I am new to Symfony2, i am using FOSUserBundle. I overwrote the default template for the FOSUserBundle, and doing so i have lost the 'loggout' link on the profile page at
resource: "#FOSUserBundle/Resources/config/routing/profile.xml"
Is there a way to loggout from the symfony tool bar at the bottom?
and what is the resource link to loggout a user in fosUserBundle
I think, that the default route for logout in FOSUerBundle is "fos_user_security_logout", so you can generate a logout link anywhere in your site with this route:
<a href="{{ path('fos_user_security_logout') }}">
It is also possible to add a new block to the Symfony2 debug toolbar with the logout link.
You should have a look here: Adding Web Profiler Templates
You can extend the 'WebProfilerBundle:Profiler:layout.html.twig' layout and add the custom content to the toolbar block.
{% block toolbar %}
{# the web debug toolbar content #}
{% endblock %}
Related
is there an easy way to always show the twig vars {{dump()}} in the symfony web developer toolbar? I already have implemented a custom data collector but if I add the following lines in my collector templates:
<div class="sf-toolbar-info-piece">
<b>Twig vars</b> {{ dump() }}
</div>
only the vars given to the toolbar will be shown not the ones I added in to the render function in the controller.
Basically I want something like the output that you see if you add
{% dump %}
in your twig template.
I've created a custom Sonata page
Simple route
medapp_adminStreamCommands:
path: /admin/stream
defaults: { _controller: MedAppBundle:VideoChat/VideoChat:adminStreamCommands }
Controller that returns the admin pool
public function adminStreamCommandsAction(Request $request)
{
return $this->render('#MedApp/AdminSonata/Stream/stream_commands.html.twig', array(
'admin_pool' => $this->get('sonata.admin.pool')));
}
Plain view template
{% extends '#MedApp/AdminSonata/standard_layout.html.twig' %}
{% block content %}
foobar
{% endblock content
This works, I can access it on my website with /admin/foo and I get a page which has the Sonata admin template with my 'foobar' content.
My question is, how can I add this route to the left and top navbar without having to modify the default template?
That is because the left menu is rendered by a KNP menu:
{% block side_bar_nav %}
{% if app.user and is_granted('ROLE_SONATA_ADMIN') %}
{{ knp_menu_render('sonata_admin_sidebar', {template: admin_pool.getTemplate('knp_menu_template')}) }}
{% endif %}
{% endblock side_bar_nav %}
And I somehow need to add my new page to be rendered by this menu.
Normally, a page is added through a service, but these are built on top of an entity:
servicename:
class: Bundle\Class
arguments: [~, Bundle\Entity\Entityname, ~]
tags:
- { name: sonata.admin, manager_type: orm, group: admin, label: CustomName}
My page is not using an entity, though, just static content or content that is not dependant on an entity.
I know already that I can modify the blocks that generate the menus, but I was thinking that the best way would be to add my class as a service tagged as sonata.admin that doesn't have an orm manager_type, in other words, is not an Entity. How can that be done?
You should override standard_layout and modify content of side_bar_nav block. This is simple and fast way. Or you can dig into sonata code to find how to inject something into admin_pool.dashboardgroups - have fun :)
I don't think that's possible, you have to create a new layout, copy the sonata admin layout and customize it to your need.
You can change the layout used by changing the yml configuration for sonata_admin (templates -> layout) or extending the SonataAdmin bundle and creating your own layout.html.twig.
I need to apply a purchased template to our dashboard. In this template, the login, register and forgot password forms are all under the same view, and switching between them using simple JQuery.
I have been looking for a nice, not-too-flashy way of combining all three forms into one, but I came up empty.
My standing options (as I see them), and why I don't like any of them:
Take the views from the fos bundle, copy them to /app/Resources/FOSUserBundle/views/, remove the {% extend %} part and {% include %} them in my own login view. Reason for dislike: to me this looks a little like a quick-n-dirty fix - "that part's not working? Let's break it off!" :)
Extend the fos bundle, accept an extra parameter in the LoginAction and RegisterAction, use {% render %} with parameters in my own login view. Reason for dislike: extending a whole bundle and modifying two different controllers just to change the way it renders feels like bad MVC.
XHR load everything. Reason for dislike: this approach makes sense when using inner pages, but for pages that reload anyway it just doesn't make sense.
TL;DR version: I'm looking for a non-hack way of including the login, register and forgot password form in one page.
Any help would be greatly appreciated!
I found a solution with which I am comfortable with for my current project. The advantages and disadvantages of the proposed solution upfront:
Advantages:
few LOC to implement
FOSUserBundle update proof (does not override the view scripts*)
Disadvantages:
performance overhead due to subrequests
only forms can be displayed, form submission (and subsequently error handling upon submission) will always go to the pages provided by FOSUserBundle
still feels like a quick-n-dirty fix, but better than other options
* only needs to override the layout.html.twig file
With that being said, here is what I have done:
Render the form in your template
Use embedded controllers to render the forms you need:
<div>
<h2>Login</h2>
{{ render(controller('FOSUserBundle:Security:login', { embeddedForm: true})) }}
</div>
<div>
<h2>Reset</h2>
{{ render(controller('FOSUserBundle:Resetting:request', { embeddedForm: true})) }}
</div>
Override FOSUserBundle layout
As I use the routes provided by the bundle, I had to override the FOSUserBundle layout template file to extend the standard layout of my application. As the overriden FOSUserBundle layout file extends the main applications layout file the layout would be repeated for each call {{ render ... }}. To prevent that, we need to dynamically disarm the extended layout file. Here is what the overriden layout file looks like:
{# app/Resources/FOSUserBundle/views/layout.html.twig #}
{% if app.request.get('embeddedForm') %}
{% set layout = 'AcmeBundle::layout-content.html.twig' %}
{% else %}
{% set layout = 'AcmeBundle::layout.html.twig' %}
{% endif %}
{% extends layout %}
{% block content %}
{% block fos_user_content %}{% endblock %}
{% endblock %}
Create the AcmeBundle::layout-content.html.twig file
This layout should only render the content block of the FOSUserBundle view scripts and is such short and simple:
{# src/Acme/DemoBundle/Resources/views/layout-content.html.twig #}
{% block content %}{% endblock %}
Now the forms will render nicely with all dependencies (CSRF and so forth). Submitting the form will however take you to the FOSUserBundle actions.
Alternative solution:
This answer describes how to manually implement the forms and link them to the FOSUserBundle controller.
On my eZ publish 5 site I have all my templates in Twig, in the vendor/ezsystems/demobundle/EzSystems/DemoBundle/Resources/views/ subfolders. They are all being used throughout my whole site, no problems there. With one exception: 404 pages. If I go to mysite/nonexistingurl, it gives me a kernel (20) / Error page, with status 404. The template being used for this is the 20.tpl somewhere in eZ publish/symfony, I don't want that, I want to use my own Twig template for this.
How can I achieve this? I added a vendor/ezsystems/demobundle/EzSystems/DemoBundle/Resources/views/Exception/error.html.twig page, but this one is not being called
first add this configuration parameter
parameters:
ezpublish_legacy.default.module_default_layout: 'YourBundle::pagelayout_legacy.html.twig'
you may add it in the parameters.yml file located in path/to/yourezpublishinstall/ezpublish/config, the parameters.yml is usually imported in the config.yml located in the same folder
this would define the twig template located in path/to/yourbundle/Resources/views/pagelayout_legacy.html.twig as the parent template for legacy stack modules templates
inside the pagelayout_legacy.html.twig template, you may use this code
{% extends 'YourBundle::pagelayout.html.twig' %}
{% block content %}
{# module_result variable is received from the legacy controller. #}
{% if module_result.errorCode is defined %}
<h1>{{ module_result.errorMessage }} ({{ module_result.errorCode }})</h1>
{% else %}
{{ module_result.content|raw }}
{% endif %}
{% endblock %}
note in the code, the template extends the pagelayout.html.twig template, that should here define a block named content, the pagelayout.html.twig may usually be the main base layout for your ez publish 5 website
you may modify the pagelayout_legacy.html.twig template to your needs
reference:
http://share.ez.no/forums/developer/overriding-legacy-error-pages-templates
I'm using Symfony2. I have a main twig template with a navbar where I want to have a pull-down menu with the options 'log in' or 'log out' depending on whether the user is logged in or not.
What is the best approach to accomplish this?
-By making a different, static main template for /admin/* with the log out option
-By checking whether the user is logged in or not inside each controller being called by the routing system and then passing the correct information to be shown to the twig template when it's time to render it
-By calling a specific controller like buildNavbarAuthOption() from inside the template
Last option doesn't seem the best one when trying to program using MVC, right?
Just use is_granted in your views:
{% if is_granted('IS_AUTHENTICATED_FULLY') %}
Logout
{% else %}
Login
{% endif %}
In your main template you could test if the variable app.user exists like this:
{% if app.user %}
{# display logout #}
{% else %}
{# display login #}
{% endif %}
Or better yet, test if the current visitor has the minimum security role:
{% if is_granted('IS_AUTHENTICATED_REMEMBERED') %}