I want to integrate processes, starting them etc., outside of the standard viewflow.frontend. To do I have been trying to create a simple page where I can start a new process, but have been struggling to find a way to implement it.
One approach was to defined a url to url('^start_something, CustomStartView.as_view()) with CustomStartView subclassing CreateProcessView from viewflow.flow.views.start. This ended in getting error after error, whenever I fixed one. I am quite sure now that this isn't the right way to do it, also because the View is used as a parameter of the Flow class itself and probably needs to be used differently than a common view.
What is the right way to do it?
The Viewflow views need to be parametrized by flow_class and flow_task instance. So you can include a start view as following::
url('^start/', CreateProcessView.as_view(), {
'flow_class': MyFlow,
'flow_task': MyFlow.start})
To add a task view URL config, you need to use process_pk and task_pk parameters
url('^(?P<process_pk>\d+)/approve/(?P<task_pk>\d+)/', ApproveView.as_view(), {
'flow_class': MyFlow,
'flow_task': MyFlow.approve
})
For each node, you also can enable detail, and various actions URLs, ex:
url('^(?P<process_pk>\d+)/approve/(?P<task_pk>\d+)/detail/', DetailTaskView.as_view(), {
'flow_class': MyFlow,
'flow_task': MyFlow.approve
}),
url('^(?P<process_pk>\d+)/approve/(?P<task_pk>\d+)/cancel/', CancelTaskView.as_view(), {
'flow_class': MyFlow,
'flow_task': MyFlow.approve
}),
All of that's a big cumbersome.
Recommended way
You can just include Flow.instance.urls that contains all URLs collected and ready for inclusion into an URL config.
url('^myflow', include(MyFlow.instance.urls, namespace='myflow'))
And at the last, to enable task list views, you can put URL entries manually, ex
url('^myflow/inbox/', TaskListView.as_view(), {
'flow_class': MyFlow}, name="tasks")
or just use as viewflow.flow.viewset.FlowViewSet class
myflow_urls = FlowViewSet(MyFlow).urls
urlpatterns = [
url(r'^myflow/', include(myflow_urls, namespace='myflow'))
]
This is the recommended way to include viewflow URLs into a django URL Config. To customize views used in that URL config, you can subclass the FlowViewSet class and provide views in the nodes definition, ex
class MyFlow(Flow):
start = flow.Start(detail_view_class=MyDetailTaskView)
For the sample usage, you can checkout the Viewflow Custom UI cookbook example - https://github.com/viewflow/cookbook/tree/master/custom_ui
Related
According to the documentation here: https://djangobook.com/syndication-feed-framework/
If link doesn’t return the domain, the syndication framework will
insert the domain of the current site, according to your SITE_ID
setting
However, I'm trying to generate a feed of magnet: links. The framework doesn't recognize this and attempts to append the SITE_ID, such that the links end up like this (on localhost):
<link>http://localhost:8000magnet:?xt=...</link>
Is there a way to bypass this?
Here's a way to do it with monkey patching, much cleaner.
I like to create a separate folder "django_patches" for these kinds of things:
myproject/django_patches/__init__.py
from django.contrib.syndication import views
from django.contrib.syndication.views import add_domain
def add_domain_if_we_should(domain, url, secure=False):
if url.startswith('magnet:'):
return url
else:
return add_domain(domain, url, secure=False)
views.add_domain = add_domain_if_we_should
Next, add it to your INSTALLED_APPS so that you can patch the function.
settings.py
INSTALLED_APPS = [
'django_overrides',
...
]
This is a bit gnarly, but here's a potential solution if you don't want to give up on the Django framework:
The problem is that the method add_domain is buried deep in a huge method within syndication framework, and I don't see a clean way to override it. Since this method is used for both the feed URL and the feed items, a monkey patch of add_domain would need to consider this.
Django source:
https://github.com/django/django/blob/master/django/contrib/syndication/views.py#L178
Steps:
1: Subclass the Feed class you're using and do a copy-paste override of the huge method get_feed
2: Modify the line:
link = add_domain(
current_site.domain,
self._get_dynamic_attr('item_link', item),
request.is_secure(),
)
To something like:
link = self._get_dynamic_attr('item_link', item)
I did end up digging through the syndication source code and finding no easy way to override it and did some hacky monkey patching. (Unfortunately I did it before I saw the answers posted here, all of which I assume will work about as well as this one)
Here's how I did it:
def item_link(self, item):
# adding http:// means the internal get_feed won't modify it
return "http://"+item.magnet_link
def get_feed(self, obj, request):
# hacky way to bypass the domain handling
feed = super().get_feed(obj, request)
for item in feed.items:
# strip that http:// we added above
item['link'] = item['link'][7:]
return feed
For future readers, this was as of Django 2.0.1. Hopefully in a future patch they allow support for protocols like magnet.
Morning,
With my django website, I want to make the urls as short as possible.
So instead of,
/user/john
/user/ronald
I just want it like /john and /ronald
So in my routes I have it configured that all the requests go to one
urlpatterns = patterns('',
....
(r'^about/$', 'frontend.views.about'),
(r'^(.*?)/$', 'users.views.index')
)
which means basically that all requests will be handled by the users controller if not handled else where, which isn't to bad.
But I want to do the same cakes.
so instead of /cakes/chocolate-coated-cake just have /chocolate-coated-cake
So really, it'd be nice if in my users method, instead of raising a 404 I could just some how call try next route, so it's conditional on a DB field.
Make sense?
In this case I'll prefer to have a separate dispatcher view (not related to users because it is not about users). There you could set up order of models in a list and iterate through it until the first sucess and call the appropriate view (user view, cake view) with this result as an argument).
P.S. Hope that you will not have a user and a cake with the same name :)
I've got my mind set on dynamically creating URLs in Django, based on names stored in database objects. All of these pages should be handled by the same view, but I would like the database object to be passed to the view as a parameter when it is called. Is that possible?
Here is the code I currently have:
places = models.Place.objects.all()
for place in places:
name = place.name.lower()
urlpatterns += patterns('',
url(r'^'+name +'/$', 'misc.views.home', name='places.'+name)
)
Is it possible to pass extra information to the view, without adding more parameters to the URL? Since the URLs are for the root directory, and I still need 404 pages to show on other values, I can't just use a string parameter. Is the solution to give up on trying to add the URLs to root, or is there another solution?
I suppose I could do a lookup on the name itself, since all URLs have to be unique anyway. Is that the only other option?
I think you can pass a dictionary to the view with additional attributes, like this:
url(r'^'+name +'/$', 'misc.views.home', {'place' : place}, name='places.'+name)
And you can change the view to expect this parameter.
That's generally a bad idea since it will query the database for every request, not only requests relevant to that model. A better idea is to come up with the general url composition and use the same view for all of them. You can then retrieve the relevant place inside the view, which will only hit the database when you reach that specific view.
For example:
urlpatterns += patterns('',
url(r'^places/(?P<name>\w+)/$', 'misc.views.home', name='places.view_place')
)
# views.py
def home(request, name):
place = models.Place.objects.get(name__iexact=name)
# Do more stuff here
I realize this is not what you truly asked for, but should provide you with much less headaches.
I am trying to de-couple two apps:
Locations - app containing details about some location (town, country, place etc)
Directory - app containing details of places of interest (shop, railway station, pub, etc) - all categorised.
Both locations.Location and directory.Item contain lat/lng coords and I can find items within a certain distance of a specific lat/lng coord.
I'd like to use the following URL structure:
/locations/<location_slug>/directory/<category_slug>/
But I don't want to make my directory app reliant on my location app.
How can I translate this url to use a view like this in my directory app?
items_near(lat, lng, distance, category):
A work around would be to create a new view somewhere that translates this - but where should I put that? if it goes in the Directory app, then I've coupled that with my Location app, and vice versa.
Would a good idea be to place this workaround code inside my project URLs file? Thus keeping clear of both apps? Any issues with doing it like this?
For your urlpattern to work, the view function invoked has to be aware of both Locations and Directories. The short answer is you can put this view function anywhere you want - it's just a python function. But there might be a few logical places for it, outside of your Directory or Location app, that make sense.
First off, I would not put that view code in in your top-level urls.py, as that file is intended for URLconf related code.
A few options of where to put your view:
Create a new view function in a file that lives outside of any particular app. <project_root>/views.py is one possibility. There is nothing wrong with this view calling the item_near(..) view from the Directory app.
# in myproject/urls.py
urlpatterns = (
...
(r'/locations/(?P<location_slug>[\w\-]+)/directory/(?P<category_slug>[\w\-]+)/',
'myproject.views.items_near_from_slug')
)
# in myproject/views.py
from directory.views import items_near
def items_near_from_slug(request, location_slug, category_slug):
location = get_object_or_404(Location, slug=location_slug)
distance = 2 # not sure where this comes from
# And then just invoke the view from your Directory app
return items_near(request, location.lat, location.lng, distance, category_slug)
Create a new app and put the code there, in <my_new_app>/views.py. There is no requirement that a Django app need to have a models.py, urls.py, etc. Just make sure to include the __init__.py if you want Django to load the app properly (if, for instance, you want Django to automatically find templatetags or app specific templates).
Personally, I would go with option 1 only if the project is relatively simple, and <project_root>/views.py is not in danger of becoming cluttered with views for everything. I would go with option 2 otherwise, especially if you anticipate having other code that needs to be aware of both Locations and Directories. With option 2, you can collect the relevant urlpatterns in their own app-specific urls.py as well.
From the django docs here if you're using django >=1.1 then you can pass any captured info into the included application. So splitting across a few files:
# in locations.urls.py
urlpatterns = ('',
(r'location/(?P<location_slug>.*?)/', include('directory.urls')),
#other stuff
)
# in directory.urls.py
urlpatterns = ('',
(r'directory/(?P<directory_slug>.*?)/', 'directory.views.someview'),
#other stuff
)
# in directory.views.py
def someview(request, location_slug = None, directory_slug = None):
#do stuff
Hope that helps. If you're in django < 1.1 I have no idea.
Irrespective of how much ever "re-usable" you make your app, inevitably there is a need for site-specific code.
I think it is logical to create a "site-specific" application that uses the views of the reusable and decoupled apps.
I come from a Cake background, and I'm just starting to learn Django now. I'm liking it quite a bit, but I kinda wish it used convention over configuration like cake does. So,
How can I get Cake-style URLs automatically? For example, if I went to mysite.com/posts/view/5 it would load up mysite.posts.views.view and pass an argument 5 to it? I was thinking I could add something like (r'^(.*)/(.*)', 'mysite.$1.$2'), to urls.py, but of course, that won't work.
How can I automatically load up a template? Each view function should automatically load a template like templates/posts/view.html.
Is this even possible, or do I have to hack the core of Django?
Here's my solution, based on what Carl suggested:
urlpatterns = patterns('',
# url pats here
url(r'^(?P<app>\w+)/(?P<view>\w+)/(?P<args>.*)$', 'urls.dispatch')
)
def dispatch(req, app, view, args): # FIXME: ignores decorators on view func!
func = get_callable(app+'.views.'+view)
if args:
ret = func(req, *args.split('/'))
else:
ret = func(req)
if type(ret) is dict:
return render_to_response(app+'/'+view+'.html', ret)
else:
return ret
Seems to be working pretty well with initial tests. Solves both problems with a single function. Probably won't support GET-style arguments tho.
Those points are both implementable without hacking Django core, but either one will require a non-trivial level of familiarity with advanced Python techniques.
You can do the generic URL pattern with a pattern like this:
url(r'^(?P<appname>\w+)/(?P<viewfunc>\w+)/(?P<args>.*)$', 'myresolverfunc')
Then define a 'myresolverfunc' "view" function that takes "appname", "viewfunc", and "args" parameters, and implement whatever logic you want, splitting args on "/" and dynamically importing and dispatching to whatever view function is referenced. The trickiest part is the dynamic import, you can search Django's source for "importlib" to see how dynamic imports are done internally various places.
The automatic template loader can be implemented as a view function decorator similar to the various "render_to" decorators out there, except you'll generate the template name rather than passing it in to the decorator. You'll have to introspect the function object to get its name. Getting the app name will be trickier; you'll probably just want to hardcode it as a module-level global in each views.py file, or else work in conjunction with the above URL dispatcher, and have it annotate the request object with the app name or some such.
I don't you'll need to hack the core of Django for this. It sounds like you might be in need of generic views. Also check out the Generic Views topic guide.
The first example given in the generic views documentation sounds like your first bullet point:
Example:
Given the following URL patterns:
urlpatterns = patterns('django.views.generic.simple',
(r'^foo/$', 'direct_to_template', {'template':'foo_index.html'}),
(r'^foo/(?P<id>\d+)/$', 'direct_to_template', {'template':'foo_detail.html'}),
)
... a request to /foo/ would render the template foo_index.html, and a request to /foo/15/ would render the foo_detail.html with a context variable {{ params.id }} that is set to 15.