Django：如何將數據插入字段，其中pk =？

Question

可以說我有一個事件加載到數據庫中，在description和status字段中有信息，但是action_taken保留為NULL。

class Incident(models.Model):
    description = models.TextField()
    status = models.ForeignKey(Status, default="open")
    action_taken = models.TextField()

如何使用此表單和視圖將信息加載到action_taken字段中？

表格

class ResolveForm(forms.Form):
    action_taken = forms.CharField(widget=forms.Textarea)

views.py

def detail(request, incident_id):

    incident = get_object_or_404(Incident, pk=incident_id)
    template = "incidents/detail.html"

    if request.method == 'POST':    
        form = ResolveForm(request.POST or None)
        if form.is_valid():     
            action_taken = (form.cleaned_data['action_taken'])

            ######### MY EFFORTS #########################
            q = Incident(action_taken=action_taken)
            q.save()
            print(incident.id)
            #new_incident, created = Incident.objects.get_or_create(action_taken)
            ##############################################

            return render(request, template, {'form': form})
    else:
        form = ResolveForm()        
        context = { 'incident': incident,
                    'form': form}

        return render(request, template, context)

錯誤

incident.action_taken = action_taken

錯誤：未定義名稱“ action_taken”

Answer 1

如何使用此表單和視圖將信息加載到action_taken字段中？

我看到您已經發生了模型實例incident ，所以應該這樣做

incident = get_object_or_404(Incident, pk=incident_id)
incident.action_taken = action_taken
incident.save()

如果在更新中，您不想觸摸其他字段：

incident = get_object_or_404(Incident, pk=incident_id)
incident.action_taken = action_taken
incident.save(update_fields=['action_taken'])

批評您嘗試了什么：

q = Incident(action_taken=action_taken)
q.save()

這不會獲取您要更新的對象，而是創建一個新對象並保存（而不是您想要的對象）