I looked at other similar questions on Stackoverflow, but those situations do not apply to me.
I have a form with a Queue field that is a required field. This form is used in multiple places and in one such instance, I don't want the Queue field to be shown to the user. So, I simply did not render it on the template. But because this a required field, the form won't submit. How do I pre-populate this field while at the same time hiding it from the user?
I cannot make changes to the model or the form's save methods because this form is also used at other places.
forms.py
class PublicTicketForm(CustomFieldMixin, forms.Form):
queue = forms.ChoiceField(
widget=forms.Select(attrs={'class': 'form-control'}),
label=_('Queue'),
required=True,
choices=()
)
views.py:
def no_queue(request):
if request.method == 'POST':
form = PublicTicketForm(request.POST, request.FILES)
form['queue'] = 9 # Tried to assign queue value to field, did not work
if form.is_valid():
if text_is_spam(form.cleaned_data['body'], request):
# This submission is spam. Let's not save it.
return render(request, template_name='helpdesk/public_spam.html')
else:
form.save()
else:
form = PublicTicketForm(initial={'queue': 9}) # tried this one too, did not work either
return render(request, 'helpdesk/no_queue.html', {'form': form})
The choices for this form were populated in the views, but because I'm not rendering it in the template, I did not do it.
You can use formsets to be able to assign a specific value and show specific inputs to the user like this
Don’t make a hidden field. Create another form class that doesn’t contain the field (subclassing will prevent repetition), then set the value in the view.
instance = form.save(commit=False)
instance.queue = 9
instance.save()
You can override POST data.
if request.method in ('POST', 'PUT'):
data = request.POST.copy()
data['queue'] = 9
form = PublicTicketForm(data, request.FILES)
I was able to do something like this in the template and that worked!
<input type="hidden" name="queue" value="9" />
I just did this last night!
In forms.py declare the field with the HiddenInput widget (be sure to render it):
scope = CharField(max_length=60,widget=HiddenInput())
Then, in views.py you can apply the initial value:
form = MyForm(initial={'scope': 'public'})
Or using a CBV:
initial = {'scope':'public'}
Related
I made a model form in Django that enables user to create an instance of that model. All is well save one thing. Built-in error lists of Django is annoying. I want those errors to show itself when user actually made that error. When I open my page it looks like this
Please help me how do I remove these errorlist temporarily so that when user left input fields empty. It pop ups itself?
I don't know if you need these codes to solve this problem, but here it is:
views.py file here
#login_required(login_url='/accounts/login/')
def article_create(request):
article_form = ArticleForm(request.POST or None, request.FILES)
if article_form.is_valid():
instance = article_form.save(commit=False)
instance.author = request.user
article_form.save()
return render(request, 'blog/article-create.html', {'form': article_form})
Thank you
You use the request.POST or None idiom - which I dislike, but never mind - for the data parameter, but not for the files parameter. Since you pass an (empty) dict for the files, Django takes your form as bound, and therefore tries to validate it.
If you insist on using this idiom, you need to do it for both:
article_form = ArticleForm(request.POST or None, request.FILES or None)
But really, it's much more Pythonic to be explicit:
if request.method == 'POST':
article_form = ArticleForm(request.POST, request.FILES)
if article_form.is_valid():
instance = article_form.save(commit=False)
instance.author = request.user
article_form.save()
return redirect('/')
else:
article_form = ArticleForm()
return render(request, 'blog/article-create.html', {'form': article_form})
Note, you always need to redirect after a successful POST, as I have done above.
I am trying to use an inline formset to create company open and close hours i.e Monday to Friday with different open and close times for each day.
The company profile form is a regular modelform.
The company hours i load below it is a modelformset with id passed from company profile form.
The modelform and the modelformset load fine to create.
The modelform loads with pre-selected values on edit but not modelformset.
I don't know if i am saving anything on create because i get an empty modelformset when i try to edit the instance.
What should i change here to make sure that the modelformset saves selected choices or loads pre-selected choices on edit?
def addprofile(request):
current_user = request.user
company = Company() ##To create new instance
#company = Company.objects.get(id= request.session['my_ids']) ## To get old instance
OpeningHourslineFormSet = inlineformset_factory(Company, OpeningHours, form=OpeningHoursForm, extra=7 )##fields=("weekday", "fromHour","fromMinute", "toHour", "toMinute")
if request.session['entry_count'] > 1:
messages.success( request, 'You can only create two business profiles now' )
return HttpResponseRedirect( reverse('home') )
else:
if request.method == 'POST':
hourformset = OpeningHourslineFormSet(request.POST, request.FILES, instance=company)
form = CompanyForm(request.POST)
###### deal with hourformset here
if form.is_valid():
model_instance = form.save(commit=False)
model_instance.pub_date= timezone.now()
model_instance.user= current_user.id
model_instance.save()
else:
print("companyform not saved")
###################
if hourformset.is_valid():
hourformset.save(commit=False)
for product in hourformset:
if product.is_valid():
product.save(commit=False)
product.company = model_instance.id
product.save()
instances = hourformset.save()
else:
print(" modelform not saved")
return HttpResponseRedirect('/bizprofile/success')
else:
hourformset = OpeningHourslineFormSet(instance=company)
form = CompanyForm()
context = {'hourformset': hourformset, 'form': form}
return render_to_response('bizprofile/addprofile.html', context, context_instance=RequestContext(request))
I think you need to give formset a queryset in order to show your saved instances:
# hours is a queryset that you got from company, pseudo code here
hours = company.hours.all()
hourformset = OpeningHourslineFormSet(request.POST,
request.FILES,
queryset=hours)
django doc.
Answer on this link:
Django modelform not saving input choices and not returning errors
Problem arose from saving different data types to different model field types.
I have a form, "results", where one of the fields, "subjectID", is many-to-many because there's more than one result for each subject. I want one of the submit buttons to let me save what I've entered, then redirect to the same form, now unbound except that the many-to-many "subjectID" field stays the same so I can enter more results for that subject.
Edit: I should have made it clear that I wanted the instance that I had selected in the subjectID field to stay the same. I posted the code below that actually seems to be working for me
from models.py
class ResultsForm(forms.Modelform):
class Meta:
model = models.Results
fields = ['subjectID', # this is the field want
# to populate the form with when I "save and add another"
'slideNum', # IntegerField
'resultType' ] # ForeignKey
from views.py
def addResults(request):
if request.method == 'POST'
form = ResultsForm(request.POST)
if form.is_valid():
form.save()
if 'Save_and_add_another' in request.POST:
subjectID = form.fields['subjectID']
prepop = {'subjectID' : subjectID}
form = ResultsForm(initial=prepop)
return render(request, 'slideAdmin/addResults.html', {'form': form})
elif 'Save_and_return' in request.POST:
return HttpResponseRedirect('/home/')
else:
form = ResultsForm()
return render(request, 'slideAdmin/addResults.html', {'form': form})
Right now when I click on "save and add another" from my addResults form, I get this error:
TypeError at /slidebox/addResults
'ModelMultipleChoiceField' object is not iterable
which happens when rendering {{ form.as_p }} in the template.
Edit: Changes I made to views.py
if 'Save_and_add_another' in request.POST:
subjectID = form.cleaned_data.get('subjectID')
form = ResultsForm(initial={'subjectID': subjectID})
return render(request, 'slideAdmin/addResults.html', {'form': form})
As far as I can tell, this change works. Thanks again
You should always use form.cleaned_data.get('subjectID') versus pulling the field directly from the post data. You need to pass in a list of the pk's for the M2M field.
Your view can also use a touch of cleanup:
from django.core.urlresolvers import reverse
def addResults(request):
form = ResultsForm(request.POST or None)
if request.method == 'POST' and form.is_valid():
form.save()
if 'Save_and_add_another' in request.POST:
subjectID = form.cleaned_data.get('subjectID', [])
if subjectID:
subjectID = subjectIDs.split(',')
form = ResultsForm(initial={'subjectID': subjectID})
elif 'Save_and_return' in request.POST:
return HttpResponseRedirect(reverse('home')) # don't hard code
return render(request, 'slideAdmin/addResults.html', {'form': form})
I'm not sure if you will be able to keep the form unbound when initialized.
Your form.fields is an ordered dict of django.forms.fields objects. You just want the ids, and not all the other info that comes across it.
Get the data straight from the POST dictionary.
subjectID = request.POST.get('subjectID', '')
If this is a true many to many model. You need to make sure the data is setup correctly for the initialization.
# We have to special-case M2Ms as a list of comma-separated PKs.
if isinstance(f, models.ManyToManyField):
initial[k] = initial[k].split(",")
Here is the initialization method from the django source code for Admin (or as I call it my super detailed and complicated Django cheat sheet, I am pedantic)
def get_changeform_initial_data(self, request):
"""
Get the initial form data.
Unless overridden, this populates from the GET params.
"""
initial = dict(request.GET.items())
for k in initial:
try:
f = self.model._meta.get_field(k)
except FieldDoesNotExist:
continue
# We have to special-case M2Ms as a list of comma-separated PKs.
if isinstance(f, models.ManyToManyField):
initial[k] = initial[k].split(",")
return initial
Some PEP8 nonsense as well
classes are camel case ex: class MyAwesomeClass(object):
everything else is lower with underscores. ex: awesome_id = awesome1245
Good Luck!!
I know how to set initial values to a form from the view. But how do I go about letting a generic view set initial values to a form? I can write a wrapper view for the generic view, but I still have no access to the form object instantiation.
The specific goal I'm trying to achieve is to have a user create a new object, using the create_object generic view. However, I'd like to set a field of the object 'user' to the currently logged in user, which is only accessible as request.user. How can the form be initialized to have this field?
Edit: I came across __new__. Could this call its own constructor with some default arguments?
Many thanks.
Unfortunately, you cannot achieve this behavior through Django's create_object generic view; you will have to write your own. However, this is rather simple.
To start off, you must create a new form class, like this:
from django import forms
class MyForm(forms.ModelForm):
class Meta:
model = MyModel # model has a user field
Then you would be able to create a view like this:
from django.shortcuts import render_to_response
from django.template import RequestContext
from django.http import HttpResponseRedirect
from django.contrib.auth.decorators import login_required
#login_required
def create_mymodel(request):
if request.method == 'POST':
# Get data from form
form = MyForm(request.POST)
# If the form is valid, create a new object and redirect to it.
if form.is_valid():
newObject = form.save()
return HttpResponseRedirect(newObject.get_absolute_url())
else:
# Fill in the field with the current user by default
form = MyForm(initial={'user': request.user})
# Render our template
return render_to_response('path/to/template.html',
{'form': form},
context_instance=RequestContext(request))
You could do this in a generic view wrapper by dynamically constructing a form class and passing it to the generic view, but that cure is probably worse than the disease. Just write your own view, and wait eagerly for this to land.
If you want all the features of the generic view then you can just create a new generic view using the original as a template.
Eg:
def create_object_with_initial(request, model=None, template_name=None,
template_loader=loader, extra_context=None, post_save_redirect=None,
login_required=False, context_processors=None, form_class=None, initial=None):
if extra_context is None: extra_context = {}
if login_required and not request.user.is_authenticated():
return redirect_to_login(request.path)
model, form_class = get_model_and_form_class(model, form_class)
if request.method == 'POST':
form = form_class(request.POST, request.FILES)
if form.is_valid():
new_object = form.save()
msg = ugettext("The %(verbose_name)s was created successfully.") %\
{"verbose_name": model._meta.verbose_name}
messages.success(request, msg, fail_silently=True)
return redirect(post_save_redirect, new_object)
else:
print "creating", form_class, " with initial data ", initial
form = form_class(initial=initial)
# Create the template, context, response
if not template_name:
template_name = "%s/%s_form.html" % (model._meta.app_label, model._meta.object_name.lower())
t = template_loader.get_template(template_name)
c = RequestContext(request, {
'form': form,
}, context_processors)
apply_extra_context(extra_context, c)
return HttpResponse(t.render(c))
This is copied from /site-packages/django/views/generic/create_update.py with only lines 3 and 21 changing to incorporate the initial data.
Then use it as you might expect:
object_info = {
'model': YourModel,
'initial': {'data' : 'Initial Value'},
'template_name': 'template.html'
}
url(r'^path/$',
login_required(create_object_with_initial),
object_info,
name='url_name'),
That should work.
Say I have a form like:
class GeneralForm(forms.Form):
field1 = forms.IntegerField(required=False)
field2 = forms. IntegerField(required=False)
And I want to show it twice on a page within one form tag each time with a different prefix e.g.,:
rest of page ...
<form ..>
GeneralForm(data,prefix="form1").as_table()
GeneralForm(data,prefix="form2").as_table()
<input type="submit" />
</form>
rest of page ...
When the user submits this, how do I get the submitted form back into two separate forms to do validation, and redisplay it?
This was the only documentation I could find and it's peckish.
You process each form as you normally would, ensuring that you create instances which have the same prefixes as those used to generate the form initially.
Here's a slightly awkward example using the form you've given, as I don't know what the exact use case is:
def some_view(request):
if request.method == 'POST':
form1 = GeneralForm(request.POST, prefix='form1')
form2 = GeneralForm(request.POST, prefix='form2')
if all([form1.is_valid(), form2.is_valid()]):
pass # Do stuff with the forms
else:
form1 = GeneralForm(prefix='form1')
form2 = GeneralForm(prefix='form2')
return render_to_response('some_template.html', {
'form1': form1,
'form2': form2,
})
Here's some real-world sample code which demonstrates processing forms using the prefix:
http://collingrady.wordpress.com/2008/02/18/editing-multiple-objects-in-django-with-newforms/
Even better, I think formsets is exactly what you're looking for.
class GeneralForm(forms.Form):
field1 = forms.IntegerField(required=False)
field2 = forms. IntegerField(required=False)
from django.forms.formsets import formset_factory
# GeneralSet is a formset with 2 occurrences of GeneralForm
# ( as a formset allows the user to add new items, this enforces
# 2 fixed items, no less, no more )
GeneralSet = formset_factory(GeneralForm, extra=2, max_num=2)
# example view
def someview(request):
general_set = GeneralSet(request.POST)
if general_set.is_valid():
for form in general_set.forms:
# do something with data
return render_to_response("template.html", {'form': general_set}, RequestContext(request))
You can even have a formset automatically generated from a model with modelformset_factory , which are used by the automated django admin. FormSet handle even more stuff than simple forms, like adding, removing and sorting items.