On 10/1/07, Joseph Heck <[EMAIL PROTECTED]> wrote:
> You could do it by having your view act as a proxy. Basically
> instantiate a new connection to the view in question from your web
> application using urllib2 or the like. Then get the results, process
> them as you wish, and send the results of that post-processing (if
> any) back to the original client request.
To be fair, that's even more overhead than is really necessary. You
can actually just do "from django.http import HttpRequest" and create
a request that way, and just pass it to the other view. There's no
need to actually involve any HTTP traffic, plus it wouldn't rely on
any particular URL scheme. It would go straight to the view, no
questions asked.
Something like the following could work, but what you add to the the
request object will depend on what the other view does with it.
from django.http import HttpRequest
from myapp.views import real_view
def proxy_view(request):
req = HttpRequest()
req.method = 'POST'
req.POST = {'arg1': 'value1', 'arg2': 'value2'}
response = real_view(req)
if response.status == 200:
return response
else:
# Handle errors here
Keep in mind that in this scenario, real_view won't have any
middleware processing, it'll just get the request exactly as you pass
it. So if the view needs request.user, for instance, you'll have to
set it manually (req.user = request.user) before passing the new
request to the view.
Of course, if all you're doing is adding POST data, you could probably
get by with just changing the method on the actual request, and adding
POST data, then triggering the other view:
from myapp.views import real_view
def proxy_view(request):
request.method = 'POST'
request.POST = {'arg1': 'value1', 'arg2': 'value2'}
return real_view(request)
I hope this helps.
-Gul
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