Short version:
Is there a simple, built-in way to identify the calling view in a Django template, without passing extra context variables?
Long (original) version:
One of my Django apps has several different views, each with its own named URL pattern, that all render the same template. There's a very small amount of template code that needs to change depending on the called view, too small to be worth the overhead of setting up separate templates for each view, so ideally I need to find a way to identify the calling view in the template.
I've tried setting up the views to pass in extra context variables (e.g. "view_name") to identify the calling view, and I've also tried using {% ifequal request.path "/some/path/" %} comparisons, but neither of these solutions seems particularly elegant. Is there a better way to identify the calling view from the template? Is there a way to access to the view's name, or the name of the URL pattern?
Update 1: Regarding the comment that this is simply a case of me misunderstanding MVC, I understand MVC, but Django's not really an MVC framework. I believe the way my app is set up is consistent with Django's take on MVC: the views describe which data is presented, and the templates describe how the data is presented. It just happens that I have a number of views that prepare different data, but that all use the same template because the data is presented the same way for all the views. I'm just looking for a simple way to identify the calling view from the template, if this exists.
Update 2: Thanks for all the answers. I think the question is being overthought -- as mentioned in my original question, I've already considered and tried all of the suggested solutions -- so I've distilled it down to a "short version" now at the top of the question. And right now it seems that if someone were to simply post "No", it'd be the most correct answer :)
Update 3: Carl Meyer posted "No" :) Thanks again, everyone.
Since Django 1.5, the url_name is accessible using:
request.resolver_match.url_name
Before that, you can use a Middleware for that :
from django.core.urlresolvers import resolve
class ViewNameMiddleware(object):
def process_view(self, request, view_func, view_args, view_kwargs):
url_name = resolve(request.path).url_name
request.url_name = url_name
Then adding this in MIDDLEWARE_CLASSES, and in templates I have this:
{% if request.url_name == "url_name" %} ... {% endif %}
considering a RequestContext(request) is always passed to the render function. I prefer using url_name for urls, but one can use resolve().app_name and resolve().func.name, but this doesn't work with decorators - the decorator function name is returned instead.
No, and it would be a bad idea. To directly refer to a view function name from the template introduces overly tight coupling between the view layer and the template layer.
A much better solution here is Django's template inheritance system. Define a common parent template, with a block for the (small) area that needs to change in each view's version. Then define each view's template to extend from the parent and define that block appropriately.
If your naming is consistent in your urls.py and views.py, which it should be, then this will return the view name:
{{ request.resolver_match.url_name }}
Be sure to apply some context to it when you call it in the template. For example, I use it here to remove the delete button from my detail view, but in my update view the delete button will still appear!
{% if request.resolver_match.url_name != 'employee_detail' %}
Since Django 1.5 you can access an instance of ResolverMatch through request.resolver_match.
The ResolverMatch gives you the resolved url name, namespace, etc.
one simple solution is :
def view1(req):
viewname = "view1"
and pass this viewname to the template context
def view2(req):
viewname = "view2"
and pass this viewname to the template context
in template access the viewname as
{{viewname}}
and also you can use this in comparisons.
I'm working on this for a help-page system where I wanted each view to correspond to a help-page in my cms with a default page shown if no help page was defined for that view. I stumbled upon this blog where they use a template context processor and some python inspect magic to deduce the view name and populate the context with it.
This sounds like the perfect example of a generic view that you can set up.
See the following resources:
Django Book - Chapter 11: Generic Views
Django Docs -Tutorial: Chapter 4
Django Docs - Generic Views
These links should help you simplify your views and your templates accordingly.
If you're using Class Based Views, you most likely have a view variable you can access.
You can use several methods from that to determine which view has been called or which template is being rendered.
e.g.
{% if view.template_name == 'foo.html' %}
# do something
{% else %}
# other thing
{% endif %}
Another option is to take out the piece of the template where you need something to change and make it into a snippet and then use {% include 'my_snippet.html' with button_type = 'bold' %} in your templates, sending arbitrary values to the snippet so it can determine what to show / how to style itself.
Most generic views — if not all — inherits the ContextMixin which adds a view context variable that points to the View instance.
In your template, you can access the current view instance like this:
{{ view }}
Define class_name method in your view
class ExampleView(FormView):
...
def class_name(self):
return self.__class__.__name__
You can get the class name of the current view in a template like this:
{{ view.class_name }}
{% if view.class_name == "ExampleView" %} ... {% endif %}
Why not trying setting up a session cookie, then read the cookie from your template.
on your views set cookies
def view1(request):
...
#set cookie
request.session["param"]="view1"
def view2(request):
request.session["param"]="view2"
then in your ONE template check something like..
{% ifequal request.session.param "view1" %}
... do stuff related to view1
{% endifequal %}
{% ifequal request.session.param "view2" %}
... do stuff related to "view2"
{% endifequal %}
Gath
Related
i cant quite find it so i hope someone can help me out.
I found the option of using the
TEMPLATE_CONTEXT_PROCESSORS = ("django.contrib.auth.context_processors.auth" )
In django (1.5). But now its not clear for me how i should use it. Should i still put the request in my views, or can i with this enabled use the user_object in my template without sending an extra variably with the Requestcontect
For example:
My view at the moment:
def user_characters(request, user_id):
characters = Character.objects.filter(user=user_id)
user = User.objects.get(id=user_id)
return render_to_response('characters.html',
{'characters': characters, "user": user},
context_instance=RequestContext(request))
My template:
{% extends "base.html" %}
{% block mainframe %}
{% if characters|length < 3 %}
<p>New Character(WN)</p>
{% endif %}
And then the rest of my view.
I notice in almost every view i make i want the user_object send with it.
Can someone please give me an example of how this works?
With kind regards
Hans
django.contrib.auth.context_processors.auth context processor is enabled by default, you don't have to add anything. When you use RequestContext(), a context variable user is available in all templates that you can use. To get id {{userd.id}}.
To check user is authenticated or not, do
{% if user.is_authenticated %}
{# handle authenticated user #}
{%else%}
{# handle anonymous non-authenticated users #}
{%endif%}
You should not expose the user id in the url, you wont need it anyway, if you use django sessions- and the authentication framework. You can always check the logged in user via request.user in your serverside view. With the context processor your should be able to access the user with user.desiredattribute, but you should not need it for the url you try to create.
The docs on this seem pretty clear to me:
https://docs.djangoproject.com/en/dev/ref/templates/api/#django.template.RequestContext
If you want context processors to function, you must ensure that you're using a RequestContext instance. You can do that by explicitly creating it in your views, as you show, or (more conveniently, in my opinion) by using the render shortcut rather than render_to_response as documented here:
https://docs.djangoproject.com/en/dev/topics/http/shortcuts/#render
With the django.contrib.auth.context_processors.auth context processor in place, the user will always be available in the context variable user. At least, assuming your template is being rendered with a RequestContext instance.
You absolutely should not trust a variable obtained from the URL to determine the user if you have any kind of controlled information. With the system you have shown, anyone can view anyone's data simply by editing the URL. That might be OK for a totally insecure application, but it's much more normal to look at request.user.
What I want to do is include a form from a separate template at the bottom of a given page, lets say; "example.com/listdataandform/".
The form-template "form.html" displays the form as it should when the view is included in the URLConf. So I can view with "example.com/form/"
What I have so far goes something like this:
{% extends "base/base.html" %}
{% block title %} page title {% endblock %}
{% block content %}
<h2>some "scene" data</h2>
<ul>
{% for scene in scenes %}
<li>{{ scene.scene }} - {{ scene.date }}</li>
{% endfor %}
</ul>
{% include "tasks/form.html"%}
{% endblock %}
The code inside "block content" works as it should, since it is defined with it's corresponding view for the url "example.com/listdataandform/".
{% include "tasks/form.html"%}: This only displays the submit button from form.html, as expected. I realize by only doing this: {% include "tasks/form.html"%}, the corresponding view method is never executed to provide the "form"-template with data.
Is there any way to this without having to define the view to a specific pattern in urls.py, so that the form can be used without going to the that specified URL..?
So I guess the more general question is; how to include templates and provide them with data generated from a view?
Thanks.
For occasions like this, where I have something that needs to be included on every (or almost every) page, I use a custom context processor, which I then add to the TEMPLATE_CONTEXT_PROCESSORS in settings.py. You can add your form to the context by using this method.
Example:
common.py (this goes in the same folder as settings.py)
from myapp.forms import MyForm
def context(request):
c = {}
c['myform'] = MyForm()
return c
You can also do any processing required for the form here.
Then add it in your settings.py file:
settings.py
.
.
TEMPLATE_CONTEXT_PROCESSORS = (
'''
All the processors that are already there
'''
"myproject.common.context",
)
.
.
I realize by only doing this: {% include "tasks/form.html"%}, the corresponding view method is never executed to provide the "form"-template with data.
Indeed. You included a template, and it really means "included" - ie: "execute in the current context". The template knows nothing about your views, not even what a "view" is.
How does this help me executing the view for the included template to provide it with form data?
It doesn't. A Django "view" is not "a fraction of a template", it's really a request handler, iow a piece of code that takes an HTTP request and returns an HTTP response.
Your have to provide the form to the context one way or another. The possible places are:
in the view
in a context processor (if using a RequestContext)
in a middleware if using a TemplateResponse AND the TemplateResponse has not been rendered yet
in a custom template tag
In all cases this will just insert the form in your template's context - you'll still have to take care of the form processing when it's posted. There are different ways to address this problem but from what I guess of your use case (adding the same form and processing to a couple differents views of your own app), using a custom TemplateResponse subclass taking care of the form's initialisation and processing might just be the ticket.
My Django app currently has URLs which are protected by 'permission_required()' functions.
This function is called in three different ways.
As a decorator in views.py, with hardcoded parameters.
As a plain function, with autogenerated parameter, in custom Class Based Generic Views.
As a function invoking views in urls.py, with hardcoded parameters.
I'm now adding a menu system to the app, and I need to make menu entries reflect whether the user has permission to request the URL of each menu entry. (Either by greying-out or hiding said entries.)
Is there a way of query the permissions required to a URL without requesting the URL?
The only solution I've thought of so far is to replace the decorator with a parameterless 'menu_permssion_required()' decorator and hardcode all of the permissions into a Python structure. This seems like a step backwards, as my custom Class Based Generic Views already autogenerate their required permissions.
Any suggestions on how to make a menu system which reflects URL permissions for the current user?
Here is an example of how to solve your problem:
First, Create a decorator wrapper to use instead of permission_required:
from django.contrib.auth.decorators import login_required, permission_required, user_passes_test
from django.core.exceptions import PermissionDenied
from functools import wraps
from django.utils.decorators import available_attrs
def require_perms(*perms):
def decorator(view_func):
view_func.permissions = perms
#wraps(view_func, assigned=available_attrs(view_func))
def _wrapped_view(request, *args, **kwargs):
for perm in perms:
return view_func(request, *args, **kwargs)
raise PermissionDenied()
return _wrapped_view
return decorator
Then, use it to decorate your views:
#require_perms('my_perm',)
def home(request):
.....
Then, add a tag to use for your menu items:
from django.core.urlresolvers import resolve
def check_menu_permissions(menu_path, user):
view = resolve(menu_path)
if hasattr(view.func, "permissions"):
permissions = view.func.permissions
for perm in permissions:
if user.has_perm(perm):
return True # Yep, the user can access this url
else:
return False # Nope, the user cannot access this url
return True # or False - depending on what is the default behavior
And finally, in your templates, when building the menu tree:
<button href="{{ some_path }} {% if not check_menu_permissions some_path request.user %}disabled="disabled"{% endif %} />
N.B. I've not tested the last part with the tag, but I hope you got the idea. The magic thing here is to add the permissions to the view_func in the decorator, and then you can access this using resolve(path). I'm not sure how this will behave in terms of performance, but after all that's just an idea.
EDIT: Just fixed a bug in the example..
Is there a way of query the permissions required to a URL without requesting the URL?
User.has_perm() and User.has_module_perms()
Any suggestions on how to make a menu system which reflects URL permissions for the current user?
I really like this question, because it concerns anyone that makes a website with django, so I find it really relevant. I've been through that myself and even coded a menu "system" in my first django project back in 2008. But since then I tried Pinax, and one of the (so many) things I learnt from their example projects is that it is completely unnecessary bloat.
So, I have no suggestion which I would support on how to make a menu "system" which respects the request user permissions.
I do have a suggestion on how to make a simple menu which respects the request user permissions, so that might not be completely unrelated.
Just make your menu in plain HTML, it's not like it's going to change so often that it has to be generated. That will also keep your Python code simpler.
Add to settings.TEMPLATE_CONTEXT_PROCESSORS: 'django.core.context_processors.PermWrapper'
Use the {{ perms }} proxy to User.has_perms.
Example:
{% if perms.auth %}
<li class="divider"></li>
{% if perms.auth.change_user %}
<li>
{% trans 'Users' %}
</li>
{% endif %}
{% if perms.auth.change_group %}
<li>
{% trans 'User groups' %}
</li>
{% endif %}
{% endif %}
{# etc, etc #}
That's how I keep navigation simple, stupid, and out of the way. But also I always include an autocomplete not far from the menus to allows the user to navigate to any detail page easily. So, that's all I know about navigation in django projects, I'm eager to read other answers !
I had a similar issue, but it went a little deeper. Instead of just permissions, I also wanted other tests based on the lidded in user (ie, is_staff, or user.units.count() > 1). Duplicating these in the view and the template seems prone to errors.
You can introspect a view object, and see all of the decorators wrapping it, and work out if they are checks (in my case: the first argument I'd u or user). If they all pass, then allow rendering the link.
Get all decorators wrapping a function describes the technique in a little more detail. You can find the app that wraps this up into a handy replacement for {% url %} at Django-menus.
Is there another way to get the request.user by not passing it from the views? I think passing request.user from all functions in views to the template is quite wrong. Is there any method or way that the template will get the user or any object in the database?
By default (I am talking about Django version 1.3) you do not need change TEMPLATE_CONTEXT_PROCESSORS. Because default value already contains *django.contrib.auth.context_processors.auth*.
So to your question: By default, you should be able to use user, messages and perms variables in your template. For example:
User: {{user.username}}
{% if perms.appname.permname %}
... do something usefull ...
{% endif %}
Context processors.
So we're writing a super simple game-app (answer questions, get points) and we want to render a high-score in the side-bar of every screen on the page.
So I understand that Django automatically (kinda) passes the authenticated user as argument (if we use render() ) but I'm not sure how to pass what basically will be:
top_list = User.objects.filter().order_by('-score')[:10]
so that we can fix a nice little top list in our base template :)
We've looked around at different questions here, like this one for example and thought about whether the solution would be to write our own context processor?
EDIT:
Might be a duplicate of this so should I perhaps write a custom template tag instead?
You can use django template tags to do it..
Under myapp/templatetags/ create a myapp_tags.py and __init__.py
In myapp_tags.py do the following:
from django import template
register = template.Library()
#register.assignment_tag
def get_top_list():
top_list = User.objects.filter().order_by('-score')[:10]
return top_list
And then use it in your templates like this:
{% load myapp_tags %}
{% get_top_list as top_list %}
{% for top_user in top_list %}
...