Django - ModelForm创建还是更新?

时间:2012-04-18 08:30:14

标签: django django-models django-forms django-views

我需要一个允许在计划中创建或添加会话的表单

模型

class Session(models.Model):
    tutor = models.ForeignKey(User)
    start_time = models.DateTimeField()
    end_time = models.DateTimeField()

表格

class SessionForm(forms.ModelForm):
    class Meta:
        model = Session
        exclude = ['tutor']

查看以呈现表单

def editor(request):
    if request.method == 'GET':
        if request.GET['id'] != '0':
            # The user has selected a session
            session = Session.objects.get(id=request.GET['id'])
            form = SessionForm(instance=session)
        else:
            # The user wants to add a new session
            form = SessionForm()
        return render_to_response('planner/editor.html',
            {'form': form,}, context_instance=RequestContext(request),)

模板 editor.html

<form action="/planner/post" method="post">{% csrf_token %}
{{ form.as_p }}
</form>

查看以发布值

def post(request):
    if request.method == 'POST':
        form = SessionForm(request.POST)
        if form.is_valid():
            form.instance.tutor = request.user
            form.save()
            obj = {'posted': True}
            return HttpResponse(json.dumps(obj), mimetype='application/json')
        else:
            return render_to_response('planner/editor.html',
                form, context_instance=RequestContext(request),)

问题

始终创建会话(永不更新)

问题

  • 在我看来post我怎么知道会话必须更新而不是创建?
  • 有没有办法简化这段代码?

4 个答案:

答案 0 :(得分:17)

如果要更新会话,则需要在绑定表单时提供实例。

如果表单有效,您可以savecommit=False一起使用,并更新导师。

form = SessionForm(instance=instance, data=request.POST)
if form.is_valid():
    instance = form.save(commit=False)
    instance.tutor = request.user
    instance.save()

答案 1 :(得分:1)

from django.shortcuts import render_to_response, get_object_or_404
from django.http import HttpResponseRedirect, Http404
from django.template import RequestContext
from application.models import Session
from application.forms import SessionForm

def allInOneView(request):
    session_id = request.POST.get('session_id')

    if session_id:
        session = get_object_or_404(Session, pk=session_id)
    else:
        session = None

    """
    A subclass of ModelForm can accept an existing model instance 
    as the keyword argument instance; 
    if this is supplied, save() will update that instance. 
    If it's not supplied, save() will create a new instance of the specified model.
    """
    form = SessionForm(instance=session)

    if request.method == 'POST':
        form = SessionForm(request.POST, instance=session)
        if form.is_valid():
            form.save()
            return HttpResponseRedirect(request.path)

    return render_to_response('planner/editor.html', {
        'form': form
    }, context_instance=RequestContext(request))

答案 2 :(得分:1)

我现在经常做的事情(遵循此处提到的建议)只使用一个视图将可选的session_id(没有创建的session_id)传递给URL调度程序。

<form action="{% url session_edit session_id=session_id|default_if_none:"" %}"
    method="post">{% csrf_token %}
{{ form.as_p }}
</form>
url('^planner/edit$', session_edit, name='session_edit'),
url('^planner/edit/(?P<session_id>\d+)$', session_edit, name='session_edit'),

我发现重新组合所有4个案例

  • 获取创作表单
  • 获取更新表单
  • 发布创建表单
  • 发布更新表格

进入一个视图更易于维护。

答案 3 :(得分:0)

在一个视图中完成所有操作。类似的东西:

def session_manager(request):

    session = None
    try:
        session = Session.objects.get(id=request.POST['id'])
    except Session.DoesNotExist:
        pass

    if request.method == "POST":
       kwargs = {
            data = request.POST
       }
       if session:
            # Update
            kwargs['instance'] session 
       form = SessionForm(**kwargs)
       if form.is_valid():
            ...
    else:
        form = SessionForm(instance=session)