I have written a simple feedback application in django. It's not particulairly complex, basically it allows authenticated users to write a shot message with a subject line and submit that message via a form. I then allows who are in a selected group to view user submitted feedback. In the future I may add more functionality but for now it does what I want.
Here comes my question, the site I'm building has multiple places where I would like to use the feedback app, for example I have a "what do you think of the site?" kind of page at /dev/feedback/ I also have one for customer support feedback at "/support/feedback/" Currently I have just copied the code from my mysite.apps.dev.feedback over to mysite.apps.support.feedback.
The problem is that this has now created two separate copies of the same code. Despite having just written the app the two versions are already starting to diverge which is annoying. My question is simply how do I create multiple instances of the same app in a django site with distinct database models?
Some resources I've found related but not helpful are https://docs.djangoproject.com/en/dev/topics/http/urls/ and Reversing namespaced URLs in Django: multiple instances of the same app The first page does not offer much on the issue and the second page provides somewhat cludgey and impractical solutions that seem to be both unrelated and more work than their worth. Is there a proper way to implement multiple instances of the same django app?
Single model approach
I'd personally try to keep this as one app and have a view that can handle being posted from multiple locations / tag them appropriately.
As S.Lott says, this is the way to go. I am providing alternatives if you're curious about methods to keep your code in one place in other situations.
For example, you could add a category field to your model, set up a single url conf which accepts an argument in the URL such as /(?P<category>\w+/feedback/$ and have the view simply tag the feedback with the appropriate category.
class MyForm(forms.ModelForm):
class Meta:
model = Feedback
def my_view(request, category):
form = MyForm(request.POST or None)
if request.method == 'POST':
if form.is_valid():
feedback = form.save(commit=False)
feedback.category = category
feedback.save()
return http.HttpResponse("Thanks for posting!")
return render(request, "mytemplate.html", {'form': form})
# urls.py
(r'^(?P<category>\w+)/feedback/$', 'my_view')
# user can visit dev/feedback or support/feedback and the feedback will be tagged appropriately
Abstract base class
Another solution is to build an abstract base class, then create subclasses for your distinct tables. That should solve the issue with your code getting out of sync.
You'd have a single abstract model (which has no tables) from which your "real" models in your separate apps would be based on.
Dynamically generated views
If you must have separate models, you could potentially write a dynamically constructed view.
def view_generator(model_class):
class MyForm(forms.ModelForm):
class Meta:
model = model_class
def my_view(request):
form = MyForm(request.POST or None)
if request.method == 'POST':
if form.is_valid():
form.save()
return http.HttpResponse("Thanks for posting!")
return render(request, "mytemplate.html", {'form': form})
return my_view
# urls.py
from foo import view_generator
(r'^my_first_feedback_form', view_generator(Model1))
(r'^my_second_feedback_form', view_generator(Model2l))
how do I create multiple instances of the same app in a django site with distinct database models?
You shouldn't.
You simply use the feedback app model in the other two apps with a simple from feedback.models import Feedback.
Then your support app can create, retrieve, update and delete Feedback objects.
Your dev app, similarly, can create, retrieve, update and delete Feedback objects because it imported the model.
That's all that's required: import.
Thanks Yuji Tomita for a very thorough answer, my final solution is derived very closely from his suggestion, but is different enough that I thought I would post it as another option if someone else runs into the same situation that I am in.
Firstly in my mysite.apps.feedback.models file I put
class Feedback( models.Model ):
subject = models.TextField( max_length=100 )
body = models.TextField( max_length=100 )
# Some other stuff here...
# Finally I used the suggestion above and created a field which I
# use to label each entry as belonging to a specific instance of the app.
instance_name = models.TextField( max_length=20 )
In my mysite.apps.feedback.views file I put
def save_message( request, instance_name ):
if request.method == 'POST':
form = FeedbackFrom( request.POST )
if form.is_valid():
form.instance.instance_name = instance_name
form.save()
return render("feedback/thanks.html")
else:
return render("feedback/submit.html", {'form':form })
else:
return render("feedback/submit.html",{'form':FeedbackForm()})
#user_passes_test( is_staff )
def all_messages( request, instance_name ):
messages = Feedback.objects.filter( instance_name = instance_name )
return render("feedback/view_all.html",{'feedback':messages} )
In my mysite.apps.dev.urls file I put
url(r'^feedback/', include('mysite.apps.feedback.urls'),
{'instance_name':'dev'}),
In my mysite.apps.support.urls file I put
url(r'^feedback/', include('mysite.apps.feedback.urls'),
{'instance_name':'support'}),
This will separate feedback messages by app instance. Note that my actual code is more complex but this should be good enough for anyone with a similar problem to get a solution up and running pretty quickly. Hope this is useful to anyone in a similar situation. Thanks again to Yuji Tomita for the suggestions upon which this solution is based.
Related
I have 3 forms with Checkboxes to configure the desired form (Final_Form). After the user chooses the desired fields (in form1, form2 and form3), i want to delet all fields that are not required in the final form and render the final form. The reason for that is, that i have 3 Subkategories with around 12 possible values, in each form (form1-form3) the user can choose one ore more subkategories. The subcategories are standardized and are used to describe a clinical incident. The users wished to have the subcategories (1-3; form1-form3) seperated and always with an example (right-side of the screen in an anther bootstrap col).
The finalform is than a combination of the the subcategories that matches best to describe the clinical incident. All fields in the Final_Form are TextAreaFields. The Input for the TextAreaFields is stored in a sqlite-db.
Here is how i tried it:
app.py:
if request.method == 'POST' and form1.form1Submit.data:
OnePointOne = form1.OnePointOne.data
if not OnePointOne:
del Final_Form.OnePointOne
return render_template('Form2.html', form2 = form2)
if request.method == 'POST' and form2.form2Submit.data:
TwoPointTwo = form2.TwoPointTwo.data
if not TwoPointTwo:
del Final_Form.TwoPointTwo
return render_template('Form3.html', form3 = form3)
if request.method == 'POST' and form3.form3Submit.data:
ThreePointThree = form3.ThreePointThree.data
if not ThreePointThree:
del Final_Form.ThreePointThree
return render_template('Final.html', Final_Form = Final_Form)
forms.py:
class form1(FlaskForm):
OnePointOne = BooleanField('Effect')
form1Submit = SubmitField('Submit Category')
class form2(FlaskForm):
TwoPointTwo = BooleanField('Measure')
form2Submit = SubmitField('Submit Category')
class form3(FlaskForm):
ThreePointThree = BooleanField('Result')
form3Submit = SubmitField('Submit Category')
class Final_Form(FlaskForm):
OnePointOne = TextAreaField('Example Effect')
TwoPointTwo = TextAreaField('Example Measure')
ThreePointThree = TextAreaField('Example Result')
Final_FormSubmit = SubmitField('Submit incident')
The problem is, that the formfields of the Final_Form objects dont get deleted (only inside the if statements). I am very thankful for every hint or explanation.
As you are showing three separate pages, there are three separate requests.
You Final_Form object cannot be simply kept between these requests.
I don't fully understand why you configure your third form this way, it would be helpful to explain your use-case for better advice.
Without more information, I'm thinking of some ways to do this:
You make it one page/request, where you go from form to form using AJAX.
You make it one page with all forms, controlling visualisation with JS + CSS
You save your desired value somewhere
probably you can keep it in cookie (session object)
or in database, if that makes sense in your context
Also, please include whole code of this function - it's not clear how you create those forms you use.
I would like to get some advice from the community.
I have recenlty started learning Django and have a question regarding the structure of the application.
I have a URL http://127.0.0.1:8000/asset/2/, a DetailView for my Asset model which also has two card blocks that houses data for two other models Tenant and Service. Check the screenshot below.
I am generating the above view from the asset/views.py file. Code as below.
class AssetMultipleDetailView(LoginRequiredMixin, UserPassesTestMixin, DetailView):
model = Asset
context_object_name = 'asset'
template_name = 'asset/asset_multiple_detail.html'
def test_func(self):
asset_multiple = self.get_object()
if self.request.user == asset_multiple.owner:
return True
return False
def get_context_data(self, **kwargs):
context = super(AssetMultipleDetailView, self).get_context_data(**kwargs)
context['tenants'] = Tenant.objects.filter(asset=context['asset']).order_by('created')
context['services'] = Service.objects.filter(asset=context['asset']).order_by('created')
return context
When you click on the Add New Tenant button, I use the below URL in tenant/urls.py
path('new/asset/<int:pk>/', TenantAssetCreateView.as_view(), name='tenant_asset_create'),
This URL generates a CreateView for Tenant. I use the primary key of the asset in the URL to load up only the right asset to the Asset selection field. Please see the image below.
Everything works well.
I would like to know whether is this the best way to achieve this? Will this be easily maintainable as there are more views similar to this upcoming in the application.
Any advice is much appreciated. Thank you in advance.
I am not quite sure what your models look like. Does tenant have a manytomany relation to asset (a tenant can be related to any amount of assets)? Or does tenant have a foreign key to asset in your design (a tenant has exactly one related asset)? Based on the screenshot I assume the latter.
Or do you want an asset to only have one tenant (foreign key on asset to tenant)?
Loading the correct asset from the URL is perfectly valid. You should maybe make asset in the form disabled, so it can not be manipulated.
In the CreateView you could override form_valid(self,form) to set self.object.asset to the one you need.
I am using Django to create an app that allows recording of medical information however I am having problems with seperating the user accounts so currently all users see the same information entered. Anyone familiar with django knows how to set the proper permissions and roles and is willing to help a newby out?
I want the user to only access to the account the user creates and the records that the user create.
This is my github link
If you are able to to help I would really appreciate it.
If you want to list only the user's records in your /home . You only need to change the query in your home/views.py, from Identity_unique.objects.all() to Identity_unique.objects.filter(user=request.user)
class Identity_view(TemplateView):
def get(self, request):
form = Identity_form()
Identities = Identity_unique.objects.filter(user=request.user)
var = {'form': form, 'Identities': Identities}
return render(request, self.template_name, var)
Or if you want to filter objects in your Django Admin panel you should read this:
Django Documentation: ModelAdmin.get_queryset(request)
Create a custom user model with an extra field user_type.
https://github.com/samimsk/debatehub/blob/master/devdebatehub/UserApp/models.py
Implemented here.
The Problem
We have the following setup.
Pretty standard Django class based view (inherits from CreateView, which is what I'll call it form now on).
After a successful POST and form validation, the object is created, and the user is redirect_to'd the DetailView of the created record.
Some users decide that they are not happy with the data they entered. They press the back button.
The HTML generated by the CreateView is fetched form browser cache, and repopulated with the data they entered.
To the user, this feels like an edit, so they change the data and submit again.
The result is 2 records, with minor differences.
What have we tried?
At first I thought the Post-Redirect-Get (PRG) pattern that Django uses was supposed to prevent this. After investigating, it seems that PRG is only meant to prevent the dreaded "Do you want to resubmit the form?" dialog. Dead end.
After hitting the back button, everything is fetched from cache, so we have no chance of interacting with the user from our Django code. To try and prevent local caching, we have decorated the CreateView with #never_cache. This does nothing for us, the page is still retrieved form cache.
What are we considering?
We are considering dirty JavaScript tricks that do an onLoad check of window.referrer, and a manual clean of the form and/or notice to user if the referrer looks like the DetailView mentioned earlier. Of course this feel totally wrong. Then again, so do semi-duplicate records in our DB.
However, it seems so unlikely that we are the first to be bothered by this that I wanted to ask around here on StackOverflow.
Ideally, we would tell the browser that caching the form is a big NO, and the browser would listen. Again, we already use #never_cache, but apparently this is not enough. Happens in Chrome, Safari and Firefox.
Looking forward to any insights! Thanks!
Maybe don't process the POST request when it's coming from a referrer other than the same page?
from urllib import parse
class CreateView(...):
def post(self, *args, **kwargs):
referer = 'HTTP_REFERER' in self.request.META and parse.urlparse(self.request.META['HTTP_REFERER'])
if referer and (referer.netloc != self.request.META.get('HTTP_HOST') or referer.path != self.request.META.get('PATH_INFO')):
return self.get(*args, **kwargs)
...
I know I'm late to this party but this may help anybody else looking for an answer.
Having found this while tearing my hair out over the same problem, here is my solution using human factors rather than technical ones. The user won't use the back button if after submitting from a CreateView, he ends up in an UpdateView of the newly created object that looks exactly the same apart from the title and the buttons at the bottom.
A technical solution might be to create a model field to hold a UUID and create a UUID passed into the create form as a hidden field. When submit is pressed, form_valid could check in the DB for an object with that UUID and refuse to create what would be a duplicate (unique=True would enforce that at DB level).
Here's example code (slightly redacted to remove stuff my employer might not want in public). It uses django-crispy-forms to make things pretty and easy. The Create view is entered from a button on a table of customers which passes the customer account number, not the Django id of its record.
Urls
url(r'enter/(?P<customer>[-\w]+)/$', JobEntryView.as_view(), name='job_entry'),
url(r'update1/(?P<pk>\d+)/$', JobEntryUpdateView.as_view(), name='entry_update'),
Views
class JobEntryView( LoginRequiredMixin, CreateView):
model=Job
form_class=JobEntryForm
template_name='utils/generic_crispy_form.html' # basically just {% crispy form %}
def get_form( self, form_class=None):
self.customer = get_object_or_404(
Customer, account = self.kwargs.get('customer','?') )
self.crispy_title = f"Create job for {self.customer.account} ({self.customer.fullname})"
return super().get_form( form_class)
def form_valid( self, form): # insert created_by'class
#form.instance.entered_by = self.request.user
form.instance.customer = self.customer
return super().form_valid(form)
def get_success_url( self):
return reverse( 'jobs:entry_update', kwargs={'pk':self.object.pk, } )
# redirect to this after entry ... user hopefully won't use back because it's here already
class JobEntryUpdateView( LoginRequiredMixin, CrispyCMVPlugin, UpdateView):
model=Job
form_class=JobEntryForm
template_name='utils/generic_crispy_form.html'
def get_form( self, form_class=None):
self.customer = self.object.customer
self.crispy_title = f"Update job {self.object.jobno} for {self.object.customer.account} ({self.object.customer.fullname})"
form = super().get_form( form_class)
form.helper[-1] = ButtonHolder( Submit('update', 'Update', ), Submit('done', 'Done', ), )
return form
def get_success_url( self):
print( self.request.POST )
if self.request.POST.get('done',None):
return reverse('jobs:ok')
return reverse( 'jobs:entry_update',
kwargs={'pk':self.object.pk, } ) # loop until user clicks Done
I have a model form:
class SnippetForm(ModelForm):
class Meta:
model = Snippet
exclude = ['author', 'slug']
and I want to be able to edit a particular instance by using this:
def edit_snippet(request, snippet_id):
#look up for that snippet
snippet = get_object_or_404(Snippet, pk=snippet_id)
if request.user.id != snippet.author.id:
return HttpResponseForbidden()
if request.method == 'POST':
form = SnippetForm(data=request.POST, instance=snippet)
if form.is_valid():
form.save()
return HttpResponseRedirect(snippet.get_absolute_url())
else:
form = SnippetForm(instance=snippet)
return render_to_response(SNIPPET_EDIT_TEMPLATE,
{'form':form, 'add':False, 'user':request.user},
RequestContext(request))
Notice that at the line
form = SnippetForm(data=request.POST, instance=snippet)
, I created a form that use the data supplied from the user, and bound it with the instance found using the primary key (received from the url). According to django documentation, when I call save() the existing instance should be updated with POSTED data. Instead, what I see is a new object is created and saved into the database. What went wrong? Thanks a lot.
[Edit] This is really embarrassed. The code indeed has nothing wrong with it. The only thing that messed up the whole thing was the action I put in the template (as I use a same template for add and edit a snippet)....Thanks a lot for your help, really appreciate that.
I don't see why it would happen. What version of django is it?
In any case, you can manually force update passing the corresponding argument.
form = SnippetForm(data=request.POST, instance=snippet, force_update=True)