Django

Question

我正在尝试创建一个表单，其中创建的对象（项目）与另一个模型（通道）有关系。问题是我不知道如何调用项目关系的频道主键。

Models.py:

class Project(models.Model):
    channel = models.ForeignKey(
        'Channel',
        on_delete=models.CASCADE,
    )

Views.py:

class ProjectCreate(CreateView):
    model = Project
    fields = ['name', 'description']

    def form_valid(self, form):
        Project = form.save(commit=False)
        form.instance.channel = Channel.objects.get(id=self.kwargs['channel'])
        Project.channel = channel

        return super(ProjectCreate, self).form_valid(form)

我认为 forms.py 文件还需要添加其他内容：

Forms.py:

class ProjectForm(forms.Form):
    name = forms.CharField(max_length=50)
    description = forms.CharField(widget=forms.Textarea)

Answer 1

首先，你应该使用一个ModelForm，这样你就可以保存它来创建实例。不要在字段中包含 channel，因为您将在视图中设置它。

class ProjectForm(forms.ModelForm):
    class Meta:
         model = Project
         fields = ['name', 'description']

然后，假设您的 url 模式已正确配置为包含频道，您需要做的就是在表单实例上设置 channel 并调用父 class ' form_valid 方法。

def form_valid(self, form):
    form.instance.channel = Channel.objects.get(id=self.kwargs['channel'])
    return super(ProjectCreate, self).form_valid(form)

Django - Createview form_valid object.id 错误

Django - Createview form_valid object.id error

django-models

django-forms

django-views

django-class-based-views