends('layout.main')
#section('content')
#foreach($array_of_all_comments as $comment)
comment {{ $comment->data }},
//here is what i need to load reply according to the current data;
//need to do something like this below
#include('dashboard.replys', $comment->lavel_2) //<--just for demo
.................
#stop
and in replys also got
#extends('layout.main')
#section('content')
// dashboard/replys.blade.php
#foreach($array_of_all_replys as $reply)
You got a reply {{ $reply->data }},
...........
#stop
is there any way i can achieve this on laravel 4?
Please help me i wanted to load both comments and replays in one go and later need to access them individually via ajax also
please help me thank you very much in advance
halo i found the solution here
all we need is to use App::make('DashboardController')->reply();
and remove all #extends and #sections from including view file
the change are like this
// dashboard/comments.blade.php
#extends('layout.main')
#section('content')
#foreach($array_of_all_comments as $comment)
comment {{ $comment->data }},
//<-- here is the hack to include them
{{-- */echo App::make('DashboardController')->reply($comment->lavel_2);/* --}}
.................
#stop
.............
and in replys is now changed to
// dashboard/replys.blade.php
#foreach($array_of_all_replys as $reply)
You got a reply {{ $reply->data }},
...........
#endforeach
-------------
thanks
You probably want to rework your views and normalise your data. Comments and replies are (probably) the same.
If you make a Comment model which belongsTo "parent" (another Comment model) and hasMany "children" (many Comment models), then simply set parent_id to 0 for a top-level Comment, and set it to the ID of another Comment to make it a Reply.
Then your Blade views do something like:
comments.blade.php
#foreach ($comments AS $comment)
#include( 'comment', [ 'comment' => $comment ] )
#endforeach
comment.blade.php
<div>
<p>{{{ $comment->message }}}</p>
#if( $comment->children->count() )
<ul>
#foreach( $comment->children AS $comment )
<li>
#include( 'comment', [ 'comment' => $comment ] )
</li>
#endforeach
</ul>
#endif
</div>
I have a controller where I am creating a form witg two dropdown list inside.
When I am rendering my view, I would like to have the same form elements on the top and the bottom of the page. The problem is that the form elemetns (dropdownlists) are displayed only on the top of the page, even if I am asking twig to put them also on the bottom.
Here is what I would like to have:
The 1 and 2 are the dropdownlists. And I would like to duplicate this on the top and on the bottom of the page.
Any Idea on how can this be done?
The top content and the bottom content, where the two dropdownlists are inside are in a single sseparate twig file (searchPanel.html.twig) and this file is included in the page
{% include "MyBundle:Search:searchPanel.html.twig" %}
Here is the searchPanel.html.twig
<div class="searchPanel">
<form action="{{ path }}" method="POST" {{ form_enctype(form) }}>
Papers per page
{{ form_widget(form.papers_per_page, { 'class': 'ppp'}) }}
/ Sort by
{{ form_widget(form.sort_by, { 'class': 'sort'}) }}
{{ form_rest(form) }}
/ Papers ({{ papers_number }} results)
<input type="submit" class="updateSearchResults" value="Update"></input>
</form>
A problem in your approach is that Symfony's Form-component will render the form elements with id's which would be duplicated if you rendered the same form twice on your page. You might also run in trouble with the csrf_token. The gist being that forms are not intended to be duplicated.
Here is what I would do. Create a twig-template containing your paginator form without using Symfony\Form, i.e. create all form elements statically and pass it the paginator-object (or array) to get the data instead of using form_widget(). Something like this:
<form action="{{ path(app.request.attributes.get('_route') }}" method="POST">
<select name="paginator[per_page]">
{% for per_page in paginator.papers_per_page %}
<option value=""{{ per_page }}">{{ per_page }}</option>
{% endfor %}
</select>
</form>
The form action will automatically submit the data to your current route, so you can embed it in different actions and it will submit the data to the same action. On POST you can just create a paginator-object with the post-data and then add it as the form's data. After that you just use isValid() as usual.
In your controller you can get the data like this:
use Symfony\Component\HttpFoundation\Request;
// ...
public function PaperController()
{
public function listAction(Request $request)
{
if ($request->getMethod() == 'POST') {
$data = $request->request->get('paginator');
$paginator = new Paginator($data);
$form = new PaginatorFormType();
$form->setData($paginator);
if ($form->isValid()) {
// ...
}
}
}
}
You can easily embed the form in your view like this:
{{ include 'AcmeDemoBundle:Form:paginator.html.twig' with { 'paginator': paginator } }}
Basically you just use the Form-component in your controller for validation purposes. If you want to set some default values or add additional arguments you might want to create a macro from that template, but for your use case this should suffice. Maybe someone else has a better solution but this is how I went with a similar problem in one of my projects.
another option is user the render twig helper. That way is is possible render the same form in the page as many time as you want. A difference is that using this helper, is also necessary to treat the form renderization as an independent controller Action namely:
in every place in your twig template you want to render the form in the helper to invoke the form there's must be something like this:
{{ render(controller('BundleNameBundle:Controller:Action', {
'param': paramId
})) }}
Thereafter is just a matter of creating the controller...
Another option is in the controller to create 2 forms:
//first form
$form = $this->createForm(new MyFormType(), ...);
//second form: must have different form name that default,
//to render in twig the fields with different ids
$formType = new MyFormType();
$formType->setName('second_form_name');
$formSecond = $this->createForm($formType, ...);
Send both when rendering the twig form:
return $this->render( ...
'form' => $form->createView(), 'formSecond'=>$formSecond->createView()));
Then define the second with name as
formSecond
, and it will conflict with the first.