从一个视图编辑两个表的值.Django

Question

在父表中，我有很多对象。用户有一种表格，他可以选择一个父对象之一。看起来这样：

class ChildForm(forms.ModelForm):
    class Meta:
        model = OrderingMassage
        fields = ('parent',
                  'name')

现在，我想为用户在父表中选择的每个对象"父"，" on_off button"值更改为false。我该如何恢复？我可以使用什么？我可以使用一种表格在视图中进行吗？

例如：

models.py

class Parent(models.Model):
    name = models.CharField(max_length=15)
    on_off_button = models.BooleanField(deflaut=True)
class Child(models.Model):
    parent = models.ForeignKey(Parent, on_delete=models.CASCADE)
    name = models.CharField(max_length=15)

views.py

if request.method == 'POST' and 'child_btn' in request.POST:
    child_form = ChildForm(request.POST)
    if child_form.is_valid():
        child = child_form.save(commit=False)
        name = child_form.cleaned_data['name']
        parent = child_form.cleaned_data['parent']
        # Can I add an element here that will change the value parent.id on False
        child.name = name
        child.parent = parent
        child.save()
else:
    child_form = ChildForm()

任何帮助将不胜感激。

Answer 1

在您的视图中，您可以做类似：

的事情

if request.method == 'POST' and 'child_btn' in request.POST:
    child_form = ChildForm(request.POST)
    if child_form.is_valid():
        child = child_form.save(commit=False)
        name = child_form.cleaned_data['name']
        parent = child_form.cleaned_data['parent']
        child.name = name
        child.parent = parent
        child.save()
        #get the parent object related to the parent selected by the user
        parent = Parent.objects.get(id=parent.id)
        parent.on_off_button = False
        parent.save()
        #or can you try this method to check
        parent.on_off_button=False
        parent.save()
else:
    child_form = ChildForm()