从一个视图编辑两个表的值。 Django

时间:2019-02-22 06:58:40

标签: django forms model

在“父表”中,我有很多对象。用户具有一种可以选择父对象之一的表单。看起来像这样:

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

现在,我想为用户在Parent表中选择的每个对象“ parent”将“ 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来查看吗?

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。

1 个答案:

答案 0 :(得分:1)

在您看来,您可以执行以下操作:

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()