Question

我想提出一个问题,即目前的用户在示范表格中被用作过滤器:

class BookSubmitForm(ModelForm):
    book = forms.ModelChoiceField(queryset=Book.objects.filter(owner=request.user),)
...

Django是否将申请通过表格? 这种做法是否好? 我如何利用请求? (当然,姓名要求没有界定)

Edit:

我尝试了另一种解决办法,即把形式称作通过请求:

form = BookSubmitForm(request)

以我的形式使用:

class BookSubmitForm(ModelForm):
    def __init__(self, request, *args, **kwargs):
        super(BookSubmitForm, self).__init__(*args, **kwargs)
        self.fields["library"].queryset = Library.objects.filter(owner=request.user)

该守则是行之有效的,其形式是。现在我不敢肯定它会找到最佳解决办法,能否改进?

Answer 1

否,请求不通过《示范公约》。你认为有必要做这样的事情:

form = BookSubmitForm()
form.fields[ book ].queryset = Book.objects.filter(owner=request.user)
# pass form to template, etc

正如你所说的那样,在表格标本中概括这一点往往比较清洁,特别是如果你有几个需要过滤的田地。为此,必须填写以下表格:init__(),并接受request:

class BookSubmitForm(ModelForm):
    def __init__(self, *args, **kwargs):
        self.request = kwargs.pop("request")
        super(BookSubmitForm, self).__init__(*args, **kwargs)
        self.fields["book"].queryset = Book.objects.filter(owner=self.request.user)
        self.fields["whatever"].queryset = WhateverModel.objects.filter(user=self.request.user)

在你看来,只要你即刻通过<代码>Book SubmitForm,你就认为:

def book_submit(request):
    if request.method == "POST":
        form = BookSubmitForm(request.POST, request=request)
        # do whatever
    else:
        form = BookSubmitForm(request=request)
    # render form, etc

Answer 2

将AdamKG的回答扩大到基于类别的观点――优先于get_form_kwargs 方法:

class PassRequestToFormViewMixin:
    def get_form_kwargs(self):
        kwargs = super().get_form_kwargs()
        kwargs[ request ] = self.request
        return kwargs

from django.views.generic.edit import CreateView
class BookSubmitCreateView(PassRequestToFormViewMixin, CreateView):
    form_class = BookSubmitForm
# same for EditView

之后为:

from django.forms import ModelForm
class BookSubmitForm(ModelForm):
    def __init__(self, *args, **kwargs):
        self.request = kwargs.pop("request")
        super().__init__(*args, **kwargs)
        ...

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