I have been able to build a simple CRUD app for a project using the Symfony EasyAdminBundle and it has worked great for the normal entity based use cases. I have some additional use cases though where I want to do things like rebuilding data. For these I have to capture certain request attributes, pass over to a controller and then delegate to a backend API call to a remote service.
This can all be done in Symfony but I am running into trouble with how to wire this into the EasyAdmin view/method of working. Ideally I want this to be a page inside easy admin and not lose the left menu etc. So far, the only way I have found to do this is to create a Model class what is using one of the existing tables but has only some properties I would need to drive into the API. I then override the controller actions so rather than do a default save, I handle against that remote API.
The problem with this approach is that obviously I am now bound to Doctrine entities and that would be problematic for requests that were not mappable to the database.
Is there a way to define a logical entity that would let me leverage associations so I can have lookups etc, that will wire into the bundle seamlessly, but are not actually tied to a backend database table or view?
I am adding my response for people who may still face this issue in the future.
How I solved this without creating an entity :
Create a custom controller: symfony console make:controller
Edit the controller's view to inherit the EasyAdmin layout :
{# ./src/templates/home/index.html.twig #}
{% extends '#EasyAdmin/Default/layout.html.twig' %}
{# Let\'s remove/empty the header #}
{% block content_header_wrapper %} {% endblock content_header_wrapper %}
{# The main page content block #}
{% block main %}
**PUT YOUR CODE HERE**
{% endblock main %}
{# Let\'s remove/empty the footer #}
{% block content_footer_wrapper %} {% endblock content_footer_wrapper %}
Add your page to the side navigation
design:
menu:
- {route: 'home', label: 'Home', default: true, icon: 'home'}
- {entity: 'MyEntity', label: 'My Relevant Entity', icon: 'briefcase'}
I'd solve this problem creating a custom action as explained here (probably you want a route-based action) and then use a template that extends from #EasyAdmin\default\layout.html.twig or any other default template similar to what you want to achieve.
Here is solution:
{# easy_admin/form.html.twig #}
{% block _product_custom_title_widget %}
{# ... #}
More information
{% endblock %}
Finally, add this custom theme to the list of themes used to render backend forms:
easy_admin:
# ...
design:
form_theme:
- 'horizontal'
# the following Twig template can be located anywhere in the application.
# it can also be added to the twig.form_themes option to use it in the
# entire application, not only the backend
- 'easy_admin/form.html.twig'
Here is link for more information: https://symfony.com/doc/master/bundles/EasyAdminBundle/book/edit-new-configuration.html
Easyadmin is a bundle and you can customize all the pages of any bundle.
This logic applies to any template that lives in a bundle: just follow the convention: app/Resources/{BUNDLE_NAME}/views/{PATH/TO/TEMPLATE.html.twig}.
Suppose you've installed an imaginary open-source AcmeBlogBundle in your project. And while you're really happy with everything, you want to override the template for a blog list page. Inside the bundle, the template you want to override lives at Resources/views/Blog/index.html.twig.
To override the bundle template, just copy the index.html.twig template from the bundle to app/Resources/AcmeBlogBundle/views/Blog/index.html.twig (the app/Resources/AcmeBlogBundle directory won't exist, so you'll need to create it). You're now free to customize the template.
Reference: https://symfony.com/doc/3.4/templating/overriding.html
Related
I've built a site where I can create new posts (essays) by the admin panel. The output is visible to users. But when I place some HTML as content in the form it doesn't render itself on the page.
example:
Output on the page (with marked unrendered HTML):
I would like to know how to fix it and also, how to name the topic I want to know ( I couldn't find anything related to my problem, probably because I don't know how to express it).
Additionally, I just start to wonder if there is one more problem nested inside. How to link CSS from the static folder having this HTML mentioned above?
Django offer the autoescape template in the builtins tags
{% autoescape off %}
{{ myhtml }}
{% endautoescape %}
But your logic seems wrong, you don't need to create a new page with the doctype, just create a base template and use the block content tag to insert your article.
In your base template replace the description and title of your page by variables that will be populated by the article data.
You need to learn the basic of Django https://docs.djangoproject.com/en/4.1/ trust me you won't regret it !
I am building an application with Django as the backend and React js for making the interface of the application.
I have a set of Posts which I want to display.
Currently, the approach which I am following is -
Get the template having the compiled js code linked to it.
Then again make get call to get the posts
My question is - In this current approach I am making 2 GET calls to the backend, one for rendering the template and then again for getting the Post.
What is the best way to achieve this? Is this the usual flow how applications are built using Django and React JS?
First off: I don't see anything wrong with doing this in two requests, because one loads the application itself and the second loads the content. To me this seperation makes sense and might turn out to be useful in the future if you want to reuse say the Endpoint, that yields the the posts(i.e. the content).
Answering your question: If, for whatever reason, you absolutely want to load everything with a single GET, a good way of doing so, would be to pass a list of posts to the context as a JSON-serialized object and then load these into the JS-context within the Django-template.
i.e. in the view.py:
from json import dumps
def view(request):
context = {
'posts':get_posts(),
}
render_to_response('django_template.html', context,
context_instance=RequestContext(request))
def get_posts():
qs = Posts.objects.all()
return dumps({'posts': qs })
in the django_template:
{% block content %}
<div id="root"></div>
{% endblock %}
{% block javascript %}
<script>
var INITIAL_POSTS = {{ posts|safe }};
</script>
<script type="text/javascript"
src="PATH_TO_STATIC_REACT_ASSET.JS"></script>
{% endblock %}
you should now have your posts in your JS context and can load them in your React component. Once again: I would agree with Daniel Rosemans comment
Is there any way to make global placeholder in my base template? I need it to be the same on every page (banners list).
How can I do that?
I usually create a page in my CMS that isn't published, but contains placeholders that I would like to use elsewhere (footer/headers) etc.
Make a new template extra_placeholders.html:
{% extends "base.html" %}
{% load cms_tags %}
{% block content %}
{% placeholder "Banner-List" %}
{% endblock %}
add it to your settings:
CMS_TEMPLATES = (
('my/path/extra_placeholders.html', 'Extra Placeholder Page'),
...
)
now go to the admin and create the placeholder with whatever plugin you want. Then go to you base template (*base.html probably) from which all your other pages inherit and add this wherever you want the placeholder to appear:
{% load cms_tags %}
...
{% show_placeholder "Banner-List" "extra_placeholders" %}
You can read more about it in the docs
EDIT
As #José L. Patiño has mentioned in the comments, this solution is only necessary for those using django-cms < 3.0. For the newer version you can simply use the static_placeholder template tag
There is the "static_placeholders" now, http://docs.django-cms.org/en/latest/reference/templatetags.html#static-placeholder
Sounds like it's what you needed way back when.
You can use the following ways to create a global palceholder for all pages.
create a place holder on the base page. {% Placeholder "footer"%}
make the contents of the placeholder through django cms as home page
then to display the same for each placeholder page, add {% show_placeholder "footer" "home"%}, this means displaying the newly created footer placeholder earlier from the home page,
This will display the entire contents existing footer placeholders on the home page of all pages that use the template.
but for the home page there will be two footer is displayed, to mengilangkannya, please do modifications to use CSS to hide the master placeholder.
I think I'm missing a basic concept here. In the stereotypical Django project, you'd have two apps responding to different urls:
http://localhost/myproj/app1/33
http://localhost/myproj/app2/newcomment.html
But what mechanisms exist to handle cases where the two apps are complementary - say one provides content, and the other provides presentation? Or maybe one is content and the other is a kind of static, side-wide content that should appear on every page.
In particular, I don't understand how both apps can use template inheritance to extend the same template. Imagine there's a base app "baseapp" with a template "base.html":
...
<div blah blah>
{% block content %}
{% endblock %}
...
App1 extends it:
{% extends "baseapp/templates/base.html" %}
{% block content %}
... here's the actual content...
{% endblock %}
App2 adds a little banner or something:
{% extends "baseapp/templates/base.html" %}
{% block content %}
<div class="banner">Please support our site!</div>
{{ block.super }}
{% endblock %}
So what are the ways that both templates can get displayed? I can think of:
app1 could extend app2's templates. But this seems wrong: app1 is the content provider, and shouldn't be dependent on something as trivial as app2.
app2 could extend app1's templates. But this seems wrong: now the URL scheme would have to funnel every URL through app2 (if I understand correctly)
middleware?
As I said, I'm probably missing something very basic. Or I'm making some very faulty assumptions that I don't know about. (This question is my third attempt, after Embed an optional Django application in another page, if that app is present and How to capture and display information external to my webapp, but relevant to users of it? - I'm having trouble framing the issue.)
App doesn't respond to an URL, a view does. View is a function that can use models, forms and other object from any app. There isn't any problem here.
If you want to add something to template, inheritance isn't the only way. You'd better use custom context processor or custom template tag.
I think what I was actually missing here:
Apps can override templates just by including a template of the right name in the right subdirectory. The Django docs don't make this very clear, that I can see: they refer to this functionality in the context of Admin templates
When overriding a template as above, you can't extend it, but:
This snippet lets you both override a template and extend it: http://djangosnippets.org/snippets/1376/
Here's a closely related question: Django: Overriding AND extending an app template
I have template that displays object elements with hyperlinks to other parts of my site. I have another function that displays past versions of the same object. In this display, I don't want the hyperlinks.
I'm under the assumption that I can't dynamically switch off the hyperlinks, so I've included both versions in the same template. I use an if statement to either display the hyperlinked version or the plain text version. I prefer to keep them in the same template because if I need to change the format of one, it will be easy to apply it to the other right there.
The template extends framework.html. Framework has a breadcrumb system and it extends base.html. Base has a simple top menu system.
So here's my dilemma. When viewing the standard hyperlink data, I want to see the top menu and the breadcrumbs. But when viewing the past version plain text data, I only want the data, no menu, no breadcrumbs. I'm unsure if this is possible given my current design. I tried having framework inherit the primary template so that I could choose to call either framework (and display the breadcrumbs), or the template itself, thus skipping the breadcrumbs, but I want framework.html available for other templates as well. If framework.html extends a specific template, I lose the ability to display it in other templates.
I tried writing an if statement that would display a the top_menu block and the nav_menu block from base.html and framework.html respectively. This would overwrite their blocks and allow me to turn off those elements conditional on the if. Unfortunately, it doesn't appear to be conditional; if the block elements are in the template at all, surrounded by an if or not, I lose the menus.
I thought about using {% include %} to pick up the breadcrumbs and a split out top menu. In that case though, I'll have to include it all the time. No more inheritance. Is this the best option given my requirement?
You can put your hyperlinks inside a block that is overridden by the loading templates.
Let's say you have your framework.html like this:
{% extends "base.html" %}
<html>...<body>...
{% block hyperlinks %}
your hyperlinks here
{% endblock %}
rest of the code
</body></html>
You can then create something of a nolinks.html template and use it
{% extends "framework.html" %}
{# here you'll have everything from framework
but now we disable the breadcrumbs #}
{% block hyperlinks %}{% endblock %}
If you're getting the past data you can then use nolinks to render instead of framework.
I hope this helps.
From here: Any way to make {% extends '...' %} conditional? - Django
It can be done like this :
{% extends ajax|yesno:"ajax_base.html,main_base.html" %}
Or:
{% extends a_variable_containing_base_template_name %}
Which ever best suited for you.
Regards;