更新表单在 Django 中创建新对象

Question

我有一个 Trip 模型并创建了一个表格来更新现有行程。

models.py:

class Trip(models.Model):
    trip_id = models.CharField(max_length=20, verbose_name="Ref. Viagem")
    destination = models.CharField(max_length=200, null=True, verbose_name='Destino')
    client = models.ForeignKey(Clients, null=True, on_delete=models.CASCADE, verbose_name="Cliente")
    out_flight = models.ForeignKey(Flight, related_name="outbound_flight" ,null=True, on_delete=models.SET_NULL, verbose_name="Voo Ida")
    hotel = models.ForeignKey(Hotels, null=True, on_delete=models.SET_NULL, verbose_name="Hotel")
    in_flight = models.ForeignKey (Flight, related_name="inbound_flight", null=True, on_delete=models.SET_NULL, verbose_name="Voo Regresso")

forms.py:

class UpdateTrip(ModelForm):

    def __init__(self, *args, **kwargs):
        super(UpdateTrip, self).__init__(*args, **kwargs)
        self.fields['trip_id'].widget = TextInput(attrs={'class': 'form-control'})
        self.fields['destination'].widget = TextInput(attrs={'class': 'form-control'})
        self.fields['client'].queryset = Clients.objects.all()


    class Meta:
        model = Trip
        fields = ('trip_id', 'destination', 'client', 'out_flight', 'hotel', 'in_flight')

这是它的views.py：

def trip_upd(request, trip_id):
    if request.method == 'POST':
        form = UpdateTrip(request.POST)
        if form.is_valid():
            form.save()
            return redirect('trips')
    else:
        form = UpdateTrip()
    return render(request, 'backend/trip_update.html', {'form': form})

我在测试时使用准系统形式：

<form method="post">
    {% csrf_token %}
    {{ form }}
    <input type="submit" value="Submit">
</form>

现在，这不会更新对象，而是会创建一个新对象。

Answer 1

您需要通过instance=…参数传递您要更新的实例：

def trip_upd(request, trip_id):
    if request.method == 'POST':
        <b>trip =</b> Trip.objects.get(trip_id=trip_id)
        form = UpdateTrip(request.POST<b>, instance=trip</b>)
        if form.is_valid():
            form.save()
            return redirect('trips')
    else:
        form = UpdateTrip()
    return render(request, 'backend/trip_update.html', {'form': form})

或者您可以使用 get_object_or_404 function [Django-doc] 来 return HTTP 404 错误，以防对象不存在：

from django.shortcuts import get_object_or_404

def trip_upd(request, trip_id):
    if request.method == 'POST':
        trip = <b>get_object_or_404(</b>Trip, trip_id=trip_id<b>)</b>
        form = UpdateTrip(request.POST, instance=trip)
        if form.is_valid():
            form.save()
            return redirect('trips')
    else:
        form = UpdateTrip()
    return render(request, 'backend/trip_update.html', {'form': form})

请注意，在表单中，您可以在 Meta 中指定小部件，这比在 __init__ 中这样写更具声明性：

class UpdateTrip(ModelForm):

    class Meta:
        model = Trip
        fields = ('trip_id', 'destination', 'client', 'out_flight', 'hotel', 'in_flight')
        <b>widgets</b> = {
            'trip_id': TextInput(attrs={'class': 'form-control'}),
            'destination': TextInput(attrs={'class': 'form-control'})
        }

更新表单在 Django 中创建新对象

Update form creates new object in Django

python

django

django-3.0