Question

Reward.objects.get()返回一个对象，但是如何在Python / Django中返回为Tastypie序列化的所有对象？

def dehydrate(self, bundle):
    res = super(SchemeResource, self).obj_update(bundle)
    rewards = Reward.objects.get()
    bundle.data['reward_participants'] = rewards
    return res

即。上面给了我<Reward object>而不是所有奖励的列表。

Answer 1

def dehydrate(self, bundle):
    res = super(SchemeResource, self).obj_update(bundle)
    rewards = Reward.objects.all()
    bundle.data['reward_participants'] = [model_to_dict(r) for r in rewards]
    return res

这对我来说就像是一种魅力：）

Answer 2

如果我理解正确，你想要这个：

rewards = Reward.objects.all()

而不是rewards = Reward.objects.get()。然后，您可以迭代奖励查询对象以访问每行中的数据（如有必要）。例如，

rewards = Reward.objects.all()
rewards = [(x.id, x.name) for x in rewards]  # returns a list of tuples for the id and name fields (if such fields exist)

在Tastypie中使用脱水来返回所有对象

2 个答案: