In my view, I want to make a request to mine.com/more/stuff/ from an arbitrary page such as mine.com/lots/of/stuff/to/use or from mine.com. Thus, I can't make this a relative request using ./ or ./../ type things. Do I have to use a context processor to do {{URL_BASE}}more/stuff/? Is there a set way to do this in Django or a best way?
why don't you use named urls? it's always works.
for example {% url 'admin:index' %} always printed as url to admin(in case if you using default django.contrib.admin app).
if you'll have in urls.py smth like
url(r'^lots/', Lots.as_view(), name='lots'),
then just use smth like
{% url 'lots' %}
Don't hardcode your urls!
Instead of a relative url, use an absolute url: /
If you're on mine.com/lots/of/stuff/to/use or mine.com, hitting a link with url: /foo/ will both go to mine.com/foo/
Related
For part of my user interface I need a URLencoded URL.
This code doesn't work, but it's intended to give an impression of what I'd like to do:
{% url 'team_event_calendar' id=team.id | urlencode %}
Basically, get a particular URL - in this case "team_event_calendar" for a particular team, and then URLencode it.
I suppose I could make a new version of the URL tag which URLencodes everything, but is there a way to do it with just Django's built-in tags?
Novice here who learned to develop a web app with python using Flask. Now I'm trying to learn django 1.9 by redoing the same app with django.
Right now I am stuck at trying to get the current URL and pass it as an argument so that the user can come back once the action on the next page is completed.
In Flask, to return to a previous URL, I would use the 'next' parameter and the request.url to get the current url before changing page.
In the template you would find something like this:
Buy punchcard :
and in the view:
redirect(request.args.get("next"))
I thought it would be about the same with django, but I cannot make it work. I did find some suggestions, but they are for older django version(older than 1.5) and do not work anymore(and they are pretty convulsed as solutions goes!)
Right now, in my view I am using
return redirect(next)
Note: The use of return redirect in django seems very recent itself if I judge by solutions on the web that always seem to use return HttpResponse(..., so I take it alot of changes happened lately in how to do things.
and in the template I have
<a href="{% url 'main:buy_punchcard' member.id next={{ request.path }} %}">Buy punchcard</p>
but this actually return an error
Could not parse the remainder: '{{' from '{{'
I did add the context_processors in settings.py
TEMPLATE_CONTEXT_PROCESSORS = (
'django.core.context_processors.request',
)
But this is only the last error in a very long streak of errors. Bottom line is, I can't make it work.
As such, anyone could point me in the right direction as to what is the way to do this in django 1.9? It look like a pretty basic function so I thought it would be easier somehow.
If you want next to be included in the query string, then move it outside of the url tag:
<a href="{% url 'main:buy_punchcard' member.id %}?next={{ request.path }}">Buy punchcard</p>
In your view, you can fetch next from request.GET, and return the redirect response using either HttpResponseRedirect or the redirect shortcut.
from django.utils.http import is_safe_url
next = request.GET.get('next', '/default/url/')
# check that next is safe
if not is_safe_url(next):
next = '/default/url/'
return redirect(next)
Note that it might not be safe to redirect to a url fetched from the query string. For example, it could link to a different domain. Django has a method is_safe_url that it uses to check next urls when logging in or out.
You don't need {{ }} there, just:
<a href="{% url 'main:buy_punchcard' member.id next=request.path %}">Buy punchcard</p>
I'm working through the Django 1.4 tutorial, and I notice that in the view layer, URLs are assembled by hand:
<form action="/polls/{{ poll.id }}/vote/" method="post">
And so on.
However, on the controller side, URLs are generated dynamically:
return HttpResponseRedirect(reverse('polls.views.results', args=(p.id,)))
Something seems a little inconsistent about this. Since the prefix for the URL (/polls in this case) is determined in the project's URLconf instead of the application's URLconf, it seems unwise to hard-code this value in the template.
Is there a helper or some other way to generate a URL in a Django template so that it takes the URLconf into account?
Yes, you can use the builtin url-tag to avoid hardcoding a path.
For example:
{% url 'myapp:view-name' %}
I've got a prob with a link in the sidebar of my django site, in the template it's like that:
<li>Profile</li>
while in the urls.py:
url(r'^(?P<user_id>\d+)/profile/$', 'auth.views.show_profile', name='profile')
When i access it from the main page with url: e.g /1001/profile/ it loads fine but when I try to access it from another subpage with url: e.g /1001/forms/profile/ i get the error: The current URL, /1001/forms/profile/, didn't match any of these. How can i fix this?
It is because "profile" is a relative URL, and a relative URL is appended to the current URL - the resulting address is not valid across the whole site. Seems like you should use an absolute URL in your case.
At the template you can try something like:
Profile
UPDATE
To get request available in templates you have to add django.core.context_processors.request to TEMPLATE_CONTEXT_PROCESSORS. I'm not sure if it is added by default.
You must have to add your second subpage url in urls like you did for /1001/profile/
url(r'^(?P<user_id>\d+)/form/profile/$', 'auth.views.show_profile', name='profile_form')
and also correct your code as #Paulo mentioned or you can also do it through reverse url.
Profile
I have problem with url tag. I want to redirect to a function that is in for eg
project_name.forum.views.function. Here is how i try to create url
{% url forum.views.function %}
it gives me this error:
Caught ViewDoesNotExist while rendering: Tried forum in module project_name.forum.views. Error was: 'module' object has no attribute 'forum'
I added this url in urls.py(I can access it directly) What am I doing wrong?
The url tag is used to reference named urls. E.g.
url(r'^$',
login_required(views.user_babies),
name='babystats_user_babies'),
Then you use {% url babystats_user_babies %} (the url pattern name, not the view name)
It sounds more like an incorrectly set up URL conf. You get that error when you specify a view that doesn't exist.
The url tag failing gives you a failure to reverse url with params... message.
What does your URL conf look like? Does project_name.forum.views.forum exist?
I mean, I find it odd you can visit the page at all, but that's the first place I'd look.
I have seen this error before with django url reversing stemming from the urlconf being set up with a root like projectname.app.views.view instead of app.views.view, so it chokes on the reverse without the name of the project.
Another common issue would be that the url takes an extra parameter that can be blank, and it needs you to pass an empty string or whatnot.