I know that we can render the content of a controller in twig file like this:
{{ render(controller('FOSUserBundle:Security:login',{"baseTemplate": true})) }}
However, I don't know if we can pass the new template so that the controller will use it instead of the default. Anyone tried to override template in this way?
I don't really understand the issue here
If you do
{{ render(controller('FOSUserBundle:Security:login',{"baseTemplate": true})) }}
You could aswell do:
{{ render(controller('FOSUserBundle:Security:login',{"template": "your_template.html.twig"})) }}
Or
{{ render(controller('FOSUserBundle:Security:login',{"templateNumber": "4"})) }}
Where templateNumber is used in a condition inside your controller ?
Related
I'm using the Timber plugin for my Wordpress site. For a little button component, i'm trying to pass the variable {{ site.link }} to use as my href.
Like so:
{% include 'component/icon-button.twig' with { href: '{{ site.theme.link }}/faq' } %}
icon-button.twig
<button class="icon-button" data-href="{{ href }}">text</button>
But that only results in {{ site.link }}/faq being output as is, as a string, not the actual URL.
How would i do this?
That's just because when you use include in twig you don't need to use double {{ again. The correct syntax would be:
{% include 'component/icon-button.twig' with { href: site.theme.link ~ '/faq' } %}
The "~" is used for string concatenation in twig. You were passing the whole thing as a string and nothing more =)
I created a custom twig function in AppExtension Class. I need to call form_label() from this new function. ¿Is it posible? I tried but does not work:
from template I call:
{{ myFunc(form.someField) }}
public function myFunc( $field )
{
$html = form_label($field);
}
The idea is to render each form field in a different order/way than the form_widget(form) twig function. The "form_label()" function it's not reconized.
Thx for any suggestion.
I feel like this is the wrong approach to handle this. Extensions are for transforming data not really to manipulate the form definition itself.
First of all the order is defined as in the form type, so you can swap those around. To render the fields differently you can use form themes, or even rendering a custom form type.
Alternatively if its a one time thing (you could also create a macro for this) you can also instead of form_widget(form) order them in the way you like.
{{ form_start(form) }}
{{ form_row(form.field3) }}
{{ form_row(form.field1, { attr: { class: 'im-different' } }) }}
{{ form_row(form.field2) }}
{{ form_end(form) }}
Or even go deeper.
{{ form_start(form) }}
{{ form_row(form.field3) }}
<div>
{{ form_label(form.field1) }}
{{ form_widget(form.field1) }}
{{ form_errors(form.field1) }}
</div>
{{ form_row(form.field2) }}
{{ form_end(form) }}
To see these functions and how they all rendered by default you can look at form_div_layout.html.twig.
I agree with Jenne van der Meer and Nico Haase that your approach isn't particularly optimal. If I had the choice, I would go a different route: Instead of rendering in your function, render in twig, then pass the result to the function (like {{ myFunc(form_label(form), form) }}). Since you omit what your function actually needs and/or does, it's hard to provide further advice. However, I'm absolutely sure, that rendering can be done in twig before or after entering your function, via a macro/block, maybe even a form theme).
However, if you really really require your function to render the form field ... the following will possibly help you. I strongly advise against doing this, there's probably a better suited solution.
The form_label function is slightly more complex than a simple function. Instead, it uses twig's compile mechanisms to generate specific php code. It will eventually call:
FormRenderer::searchAndRenderBlock(FormView $view, string $blockNameSuffix, array $variables = [])
Diving deep into the compiler, the template call form_label(form, options) would be turned into:
$this->env->getRuntime('Symfony\Component\Form\FormRenderer')->searchAndRenderBlock(
$form, 'label', $options
);
where the $this->env seems to be the twig environment. That means, to call this in your twig extension you need to have access to the proper Twig environment, and then it should already work with the recipe I just provided. Especially if you can omit the options argument, I didn't take a deeper look into how that one's assembled (but it's probably just straight forward).
So your twig function must be defined via:
public function getFunctions(): array
{
return [
new TwigFunction('myFunc', [&$this, 'myFunc'], [
'needs_environment' => true, // <--- this!
'is_safe' => ['html'],
]),
];
}
public function myFunc(\Twig\Environment $env, $field) {
// other stuff
$html = $env->getRuntime(\'Symfony\Component\Form\FormRenderer\')->searchAndRenderBlock(
$field, 'label', $options
);
return $html;
}
Is there a possibility to display file in twig template when the the file is not stored inside public directory? Like call controller which would return binary response for requested file?
If you want to embed a controller response into a twig template you can use the render function.
{{ render(path('latest_articles', {max: 3})) }}
or
{{ render(controller('App\\Controller\\BlogController::recentArticles', {max: 3})) }}
Ref: https://symfony.com/doc/current/templates.html#embedding-controllers
I would like to have the current route in Symfony 3 in a Twig view.
I already did some research, but I didn't find any solution. I try the following in my Twig view:
{{ app.request.uri }}
It returns something like: http://localhost:8000/_fragment?_path=_format%3Dhtml%26_locale%3Dfr%26_controller%3DFOSUserBundle%253ASecurity%253Alogin
{{ app.request.get('_route') }}
Always returns NULL.
{{ app.request.getpathinfo }}
Always have: /_fragment
What I need is very simple. For an URL like localhost:8000/article/5, I would like to retrieve /article/5.How to do this?
The following code snippet will do the trick:
{{ path(app.request.attributes.get('_route'), app.request.attributes.get('_route_params')) }}
app.request.attributes.get('_route') - returns current route name.
app.request.attributes.get('_route_params') - returns current route params.
path() - generates route path by route name and params.
This approach will not work for forwarded requests. In case of forward request, there is a workaround: you should pass "_route" and "_route_params" to a forwarded params list.
return $this->forward('Yourbundle:Controller:action', array(
//... your parameters...,
'_route' => $this->getRequest()->attributes->get('_route'),
'_route_params' => $this->getRequest()->attributes->get('_route_params');
));
Also you can use app.request to access any Request object functions, such as getPathInfo(), which should return current route path.
{{ app.request.pathInfo }}
I am passing an array of get arguments from controller to twig template, and then create a link:
{{ url('route_name', array_of_get_parameters) }}
It works, but what if I want to pass all BUT ONE of those parameters unchanged? Something like:
{{ url('route_name', array_of_get_parameters, {'param1': 'value'}) }}
The example above doesn't work of course...is there a way to do this?
Use twig merge filter like this:
{{ url('route_name', array_of_get_parameters|merge({'param1': 'value'})) }}
You cannot do this.
Instead override the value in the controller (<- better) or in the twig template, before the url generation.