I need to redirect while keeping the original request. Is this approach inline with Django 3 logic or am I creating some furute catastrofe?
def final_view(request):
# use some_new_variable to do stuff and then display it in the template
return render(request, another_template_name, {})
def original_view(request):
# do stuff
request.some_new_variable = "x"
return final_view(request)
Not sure exactly what you're trying to accomplish, but you could always pass "some_new_variable" via the URL
For instance:
views.py
def original_view(request):
x = # do your stuff here
return redirect('final_view_url', var=x)
def final_view(request, var)
final_var = # do some additional stuff with var here
return render(request, another_template_name)
urls.py
path('/final_view/<var>', views.final_view, name="final_view_url")
So you send your original request to "original_view", do some stuff, then pass that variable into the URL to "final_view"
Related
My web application makes API calls to Spotify. In one of my Flask views I use the same method with different endpoints. Specifically:
sh = SpotifyHelper()
...
#bp.route('/profile', methods=['GET', 'POST'])
#login_required
def profile():
...
profile = sh.get_data(header, 'profile_endpoint')
...
playlist = sh.get_data(header, 'playlist_endpoint')
...
# There are 3 more like this to different endpoints -- history, top_artists, top_tracks
...
return render_template(
'profile.html',
playlists=playlists['items'],
history=history['items'],
...
)
I do not want to make an API call during testing so I wrote a mock.json that replaces the JSON response from the API. I have done this successfully when the method is only used once per view:
class MockResponse:
#staticmethod
def profile_response():
with open(path + '/music_app/static/JSON/mock.json') as f:
response = json.load(f)
return response
#pytest.fixture
def mock_profile(monkeypatch):
def mock_json(*args, **kwargs):
return MockResponse.profile_response()
monkeypatch.setattr(sh, "get_data", mock_json)
My problem is that I need to call get_data to different endpoints with different responses. My mock.json is written:
{'playlists': {'items': [# List of playlist data]},
'history': {'items': [# List of playlist data]},
...
So for each API endpoint I need something like
playlists = mock_json['playlists']
history = mock_json['history']
I can write mock_playlists(), mock_history(), etc., but how do I write a monkeypatch for each? Is there some way to pass the endpoint argument to monkeypatch.setattr(sh, "get_data", mock_???)?
from unittest.mock import MagicMock
#other code...
mocked_response = MagicMock(side_effect=[
# write it in the order of calls you need
profile_responce_1, profile_response_2 ... profile_response_n
])
monkeypatch.setattr(sh, "get_data", mocked_response)
There's lots of documentation about Django and the reverse() method. I can't seem to locate my exact problem. Suppose I have two urlconfs like this:
url(r'ParentLists/$', main_page, name = "main_page"),
url(r'ParentLists/(?P<grade_level>.+)/$', foo, name = "foo")
and the two corresponding views like this:
def main_page(request):
if request.method == 'POST':
grade_level = request.POST['grade_picked']
return HttpResponseRedirect(reverse('foo', args = (grade_level,)))
else:
return render(request, 'index.html', context = {'grade_list' : grade_list})
def foo(request, grade_level):
grade_level = request.POST['grade_picked']
parent_list = # get stuff from database
emails = # get stuff from database
return render(request, 'list.html', context = {'grade_list' : grade_list, 'parent_list' : parent_list})
Here, list.html just extends my base template index.html, which contains a drop down box with grade levels. When the user goes to /ParentLists, the main_page view renders index.html with the drop down box as it should.
When the user picks a grade level from the drop down box (say 5th Grade), the template does a form submit, and main_page once again executes - but this time the POST branch runs and the HttpResponseRedirect takes the user to /ParentLists/05. This simply results in an HTML table pertaining to grade 5 being displayed below the drop down box.
The problem is, when the user now selects say 10th Grade, the table updates to show the grade 10 content, but the URL displayed is still /ParentLists/05. I want it to be /ParentLists/10.
Clearly, after the first selection, the main_page view never executes again. Only foo does...and so the HttpResponseRedirect never gets called. How should I reorganize this to get what I'm looking for? Thanks in advance!
As you correctly mentioned you will never redirect to foo() from foo().
So the simple way to fix this is just add similar code as in main_page() view:
def foo(request, grade_level):
if request.method == 'POST':
grade_level = request.POST['grade_picked']
return HttpResponseRedirect(reverse('foo', args = (grade_level,)))
else:
parent_list = # get stuff from database
emails = # get stuff from database
return render(request, 'list.html', context = {'grade_list' : grade_list, 'parent_list' : parent_list})
Please note that I remove grade_level = request.POST['grade_picked'] because as Nagkumar Arkalgud correctly said it is excessively.
Also instead of combination of HttpResponseRedirect and reverse you can use shortcut redirect which probably little easy to code:
from django.shortcuts redirect
...
return redirect('foo', grade_level=grade_level)
I would suggest you to use kwargs instead of args.
The right way to use the view is:
your_url = reverse("<view_name>", kwargs={"<key>": "<value>"})
Ex:
return HttpResponseRedirect(reverse('foo', kwargs={"grade_level": grade_level}))
Also, you are sending "grade_level" to your view foo using the URL and not a POST value. I would remove the line:
grade_level = request.POST['grade_picked']
as you will override the grade_level sent to the method from the url.
I have build a view and a form in Django1.5. If the POST request is successful (based on some values I set) then I need the page to redirect to another URL which is created simultaneously.
Otherwise, if the POST was not successful I need to stay on the same page. Right now I have solved the problem as following but I am quite sure this is not the best way to do it:
This is a part of my view:
def layer_create(request, template='layers/layer_create.html'):
if request.method == 'GET':
....
elif request.method == 'POST':
out = {}
...
new_table = 'something that comes from the form'
if form.is_valid():
...
try:
...
out['success'] = True
except:
...
out['success'] = False
finally:
if out['success']:
status_code = 200
# THIS IS THE PART WHICH I THINK I CAN IMPROVE
template = '/something/workspace:' + new_table + '/metadata'
else: # if form not valid
out['success'] = False
return render_to_response(template, RequestContext(request, {'form': form}))
This part of the code:
template = '/something/workspace:' + new_table + '/metadata'
seems very ugly to me. But as I am quite new in Django I am not sure how to approach this matter.
A side note first about Django 1.5 - you're highly advised to upgrade to a supported version like 1.8.
Redirecting
For redirecting you can use the redirect shortcut. (Or HttpResponseRedirect)
from django.shortcuts import redirect
# out of a view context
return redirect('/url/to/redirect/to/')
Building URLs
Indeed - as you did mention, your attempt with template = '/something/workspace:' + new_table + '/metadata' is not the cleanest way :)
Django provides a really nice way with the URL dispatcher.
A complete solution here would go too far (or definitely would require more detailed information about your project structure) - I would recommend you to dive into the Django URL dispatcher.
In short you would do something like:
# app/urls.py
urlpatterns = [
#...
url(r'^workspace/(?P<id>[0-9]+)/metadata/$', views.workspace_detail, name='workspace-detail-metadata'),
#...
]
Then you are able to reverse your URL patterns:
from django.core.urlresolvers import reverse
url = reverse('workspace-detail-metadata', kwargs={'id': 123})
# would result in:
# .../workspace/123/metadata/
After all, I have used the "reverse" method as follows:
layer = 'geonode:' + new_table
return HttpResponseRedirect(
reverse(
'layer_metadata',
args=(
layer,
)))
Where my urls.py file includes:
url(r'^(?P<layername>[^/]*)/metadata$', 'layer_metadata', name="layer_metadata"),
As described here this is the most appropriate way to do it.
I want to set a cookie inside a view and then have that view render a template. As I understand it, this is the way to set a cookie:
def index(request):
response = HttpResponse('blah')
response.set_cookie('id', 1)
return response
However, I want to set a cookie and then render a template, something like this:
def index(request, template):
response_obj = HttpResponse('blah')
response_obj.set_cookie('id', 1)
return render_to_response(template, response_obj) # <= Doesn't work
The template will contain links that when clicked will execute other views that check for the cookie I'm setting. What's the correct way to do what I showed in the second example above? I understand that I could create a string that contains all the HTML for my template and pass that string as the argument to HttpResponse but that seems really ugly. Isn't there a better way to do this? Thanks.
This is how to do it:
from django.shortcuts import render
def home(request, template):
response = render(request, template) # django.http.HttpResponse
response.set_cookie(key='id', value=1)
return response
The accepted answer sets the cookie before the template is rendered. This works.
response = HttpResponse()
response.set_cookie("cookie_name", "cookie_value")
response.write(template.render(context))
If you just need the cookie value to be set when rendering your template, you could try something like this :
def view(request, template):
# Manually set the value you'll use for rendering
# (request.COOKIES is just a dictionnary)
request.COOKIES['key'] = 'val'
# Render the template with the manually set value
response = render(request, template)
# Actually set the cookie.
response.set_cookie('key', 'val')
return response
response = render(request, 'admin-dashboard.html',{"email":email})
#create cookies
expiry_time = 60 * 60 #in seconds
response.set_cookie("email_cookie",email,expiry_time);
response.set_cookie("classname","The easylearn academy",expiry_time);
def index(request, template):
response = HttpResponse('blah')
response.set_cookie('id', 1)
id = request.COOKIES.get('id')
return render_to_response(template,{'cookie_id':id})
I have a page that can be viewed by list or grid view through jQuery, but I want the URL to display what view it is -- especially when user paginate so that they don't have to reclick what they want to view -- so something like this: /test/?grid_view or /test/?list_view.
I've been trying to use request.GET.get, but it doesn't seem to be working. Here is what I have so far:
def test (request):
grid_view = request.GET.get('grid_view')
list_view = request.GET.get('list_view')
if grid_view:
return render_to_response('grid_view.html', {}, context_instance=RequestContext(request))
elif list_view:
return render_to_response('list_view.html', {}, context_instance=RequestContext(request))
else:
return render_to_response('default_view.html', {}, context_instance=RequestContext(request))
Then in my main template I'd point the different views to list view, etc. There is probably a better way to do it so any suggestions are appreciated too.
An empty string is evaluated to False and, that is what your GET request Query String parameters will contain.
You could modify your code to:
if 'grid_view' in request.GET or 'grid_view/' in request.GET:
pass
elif 'list_view' in request.GET or 'list_view/' in request.GET:
pass
Or:
if request.GET.has_key('grid_view') or request.GET.has_key('grid_view/'):
# ...
That's not a GET parameter. Use request.META['QUERY_STRING'] to get the query string.