I'm having problem how can I able to generate links on twig view using the illuminate pagination. below is my codes.
Route:
$app->get('/', function ($request, $response) {
$data = Todo::paginate(5);
return $this->view->render($response, 'home.twig', [
'title' => 'Home',
'todolist' => $data,
]);
})->setName('homepage');
View:
{{todolist.links()}}
Error:
Call to a member function make() on null
btw I'm using Slim 3
Try this
Raw statement is to print the output as plain html (not text)
Refer to this link https://twig.sensiolabs.org/doc/2.x/filters/raw.html
{{todolist.links | raw}}
I tried on my slim 3 website and its works perfectly.
My Eloquent ver in the composer as below
"illuminate/database": "^5.2",
"illuminate/pagination": "~5.0"
If you're using Laravel pagination & Blade templates.
Try this in your blade file:
{{ $todolist->links() }}
Laravel docs: Displaying Pagination Results
Related
I have been following https://developer.shopware.com/docs/guides/plugins/plugins/storefront/add-custom-javascript to create a javascript plugin. But, I am struggling with adding options through twig template. I have something like the following:
Twig file : product.html.twig:
{% set contentModalOptions = {
cmsContentId: "some-hash",
navigationUrl: path('frontend.cms.page'),
failSafeRedirectUrl: '/some-failsafe-url/'
} %}
<div>
<a target="_self" href="#" data-content-modal="true" data-content-modal-options="{{ contentModalOptions|json_encode|escape('html_attr') }}">
help text
</a>
</div>
plugin file : custom-plugin.js:
import Plugin from 'src/plugin-system/plugin.class';
export default class ContentModalPlugin extends Plugin {
static options = {
cmsContentId: '',
navigationUrl: '',
failSafeRedirectUrl: ''
};
init() {
console.log(this);
console.log(this.options); // empty values
}
}
Notes:
In the browser, I see that values set using twig as the HTML attribute.
Plugin has been registered and works with the template.
console.log() in the plugin doesn't print any values that are set from twig. It just shows the options object that has been initialized in the plugin.
Any help is appreciated.
I assume you register the plugin like this:
PluginManager.register('ContentModal', ContentModalPlugin, '[data-content-modal]');
Within the constructor of the Plugin class this.options = this._mergeOptions(options); should then called, which in turn parses the data-${dashedPluginName}-options attribute. It should throw an error if it can't parse the json:
`The data attribute "data-${dashedPluginName}-options" could not be parsed to json: ${e.message}`
Are the any errors when you look at the console of your browsers dev tools?
For further debugging you could also try calling super._mergeOptions(this.options); from your init method.
Now, I see what the problem was. When I registered my plugin, I had the following:
PluginManager.register('ContentModalPlugin', ContentModalPlugin, '[data-content-modal]');
So, I tried passing the options with data-content-modal-options attribute through twig but it seems that the resolved plugin name in _mergeOptions() at src/plugin-system/plugin.class takes the plugin name (i.e the string that is the first argument of the register function) and not the attribute definition in the register method.
So, adding a html attribute as data-content-modal-plugin-options based on my class name resolved the problem.
I'm struggling to find the correct Twig concatenation syntax.
Here's what I've got:
{{ this.page.baseFileName|page({ (__SELF__.pageParam): page }) }}
it gives the url with the pagination number at the end of the url:
http://sites.local12/category/style/rock/2
Now I want to insert a colon before the pagination number like this:
http://sites.local12/category/style/rock/:2
How to achieve that ?
Ok I've got it. I was confused by the "page" filter (I'm new to Twig).
{{ this.page.baseFileName|page( {(__SELF__.pageParam): ':' ~ page }) }}
I've been using X-Ray to scrape website which has been working really well. I can use it bring in images very easily. The one item I run into is I don't see an easy way to scrape a background image. Say I have a div where they are setting a style attribute on that dev and then setting the URL im not sure how to get the background-image url from this. I don't think I can just pass the featured image attribute the css property such as
.featured-image.attr('background-image');
const getWebsiteContent = async (blogURL, selector) => {
try {
return await x(blogURL, selector, [{
slug: 'a#href',
featuredImage: 'img#src'
}])
.paginate(`${pagi}#href`)
.limit(200)
.then((response) => {
spinner.succeed('Got the data');
return response;
})
} catch (error) {
throw new Error('Cannot get Data from website, try checking your URL');
}
};
For anyone that wants a solution to this with X-ray scraper what I ended up doing is pulling the attribute from the selector you pass into the object.Given the html looks like the following.
<div class="img" style="background-image: url('../path-to-img.jpg')"></div>
Instead of writing .img#src you could write .img#style and this would return to you the style attribute. From there you would need to use a regex to remove the rest of the un-needed data that is not the URL of the image.
I try to call a static helper method inside Twig (Timber).
{{ function('Theme\Helpers::get_template_name') }}
Warning: call_user_func_array() expects parameter 1 to be a valid
callback, class 'ThemeHelpers' not found in
/var/www/html/wp-content/plugins/timber-library/lib/Twig.php on line
268.
Does anyone know how to call a method of a different class inside Twig?
As far as I know you can't call PHP classes directly from your twig template.
What you can do is setting up a Twig filter which communicates with your class
and returns the needed value.
You would have this in your php controller file that is responsible to load your twig template:
<?php
function twg_get_template_name() {
# edit this according to the implementation of your class:
return Helpers::get_template_name();
}
function add_to_twig($twig) {
/* this is where you can add your own fuctions to twig */
$twig->addExtension(new Twig_Extension_StringLoader());
$twig->addFilter('twg_get_template_name', new Twig_Filter_Function('twg_get_template_name'));
return $twig;
}
add_filter('get_twig', 'add_to_twig');
In your Twig template you would call the filter like this:
{{ ''|twg_get_template_name }}
Because it's a filter function it expects a value "to filter", so pass at least an empty string.
If I were in that situation I probably would determine the name of the template in your
controller and send the value to your Twig template directly instead of calling the php class
via a filter-function.
You can call static functions from a Twig file in Timber using the array notation, where first item is the name of the class and the second item the name of the static method you want to call:
{{ function( [ 'Theme\Helpers', 'get_template_name' ] ) }}
Thanks for your answer.
I tried your approach - it works. But using a filter feels a little hacky, especially when no value is passed. Why not create a timber function the same way as a filter?
Bridging own functions from plain php into twig is not great, but I also don't see another solution to this.
After playing around a little, I came up with a different approach. I now fixed my need by customizing the Timber Object and adding a template property to the post variable.
Looks something like this:
class OnepagePost extends TimberPost {
var $_template;
// Add template property to Twig Object
public function template() {
return Helpers::get_template_name( $this->custom['_wp_page_template'] );
}
}
Then inside the .php file where the Twig View gets called, I called the custom object like this:
$context['posts'] = new Timber\PostQuery( $args, 'OnepagePost' );
Timber::render('onepager.twig', $context);
Inside the Twig Template I'm able to get my custom property very easy (in my way the template):
{% for post in posts %}
{% include ["section/section-#{post.template}.twig"] %}
{% endfor %}
I'm using the Post/Redirect/Get (PRG) pattern in my Laravel controllers to prevent duplicate form submission.
It works well when I don't use layouts or when my layouts don't use any variable. The problem is my layout uses a variable named $title. When I load the view and the layout without redirect it works well, the title set in the controller is passed to the layout, but after processing a form and redirecting to the same route which uses the same layout and the same controller method I get a "Undefined variable: title" error coming from my layout file.
Here is my code:
File: app/routes.php
Route::get('contact', array('as' => 'show.contact.form', 'uses' => 'HomeController#showContactForm'));
Route::post('contact', array('as' => 'send.contact.email', 'uses' => 'HomeController#sendContactEmail'));
File: app/controllers/HomeController.php
class HomeController extends BaseController {
protected $layout = 'layouts.master';
public function showContactForm()
{
$this->layout->title = 'Contact form';
$this->layout->content = View::make('contact-form');
}
public function sendContactEmail()
{
$rules = ['email' => 'required|email', 'message' => 'required'];
$input = Input::only(array_keys($rules));
$validator = Validator::make($input, $rules);
if($validator->fails())
return Redirect::back()->withInput($input)->withErrors($validator);
// Code to send email omitted as is not relevant
Redirect::back()->withSuccess('Message sent!');
}
}
File: app/views/layouts/master.blade.php
<!DOCTYPE html>
<html>
<head>
<title>{{{ $title }}}</title>
</head>
<body>
#yield('body')
</body>
</html>
File: app/views/contact-form.blade.php
#section('body')
#if (Session::has('success'))
<div class="success">{{ Session::get('success') }}</div>
#endif
{{
Form::open(['route' => 'send.contact.email']),
Form::email('email', null, ['placeholder' => 'E-mail']),
Form::textarea('message', null, ['placeholder' => 'Message']),
Form::submit(_('Send')),
Form::close()
}}
#stop
I don't understand why after redirecting the next line of code is ignored
$this->layout->title = 'Contact form';
I've tried with Redirect::action('HomeController#sendContactEmail'); or Redirect::route('show.contact.form'); but the result is the same.
The controller in charge of rendering that view is exactly the same before the redirect than after the redirect, and it has no business logic at all, so why it only works on the first case but not in the second?
This
Redirect::back()->withSuccess('Message sent!');
should be
return Redirect::back()->withSuccess('Message sent!');
When layout attribute is set in a controller and method is not returning any response, controller try to render the layout. In your sendContactEmail() method both conditions fulfilled and controller tried to render layout before $title is set.
see callAction() in Illuminate\Routing\Controllers\controller.
http://laravel.com/api/source-class-Illuminate.Routing.Controllers.Controller.html#93-127
Have you tried using
return View::make('contact-form', array('title' => 'Contact Form'));
Instead of interacting with the layout directly?
Redirect::back() creates a 302 using the referer value of the current HTTP request. I would start by comparing the initial form request to the redirect request to see if that yields any clues. You could also try...
Redirect::route('HomeController#showContactForm')->withInput()...
I know it's less dynamic but it will generate the URL rather then rely on the referer value in the HTTP header.