python – Tastypie使用自定义detail_uri_name,不匹配的类型错误
作者:互联网
我试图覆盖get_bundle_detail_data
class MyResourse(ModelResource):
foo = fields.CharField( attribute = 'modelA__variableOnModelA' )
def get_bundle_detail_data(self, bundle):
return bundle.obj.foo
class Meta:
resource_name='resource'
使用代码行foo = fields.CharField(attribute =’modelA__variableOnModelA’),我将资源MyResource上的变量foo设置为modelA上名为variableOnModelA的变量.这很有效.
但我试图让variableOnModelA成为MyResource的标识符,这样我可以做/ api / v1 / resource / bar /来获取详细的MyResource,变量foo设置为bar.
我遇到的问题是错误:提供的资源查找数据无效(类型不匹配).这个错误说的是什么?
终极问题:我如何使用foo作为detail_uri_name?
编辑
模型:
class AgoraUser(models.Model):
user = models.OneToOneField(User, on_delete=models.CASCADE, primary_key=True, related_name='agora_user')
class Meta:
db_table = 'agora_users'
网址:
full_api = Api(api_name='full')
full_api.register(AgoraUserResourse())
urlpatterns = [
url(r'^admin/', admin.site.urls),
url(r'^', include(full_api.urls)),
url(r'^', include(min_api.urls)),
url(r'^search/', include('haystack.urls')),
url(r'^accounts/login/$', auth_views.login, {'template_name': 'login.html'}, name='login'),
]
实际资源:
class AgoraUserResourse_min(ModelResource):
username = fields.CharField(attribute = 'user__username' )
class Meta:
resource_name='user'
#detail_uri_name = 'user__username'
queryset = AgoraUser.objects.all()
allowed_methods = ['get', 'put', 'post']
authentication = AgoraAuthentication()
authorization = AgoraAuthorization()
def get_bundle_detail_data(self, bundle):
return bundle.obj.username
解决方法:
看起来您需要覆盖您的资源的detail_uri_kwargs.
我结束了这样的事情:
from tastypie import fields
from tastypie.resources import ModelResource
from tastypie.bundle import Bundle
from .models import AgoraUser
class AgoraUserResourse(ModelResource):
username = fields.CharField(attribute='user__username')
class Meta:
resource_name='user'
detail_uri_name = 'user__username'
queryset = AgoraUser.objects.all()
allowed_methods = ['get', 'put', 'post']
# authentication = AgoraAuthentication()
# authorization = AgoraAuthorization()
def detail_uri_kwargs(self, bundle_or_obj):
if isinstance(bundle_or_obj, Bundle):
bundle_or_obj = bundle_or_obj.obj
return {
'user__username': bundle_or_obj.user.username
}
def get_bundle_detail_data(self, bundle):
return bundle.obj.username
标签:python,django,model,tastypie 来源: https://codeday.me/bug/20190706/1394506.html