我正试图向tastypeapi发送一个ApiKey认证的POST请求
class Thing(models.Model):
name = models.TextField()
def __unicode__(self):
return u'%s'%self.name
class ThingResource(ModelResource):
class Meta:
queryset = Thing.objects.all()
resource_name="thing"
authentication = ApiKeyAuthentication()
authorization = DjangoAuthorization()
from django.conf.urls.defaults import patterns, include, url
from tastypie.api import Api
from myapp.api import ThingResource
mobile_api = Api(api_name='mobile')
mobile_api.register(ThingResource())
from django.contrib import admin
admin.autodiscover()
urlpatterns = patterns('',
url(r'^admin/doc/', include('django.contrib.admindocs.urls')),
url(r'^admin/', include(admin.site.urls)),
(r'^api/', include(mobile_api.urls)),
)
curl --dump-header - -H "Accept: application/json" -H "Content-Type: application/json" -d"username=vikingosegundo" -d"api_key=12345" -X POST --data "{\"name\":\"arrrg\"}" http://localhost:8000/api/mobile/thing/
{"error_message": "No JSON object could be decoded",
"traceback": "Traceback (most recent call last):\n\n
File \"/Users/vikingosegundo/Coding/Project/serverside/mysite/tastypie/resources.py\", line 178, in wrapper\n response = callback(request, *args, **kwargs)\n\n
File \"/Users/vikingosegundo/Coding/Project/serverside/mysite/tastypie/resources.py\", line 379, in dispatch_list\n return self.dispatch('list', request, **kwargs)\n\n
File \"/Users/vikingosegundo/Coding/Project/serverside/mysite/tastypie/resources.py\", line 409, in dispatch\n response = method(request, **kwargs)\n\n
File \"/Users/vikingosegundo/Coding/Project/serverside/mysite/tastypie/resources.py\", line 1077, in post_list\n deserialized = self.deserialize(request, request.raw_post_data, format=request.META.get('CONTENT_TYPE', 'application/json'))\n\n
File \"/Users/vikingosegundo/Coding/Project/serverside/mysite/tastypie/resources.py\", line 328, in deserialize\n deserialized = self._meta.serializer.deserialize(data, format=request.META.get('CONTENT_TYPE', 'application/json'))\n\n
File \"/Users/vikingosegundo/Coding/Project/serverside/mysite/tastypie/serializers.py\", line 161, in deserialize\n deserialized = getattr(self, \"from_%s\" % desired_format)(content)\n\n
File \"/Users/vikingosegundo/Coding/Project/serverside/mysite/tastypie/serializers.py\", line 305, in from_json\n return simplejson.loads(content)\n\n
File \"/System/Library/Frameworks/Python.framework/Versions/2.6/lib/python2.6/json/__init__.py\", line 307, in loads\n return _default_decoder.decode(s)\n\n
File \"/System/Library/Frameworks/Python.framework/Versions/2.6/lib/python2.6/json/decoder.py\", line 319, in decode\n obj, end = self.raw_decode(s, idx=_w(s, 0).end())\n\n
File \"/System/Library/Frameworks/Python.framework/Versions/2.6/lib/python2.6/json/decoder.py\", line 338, in raw_decode\n raise ValueError(\"No JSON object could be decoded\")\n\n
ValueError: No JSON object could be decoded\n"
}
我做错什么了?如何将tastype指向json对象?auth+auth似乎正在工作。
我想自己补充一个答案。与Josh的解决方案不同的是,我没有将凭据添加为
GET
参数,而是将它们作为自定义HTTP头进行传输。现在我们可以将
X-MYAPP-USERNAME
和X-MYAPP-APIKEY
添加到请求中 这里是一个普通的telnet会话,使用GET还有一个会议:
而且,由于在没有提供头的情况下我们也检查GET参数,这也可以工作:
对curl使用
-d
和--data
会损坏已发布的数据。在GET中包含
username
和api_key
参数应该可以解决这个问题,如下所示:相关问题 更多 >
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