Add url parameters to a url in django view - django

I have a django view that lists several urls on external sites.
When I render them I would like to add a few url parameters to each.
These urls are to an external system and thus not listed in my urls.py. Furthermore, some of the links have a hash '#' so it is not as easy as appending a few parameters to the end of the string.
Based on these requirements it seems the url template tag will not be a good fit. I was wondering if there is a custom filter out that to do this.

You don't need Django's url tag here. The url tag is to resolve to URLs that belong to your application.
However, there is the nice django-spurl library. It allows you to handle query parameters via template tags.
An example from the documentation to add query parameters:
{% spurl base="http://example.com/?foo=bar" add_query="bar=baz" %}
<!--
will result in: http://example.com?foo=bar&bar=baz
-->

Related

django url concatenation. I don't want to concatenate

when I do something in 'profile' page,
the url is concatenated to the next of 'profile'.
but I want to link to just 'signout'. not 'profile/signout'
this is my urls.py.
when ever I do something in 'profile'page,
the href link is concatenated to 'profile'url.
this is href source.
since this href source is header.html,
this page is included another pages.
and in the other pages, it works well.
only in profile page, the href url is concatenated to 'profile/1' url.
how can I fix it?
Yes, a URL not starting with a slash or a scheme is a relative URL. href="foo" is equivalent to href="./foo", i.e. it refers to the path foo relative to the current path. If you want the top-level path, you want href="/foo".
In Django you're supposed to use the {% url %} template tag to generate URLs, you don't hardcode them. Django will take care to generate the correct URL; especially if you move the app around to other environments, the URL may require a prefix or such, so you should never hardcode the URL.

Is there a way to apply a filter to a URL generated with the URL tag in Django?

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?

Is it possible to give external URLs names in django

I have just started naming my URL patterns in Django, so that if I want to change the URL pattern, I just have to change it in one place. e.g:
url(r'^$', HomeListView.as_view(), name='home'),
And referencing it in my templates like this:
{% url home %}
Is this possible with external links in case they change or I change the a Facebook page link for example. How would this look?
Thanks
One way to do this could be to write an external_url template tag, and have an ExternalURL model that stores them.
This would give you the advantage of being able to have the urls editable without redeploying changed code.
The disadvantage is that there will be database lookups to see those urls. Also, you would need to {% load external_urls %} in templates you want to use it in.
# models.py pseudo-code
class ExternalURL(models.Model):
name = models.CharField(unique=True)
url = models.URLField()
Your template tag might look something like:
# app/templatetags/external_url.py
#library.register()
def external_url(name):
try:
return ExternalURL.objects.get(name=name).url
except ExternalURL.DoesNotExist:
return ''
Another alternative could be to have a Context Processor that stores them all in the context, allowing you to not have to pass them explicitly into views: would be useful if you had several external urls that were used in many places within your system.
def external_urls(request):
return {
'google': 'http://www.google.com/',
# more here
}
The advantages of this is no database lookup, no requirement to load the template tag, but you will need to add it to your settings.CONTEXT_PROCESSORS. Also, you could inspect request to see if the current user may see all the urls.
If its external links, and possibility of the change then you should define it in settings or separate static url file and pass those variable with request context.
urls.py should be recommended to use only for your app specific urls.
In this SO thread you can see how to approach defining constant

django url parse formatted url

I'm in the design stages of a single page web app, and would like to make it so that a user can click on a formatted URL and the data requests will load in the page.
For example, a url of http://www.mysite.com/?category=some_cat will trigger the Category view with the relevant data.
My intention is to parse the URL, gather the data, then pass it to the index.html template for rendering on page load. Once the page has been loaded, a Javascript trigger setting will trigger the appropriate button to load the client view.
However, I'm having an issue setting up the URL parser, as the following settings are not matching the example url above.
from app.views import app_views, photo_views, user_views, admin_views
urlpatterns = patterns("",
url(r'^/(?P<category>\d+)/$', app_views.index)
)
You're confusing between sending information through your urls with GET and formatting you urls with arguments for the view functions. Say I am visiting a site called http://www.mysite.com/ and the page has a form that looks like this:
<form>
<input type='text' name='category' id='category'></input>
<button type='submit'>Send!</button>
</form>
upon clicking, the url will automatically change to http://www.mysite.com/?category=<value of input>. The ? marks that everything afterwards should be treated as GET data, with the syntax of <id>=<value>. You can then access them like so:
def response(request):
category = request.GET['category']
formatting urls is different, because it means looking for patterns that are part of the url. i.e. a pattern that looks like r'^/(?P<category>\d+)/$' will look for this: http://www.mysite.com/<category>/ and it will send it to the request in your views as an additional argument like so:
def response(request, category):
...
The regex is used to define how you recognize that part of the url. For example, the \d+ you're using means that category needs to be a number. You can search how to define different types of patterns according to your needs
Note that with GET you are sending the data to the same view function that rendered the page you are currently visiting, while using a different url means you tell it where to go through your urls.py (usually a different function). Does that make things a bit clearer?

View referenced by two urls and url tag

I am using url tag in my template for a view, that is used by two different urls. I am getting the wrong url in one place. Is there any way to force django to retrieve different url? Why it doesn't notify my, that such conflict occured and it doesn't know what to do (since python zen says, that is should refuse temptation to guess).
Code in template:
{% url djangoldap.views.FilterEntriesResponse Entry=entry.path as filter_url %}
Code in urls:
(r'^filter_entries/(?P<Entry>.*)/$',
'djangoldap.views.FilterEntriesResponse',
{'filter_template': 'filter_entries.html',
'results_template': 'filter_results.html'}),
(r'^choose_entries/(?P<Entry>.*)/$',
'djangoldap.views.FilterEntriesResponse',
{'filter_template': 'search_entries.html',
'results_template': 'search_results.html'}),
As you can see, those two urls use the same view, but with different templates. How I can force django to retrieve former url, rather than latter?
Name your URLs by adding another item to the tuple:
(r'^choose_entries/(?P<Entry>.*)/$',
'djangoldap.views.FilterEntriesResponse',
{'filter_template': 'search_entries.html',
'results_template': 'search_results.html'},
'sensibleprefix-choose_entries') # <-- this is the name
Then you can use the name in the URL tag.