Question

我是Django的新手，我现在已经坚持了一段时间。我想创建一个通用的Create View，它接受任何模型并从中创建一个表单，但我不知道如何将模型属性设置为作为URL参数传递给视图的模型。

urls.py:

url(r'^(?P<model>\w+)/new$', views.ModelCreate.as_view(), name='new-record')

views.py：

class ModelCreate(CreateView):
    model = ???

    fields = []
    for field in model._meta.get_fields():
        fields.append(field.name)

    template_name = 'new_record.html'

提前感谢您的帮助！

Answer 1

首先......为了创建一个通用视图，为任何模型创建模型表单，您需要在模型中传递，您尝试使用POST或GET通过请求创建表单。

然后，您需要在正确的位置设置您希望创建表单的模型。如果你深入研究Django的CreateView，你会发现父类ModelFormMixin有一个方法get_form_class（）。我会在ModelCreate类中覆盖此方法，并从您通过请求传入的变量中设置模型。如您所见，此方法负责从模型创建表单。

    def get_form_class(self):
    """
    Returns the form class to use in this view.
    """
    if self.fields is not None and self.form_class:
        raise ImproperlyConfigured(
            "Specifying both 'fields' and 'form_class' is not permitted."
        )
    if self.form_class:
        return self.form_class
    else:
        if self.model is not None:
            # If a model has been explicitly provided, use it
            model = self.model
        elif hasattr(self, 'object') and self.object is not None:
            # If this view is operating on a single object, use
            # the class of that object
            model = self.object.__class__
        else:
            # Try to get a queryset and extract the model class
            # from that
            model = self.get_queryset().model

        if self.fields is None:
            raise ImproperlyConfigured(
                "Using ModelFormMixin (base class of %s) without "
                "the 'fields' attribute is prohibited." % self.__class__.__name__
            )

        return model_forms.modelform_factory(model, fields=self.fields)

Django - 如何将模型传递给CreateView

1 个答案: