Currently I have a form that was built using the form wizard, the form is processed using a separate script that I wrote. I need to pass the current user (user currently logged in) to this script so that I can run an insert query to my database. Any suggestions on how to do this?
class QuestionWizard(SessionWizardView):
def done(self, form_list, **kwargs):
import process_form
userID = request.user.id
result = process_form.main(form_list,userID)
return render_to_response('done.html', {
#'form_data': [form.cleaned_data for form in form_list],
'data_return': result[0],
})
I believe the form wizard done function only accepts two variables.
It turns out it was a lot simpler than I anticipated. Inside of the done function I added the following:
user_id = self.request.user.id
And then passed the user_id into the process_form function.
Annoying that I overlooked it, but glad that I figured it out.
Related
I'm trying to wrap my head around an issue I'm having.
I want to present the users with a page in which they can look up a user by entering his username.
Yet I can't wrap my head around how to do this. Do I provide a function in my view in which a query is executed that retrieves a user from the database? For example retrieve his username or id and go to his profile page? I can't seem to grasp how to do this, I have been looking for examples on this but I can't find anything so I hope someone here can help me out!
Yes, at the simplest level you would have a view that accepts a form submission which takes the value of the field and does something like User.objects.get(username=my_variable) and returns the results, if any. You should wrap the call in a try/ except block so your view doesn't blow up if no such user exists with that username. Django provides a shortcut function for doing just this, so instead of the line above you could do user = get_object_or_404(User, username=my_variable). Your whole view would look something like
from django.shortcuts import render, get_object_or_404
def username_search(request):
# if there are any form values sent in a GET
if request.GET:
my_variable = request.GET.get('username', '')
user = get_object_or_404(User, username=my_variable)
else:
user = None
return render(request, 'some_template.html', {'user': user}
In reality you might also create a Django form to do some validation on the submissions and make it easier to render your form.
I have simple view in django app, which I want to show only when one of the forms is valid. I have something like:
#login_required
#require_role('admin')
def new_package(request):
invoicing_data_form = InvoicingDataForm(instance=request.user.account.company.invoicingdata)
if invoicing_data_form.is_valid():
# all here
return HttpResponse('Form valid')
else:
logger.info("Form invalid")
return HttpResponse(json.dumps(invoicing_data_form.errors)
I always get log info message that form is invalid, however, I get nothing in
invoicing_data_form.errors
It is very strange, because I am validating this form in other view using user input data and it works just fine. Any idea?
EDIT:
Just for clarification.
I am not requesting any data from user in this form.
I am using this form to validate some model instance (this form is subclassing from ModelForm).
That's because you're not "feeding" your form.
Do this:
invoicing_data_form = InvoicingDataForm(instance=invoice, data=request.POST or None)
You have an unbound form.
https://docs.djangoproject.com/en/1.7/ref/forms/api/#bound-and-unbound-forms
A Form instance is either bound to a set of data, or unbound.
If it’s bound to a set of data, it’s capable of validating that data and rendering the form as HTML with the data displayed in the HTML.
If it’s unbound, it cannot do validation (because there’s no data to validate!), but it can still render the blank form as HTML.
To bind data to a form, pass the data as a dictionary as the first parameter to your Form class constructor:
invoicing_data_form = InvoicingDataForm(request.POST or None, instance=invoice)
If you're already giving request.POST to your form using request.POST or None, but it's still invalid without errors, check that there isn't any redirect going on. A redirect loses your POST data and your form will be invalid with no errors because it's unbound.
I got this for AuthenticationForm which needs AuthenticationForm(None, request.POST) see Using AuthenticationForm in Django
I want to expand on the answer by #yuji-tomita-tomita
I typically use a CBV approach in Django, and how I'm handling forms:
def post(self, request, *args, **kwargs):
form = self.get_form()
if form.is_valid():
# do things
Reading the source code I noticed that self.get_form() using get_form_kwargs(self) to populate the form with request.POST, thus getting bound to data. So if you're overloading it like I did:
def get_form_kwargs(self):
company = self.get_company()
return {"company": company}
Make sure to call the super(), and it will finally work:
def get_form_kwargs(self):
company = self.get_company()
kwargs = super().get_form_kwargs()
kwargs.update({"company": company})
return kwargs
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.
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)
I have what I think should be a simple problem. I have an inline model formset, and I'd like to make a select field have a default selected value of the currently logged in user. In the view, I'm using Django's Authentication middleware, so getting the user is a simple matter of accessing request.user.
What I haven't been able to figure out, though, is how to set that user as the default selected value in a select box (ModelChoiceField) containing a list of users. Can anyone help me with this?
This does the trick. It works by setting the initial values of all "extra" forms.
formset = MyFormset(instance=myinstance)
user = request.user
for form in formset.forms:
if 'user' not in form.initial:
form.initial['user'] = user.pk
I'm not sure how to handle this in inline formsets, but the following approach will work for normal Forms and ModelForms:
You can't set this as part of the model definition, but you can set it during the form initialization:
def __init__(self, logged_in_user, *args, **kwargs):
super(self.__class__, self).__init__(*args, **kwargs)
self.fields['my_user_field'].initial = logged_in_user
...
form = MyForm(request.user)
I'm using Rune Kaagaard's idea above, except I noticed that formsets provide an extra_forms property: django.forms.formsets code
#property
def extra_forms(self):
"""Return a list of all the extra forms in this formset."""
return self.forms[self.initial_form_count():]
So, sticking with the example above:
formset = MyFormset(instance=myinstance)
user = request.user
for form in formset.extra_forms:
form.initial['user'] = user.pk
Saves having to test any initial forms, just provide default for extra forms.