4

我想在通过 ModelForm 创建的实体中创建实体组关系。

如何传递父实例并parent=在 ModelForm 中设置属性?

4

2 回答 2

4

我很想看看你是否有任何好的解决方案来解决这个问题。我自己的解决方案远非优雅,就是这样做:

book = models.Book(title='Foo')
chapter = models.Chapter(parent=book, title='dummy')
form = forms.ChapterForm(request.POST, request.FILES, instance=chapter)

基本上,我首先创建一个chapter具有正确父关系的虚拟对象(在本例中),然后将其作为instance参数传递给表单的构造函数。该表单将使用请求中给出的数据覆盖我用来创建虚拟对象的一次性数据。最后,为了得到真正的子对象,我做了这样的事情:

if form.is_valid():
    chapter = form.save()
    # Now chapter.parent() == book
于 2010-01-19T21:49:51.837 回答
0

我将 djangoforms.ModelForm 子类化并添加了一个创建方法*:

class ModelForm(djangoforms.ModelForm):
  """Django ModelForm class which uses our implementation of BoundField.
  """

  def create(self, commit=True, key_name=None, parent=None):
    """Save this form's cleaned data into a new model instance.

    Args:
      commit: optional bool, default True; if true, the model instance
        is also saved to the datastore.
      key_name: the key_name of the new model instance, default None
      parent: the parent of the new model instance, default None

    Returns:
      The model instance created by this call.
    Raises:
      ValueError if the data couldn't be validated.
    """
    if not self.is_bound:
      raise ValueError('Cannot save an unbound form')
    opts = self._meta
    instance = self.instance
    if self.instance:
      raise ValueError('Cannot create a saved form')
    if self.errors:
      raise ValueError("The %s could not be created because the data didn't "
                       'validate.' % opts.model.kind())
    cleaned_data = self._cleaned_data()
    converted_data = {}
    for name, prop in opts.model.properties().iteritems():
      value = cleaned_data.get(name)
      if value is not None:
        converted_data[name] = prop.make_value_from_form(value)
    try:
      instance = opts.model(key_name=key_name, parent=parent, **converted_data)
      self.instance = instance
    except db.BadValueError, err:
      raise ValueError('The %s could not be created (%s)' %
                       (opts.model.kind(), err))
    if commit:
      instance.put()
    return instance

用法很简单:

book = models.Book(title='Foo')
form = forms.ChapterForm(request.POST)
chapter = form.create(parent=book)

请注意,我没有复制/粘贴允许您在 request.POST 中指定 key_name 的代码,而是将其作为参数传递给 create。

* 代码是从原来modelform中的save方法修改的google.appengine.ext.db.djangoforms

于 2011-02-23T20:54:43.433 回答