In a django template I don't want to show some element in case if the url/path is a specific one. In pseudo:
{% if not url = account:detail %}
We can do this in two steps here:
first we resulve the url, and assign it to a variable (here url2); and
next compare the urls
So:
{% url account:detail as url2 %}
{% if url != url2 %}
<!-- ... (do something) ... -->
{% endif %}
Note however that if two urls are syntactically different (for example yourdomain.com/foo and /foo), that they per se point to something different.
If you want access to the current path, you can - like #RajaSimon says, usually use request.path (given you render the template with a RequestContext, render(..), or another way to pass the request object).
You can give your urls a name ( url_name ) in urls.py file and then you can compare directly using HttpRequest.resolver_match object available in template.
from django.urls import path
from . import views
urlpatterns = [
path('articles/<int:year>/', views.year_archive, name='news_year_archive'),
# ...
]
In template you compare it like,
{% if request.resolver_match.url_name == "news_year_archive" %}
...your stuff
{% endif %}
Related
this is my django+html code
{% extends "tasks/layout.html" %}
{% block body %}
<h1>Tasks Lists</h1>
Add a task
{% endblock %}
this is where i use varriable name add and link it to another page.
Urls.py
from django.urls import path
from . import views
app_name = "tasks"
urlpatterns=[
path("",views.index,name="index"),
path("add",views.add,name="add")
]
this is where i use the varriable name "add".
A template tag [Django-doc] is written between curly brackets, so It is {% … %}, not [% … %], you thus write this as:
Add a task
In my Django project, I have created a custom admin page for an app via the get_urls() method. I'd like to add a link to the app's main model index view that will take users to this custom page - however, I'm having some trouble creating this link element correctly and I don't seem to be able to piece together the right way to do it - I'm just left with a Reverse for 'export' not found. 'export' is not a valid view function or pattern name. error.
I've set up the admin for the app like so:
# my_project/observations/admin.py
from django.template.response import TemplateResponse
from django.urls import path
class ObservationAdmin(SimpleHistoryAdmin, SoftDeletionModelAdmin):
change_list_template = 'export_link.html'
def get_urls(self):
urls = super().get_urls()
custom_urls = [
path('export/', self.admin_site.admin_view(self.export_view), name='export')
]
return custom_urls + urls
def export_view(self, request):
context = dict(
self.admin_site.each_context(request),
)
return TemplateResponse(request, 'export.html', context)
and the two templates that are referenced:
# my_project/observations/templates/export.html
{% extends "admin/base_site.html" %}
{% block content %}
<div>
Some custom content
</div>
{% endblock %}
# my_project/observations/templates/export_link.html
{% extends 'admin/change_list.html' %}
{% block object-tools-items %}
<li>
Export
</li>
{{ block.super }}
{% endblock %}
Navigating directly to http://localhost:8000/admin/observations/observation/export/ works perfectly, I see the custom content page exactly as I want it... so the issue I'm striking is with the link template - I get the Reverse... error when I navigate to the model index page.
Perhaps the argument I'm passing to url is incorrect, or I need to register that URL elsewhere - but I don't quite know. The other examples of link elements like this that I've been able to find don't reference URLs created via the admin class' get_urls() method - so any guidance on this would be greatly appreciated.
Thanks very much, let me know if there's any other info that I can provide to help sort this out.
I think the problems is in missing namespace in your export_link.html template. Instead of:
Export
try:
Export
I would like to access the root path variable in my template, how can I do that?
So far, I have:
{% if user.username == "" and request.path != "/login/" and request.path != "/" %}
<meta http-equiv="REFRESH" content="0;url=/login/">
{% else %}
I would like to write something like this:
{% if user.username == "" and request.path != projectRoot+"/login/" and request.path != projectRoot %}
<meta http-equiv="REFRESH" content="0;url=projectRoot+/login/">
{% else %}
Should I create and send this projectRoot variable from my view, or is there already a variable for that?
Many thanks in advance,
Romain
You can get the full path of the URl using request.full_path and write a template tag to check if it contains a string but this is a very bad way of trying to do what you want (redirect when unauthorized).
Instead, you should decorate your view using #login_required:
from django.contrib.auth.decorators import login_required
#login_required
def my_view(request):
...
Although I agree with Timmy that you shouldn't be trying to do this sort of redirection in the template, note that there is an easy way of getting the full URL to a view: and that is using the {% url %} tag. That tag (and the reverse function in views) is aware of the full path to your site, which it gets from the WSGI environment, so there is no need to pass an extra variable. Since you shouldn't be hard-coding URL paths anyway, using this is a win all round.
In my Django URLs, I have many URL patterns that end with :
(redirect/(?P<redirect_to>\w+))
Which means that these URLs can be (or not) ending by /redirect/TARGET/. These URL patterns have other named arguments (mostly one : pk)
Now, I'd like, in the templates used by these URL patterns, to be able to alter the current page path, by just adding the redirect_to argument, and keeping the other arguments and URL reverse name untouched.
I was able to get the URL reverse name in the template, by adding resolve(path).url_name to the current context, and then to use that with the {% url %} template tag.
I'd like to know if there is any easy way to dynamically add the arguments (from resolve(path).kwargs) to the URL reverse tag ?
I think you should create a custom tag for this (replacing your {% url %} tag with {% url_redirect "your_new_destination" %}).
in your_app/templatetags/my_custom_tags.py:
from django.core.urlresolvers import reverse, resolve
#register.simple_tag(takes_context=True)
def url_redirect(context, new_destination):
match = resolve(context.request.path)
match.kwargs['redirect_to'] = new_destination
return reverse(match.url_name, args=match.args, kwargs=match.kwargs)
in your template:
{% load my_custom_tags %}
{% url_redirect "your_new_destination" %}
Please note that you need to add 'django.core.context_processors.request' to your TEMPLATE_CONTEXT_PROCESSORS in order for this snippet to work.
I'm sure I've seen this question on Stack Overflow before, but I couldn't find it by my life, so here goes nothing.
I have a normal Django menu which uses the {% url %} tag and static names for the menu items. Now I want to have a different style for the menu item which has been selected. But the menu is being rendered in the base template, so how do I figure out which menu item it is?
You could surely do this with some ugly template code, but a better more globally known way is to use a CSS selector. This lets CSS do all of the work automatically for you.
Here's how it works:
You simply put an id in your body depending on which page you are on.
Then in css you do something like this:
#section-aboutme #nav-aboutme,
#section-contact #nav-contact
/* ... put one of these per body/menu item ... */
{
font-color: red;
}
You put the nav-aboutme, and nav-contact ids on each of your menu items.
The style will automatically be selected by CSS depending on which body id they are inside of.
I normally do it the way Brian suggested, but to accommodate for a template which a designer gave me which used the more common class="selected" method, I wrote a {% nav %} template tag.
Your HTML navigation template will look something like:
{% block nav %}
<ul class="nav">
<li{% if nav.home %} class="selected"{% endif %}>Home</li>
<li{% if nav.about %} class="selected"{% endif %}>About</li>
</ul>
{% endblock %}
To set the navigation in a child template, do:
{% include "base.html" %}
{% load nav %}
{% block nav %}
{% nav "about" %}
{{ block.super }}
{% endblock %}
How about a custom tag which you use to generate your nav item?
The following takes the name of the url for which a nav item should be generated and the text it should display. It generates a li tag with a class of "selected" if the named url's path is the same as the current url (requires 'django.core.context_processors.request' in your TEMPLATE_CONTEXT_PROCESSORS). Within the li, it generates an a tag with the path of the url specified by the url_name. It has the contents specified by contents.
Obviously, this could be tweaked to generate different markup for the nav item, as required.
The rest can be done using CSS.
Advantages:
Easy to use
Little code required
DRY
Could be made to be more flexible
Disadvantages:
Requires 'django.core.context_processors.request'
Requires urls to be named e.g. urlpatterns = patterns('django.views.generic.simple',
...
(r'^$', 'direct_to_template', {'template': 'index.html'}, 'index'),
...
). This could potentially be done differently (e.g. pass in url).
Doesn't cope with pages not exactly equal to the specified and therefore will not apply the selected class to the li when on a page lower in the url heirarchy. For example, if I'm on /products/, it will highlight the nav item directing to /products/. If I'm on /products/myProduct/, it will not highlight the /products/ link. This could be coded around, but it would force people to use sensible urls. For example, change the additionalAttrs assignment to additionalAttrs = ' class=selected' if (context['request'].path.startswith(path) and path != '/') or (context['request'].path == path) else ''.
Code:
from django import template
from django.core.urlresolvers import reverse
register = template.Library()
class NavNode(template.Node):
def __init__(self, url_name, contents):
self.url_name = url_name
self.contents = contents
def render(self, context):
path = reverse(self.url_name)
additionalAttrs = ' class=selected' if path == context['request'].path else ''
return '<li'+additionalAttrs+'>'+self.contents+'</li>'
#register.tag
def nav_link(parser, token):
bits = token.split_contents()
if len(bits) == 3:
contents = bits.pop()
url_name = bits.pop()
else:
raise template.TemplateSyntaxError, "%r tag requires a single argument" % bits[0]
if contents[0] == contents[-1] and contents[0] in ('"', "'"):
contents = contents[1:-1]
return NavNode(url_name, contents)
You can pass request.path to your template
from django.shortcuts import render_to_response
from django.template import RequestContext
return render_to_response('templ.html', {'page':request.path}, context_instance=RequestContext(request))
then use an ugly if template tag to add a CSS class to your menu item
Let's say one has an app named "stackoverflow" and inside of that app folder one has a templates folder with a stackoverflow.html file that extends from the base template.
One way to achieve it is by defining a variable in the views.py of ones stackoverflow app, like
def stackoverflow(request):
return render(request,
'stackoverflow/stackoverflow.html',
{'section': 'stackoverflow'})
Then, in the base template
<li {% if section == "stackoverflow" %} class="selected" {% endif %}>
StackOverflow
</li>
Essentially having the variable section allows to figure the section. Note that one needs a space between section and ==... if one doesn't respect that, then one will get a Django Template Error saying
Could not parse the remainder.