django - 在 Tastypie 中序列化 django-mptt 树

Question

我如何序列化 django-mptt树木在Tastypie ?

我想用django-mptt的 cache_tree_children() .我尝试在不同的 Tastypie Hook 中应用，但它会引发错误。

Answer 1

没有 cache_tree_children您可以通过简单地连接 ToManyField 来序列化您的 child 的方法。与 full=True指着 children属性(property):

class MenuResource(ModelResource):

    children = fields.ToManyField('self', 'children', null=True, full=True)
    parent = fields.ToOneField('self', 'parent', null=True)

    class Meta:
        queryset = Menu.objects.all()

实现 cache_tree_children您可以编写自己的函数 ToManyField覆盖标准的子类 dehydrate功能。请注意，我只是非常肤浅地测试了这个解决方案:

def dehydrate(self, bundle):
    if not bundle.obj or not bundle.obj.pk:
    if not self.null:
        raise ApiFieldError("The model '%r' does not have a primary key and can not be used in a ToMany context." % bundle.obj)

        return []

    the_m2ms = None
    previous_obj = bundle.obj
    attr = self.attribute

    if isinstance(self.attribute, basestring):
        attrs = self.attribute.split('__')
        the_m2ms = bundle.obj

        for attr in attrs:
            previous_obj = the_m2ms
            try:
                the_m2ms = getattr(the_m2ms, attr, None)
            except ObjectDoesNotExist:
                the_m2ms = None

            if not the_m2ms:
                break

    elif callable(self.attribute):
        the_m2ms = self.attribute(bundle)

    if not the_m2ms:
        if not self.null:
            raise ApiFieldError("The model '%r' has an empty attribute '%s' and doesn't allow a null value." % (previous_obj, attr))

        return []

    self.m2m_resources = []
    m2m_dehydrated = []

    # There goes your ``cache_tree_children``
    for m2m in cache_tree_children(the_m2ms.all()):
        m2m_resource = self.get_related_resource(m2m)
        m2m_bundle = Bundle(obj=m2m, request=bundle.request)
        self.m2m_resources.append(m2m_resource)
        m2m_dehydrated.append(self.dehydrate_related(m2m_bundle, m2m_resource))

    return m2m_dehydrated

这种方法的主要优点之一是您不必再关心细节/ ListView 约束/差异。您甚至可以进一步参数化资源的这一方面，直到获得某种符合您需求的默认行为。基于字段，即。我认为这很酷。