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django create view filter options

Disgusted Dormouse answered on July 19, 2020 Popularity 8/10 Helpfulness 1/10

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django create view filter options

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class FolderCreate(CreateView):
    template_name = 'Form/folder_create.html'
    form_class = FolderForm

    def get_form_kwargs(self):
        kwargs = super(FolderCreate, self).get_form_kwargs()
        kwargs['user'] = self.request.user
        return kwargs

Popularity 8/10 Helpfulness 1/10 Language python

Source: stackoverflow.com

Tags: create-view django filter python

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Contributed on Jul 19 2020

Disgusted Dormouse

0 Answers Avg Quality 2/10

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class FolderForm(forms.ModelForm):
    class Meta:
       model = Folder
       fields = ['name', 'parent']

    def __init__(self, *args, **kwargs):
       user = kwargs.pop('user')
       super(FolderForm, self).__init__(*args, **kwargs)
       self.fields['parent'].queryset = Folder.objects.filter(user=user)

Popularity 8/10 Helpfulness 1/10 Language python

Source: stackoverflow.com

Tags: filter python django create-view p

Link to this answer
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Contributed on Jul 19 2020

Disgusted Dormouse

0 Answers Avg Quality 2/10

django create view filter options

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