Question

我想创建一个查询集，其中当前用户用作ModelForm中的过滤器：

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

Django是否将请求传递给表单？这是好习惯吗？我该如何使用该请求？（当然没有定义名称请求）

编辑：

我尝试了另一个解决方案，即在视图中调用表单并传递请求：

form = BookSubmitForm(request)

然后以我使用的形式：

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)

它有效并且代码在表单中。现在我不确定它是最好的解决方案，可以改进吗？

Answer 1

不，请求未传递给ModelForm。您需要在视图中执行以下操作：

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

正如您所说，将它封装在Form对象中通常更加清晰，特别是如果您有多个需要过滤查询集的字段。要执行此操作，请覆盖表单的__init__()并让其接受request的kwarg：

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)

然后只需在视图中实例化BookSubmitForm时传递请求：

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 answer扩展到基于类的视图 - 覆盖get_form_kwargs方法：

class PassRequestToFormViewMixin:
    def get_form_kwargs(self):
        kwargs = super(PassRequestToFormViewMixin, self).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

然后以表格形式：

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

如何在Django中的ModelForm中使用该请求

2 个答案: