如何将Django Model上的属性(虚拟字段)公开为TastyPie ModelResource中的字段

时间:2012-01-31 10:33:51

标签: python django django-models tastypie

我有一个Django模型中的属性,我想通过TastyPie ModelResource公开。

我的模特是

class UserProfile(models.Model):
    _genderChoices = ((u"M", u"Male"), (u"F", u"Female"))

    user = Models.OneToOneField(User, editable=False)
    gender = models.CharField(max_length=2, choices = _genderChoices)

    def _get_full_name(self):
        return "%s %s" % (self.user.first_name, self.user.last_name)

    fullName = property(_get_full_name)

我的ModelResource是

class UserProfileResource(ModelResource):
    class Meta:
        queryset = models.UserProfile.objects.all()
        authorization = DjangoAuthorization()
        fields = ['gender', 'fullName']

然而,我现在摆脱tastypie api的是:

{
    gender: 'female',
    resource_uri: "/api/v1/userprofile/55/"
}

我尝试过使用ModelResource中的fields属性,但这没有帮助。很想知道这里发生了什么。

2 个答案:

答案 0 :(得分:34)

您应该可以将其定义为field尝试:

class UserProfileResource(ModelResource):
    fullname = fields.CharField(attribute='_get_full_name', readonly=True)
    class Meta:
        queryset = models.UserProfile.objects.all()
        authorization = DjangoAuthorization()
        fields = ['gender',]

修改

您还必须在set readonly=True上添加CharField,否则TastyPie会尝试在插入或更新时设置其值。

答案 1 :(得分:3)

脱水的完整示例:

class UserResource(ModelResource):
    fullname = fields.CharField(readonly=True)

    class Meta:
        queryset = auth_models.User.objects.all()
        resource_name = 'user'

    def dehydrate_fullname(self, bundle):
        return u"{first_name} {last_name}".format(
            first_name=bundle.obj.first_name, last_name=bundle.obj.last_name)