使用Neo4J和Django创建REST API [英] Create REST API with Neo4J and Django

问题描述

我正在尝试在后端使用Neo4j和Django创建REST API.

I am trying to create a REST API with Neo4j and Django in the backend.

问题在于，即使当我使用Neo4Django拥有Django模型时，也无法使用通常将模型序列化为JSON(或XML)的类似Tastypie或Piston之类的框架.

The problem is that even when I have Django models using Neo4Django , I can't use frameworks like Tastypie or Piston that normally serialize models into JSON (or XML).

很抱歉，如果我的问题令人困惑或不清楚，我是Web服务的新手.

Sorry if my question is confusing or not clear, I am newbie to webservices.

感谢您的帮助

因此，我先从Tastypie开始，然后按照本页http://django-tastypie.readthedocs.org/en/latest/tutorial.html上的教程进行操作.我正在寻找在浏览器中显示Neo4j JSON响应，但是当我尝试访问http://127.0.0.1:8000/api/node/?format=json时，却收到此错误:

So I started with Tastypie and followed the tutorial on this page http://django-tastypie.readthedocs.org/en/latest/tutorial.html. I am looking for displaying the Neo4j JSON response in the browser, but when I try to access to http://127.0.0.1:8000/api/node/?format=json I get this error instead:

{"error_message": "'NoneType' object is not callable", "traceback": "Traceback (most recent call last):\n\n  File \"/usr/local/lib/python2.6/dist-packages/tastypie/resources.py\", line 217, in wrapper\n    response = callback(request, *args, **kwargs)\n\n  File \"/usr/local/lib/python2.6/dist-packages/tastypie/resources.py\", line 459, in dispatch_list\n    return self.dispatch('list', request, **kwargs)\n\n  File \"/usr/local/lib/python2.6/dist-packages/tastypie/resources.py\", line 491, in dispatch\n    response = method(request, **kwargs)\n\n  File \"/usr/local/lib/python2.6/dist-packages/tastypie/resources.py\", line 1298, in get_list\n    base_bundle = self.build_bundle(request=request)\n\n  File \"/usr/local/lib/python2.6/dist-packages/tastypie/resources.py\", line 718, in build_bundle\n    obj = self._meta.object_class()\n\nTypeError: 'NoneType' object is not callable\n"}

这是我的代码:

api.py文件:

class NodeResource (ModelResource): #it doesn't work with Resource neither
    class meta:
        queryset= Node.objects.all()
        resource_name = 'node'

urls.py文件:

urls.py file:

node_resource= NodeResource()

urlpatterns = patterns('',
    url(r'^api/', include(node_resource.urls)),

models.py文件:

models.py file :

class Node(models.NodeModel):
    p1 = models.StringProperty()
    p2 = models.StringProperty()

推荐答案

Django-Tastypie允许使用NoSQL数据库以及

Django-Tastypie allows to create REST APIs with NoSQL databases as well as mentioned in http://django-tastypie.readthedocs.org/en/latest/non_orm_data_sources.html.

原理是使用tastypie.resources.Resource，而不是RDBMS特有的tastypie.resources.ModelResource，然后必须重新定义主要功能，以便为JSON提供所需的参数.

The principle is to use tastypie.resources.Resource and not tastypie.resources.ModelResource which is SPECIFIC to RDBMS, then main functions must be redefined in order to provide a JSON with the desired parameters.

因此，我采用了链接中给出的示例，对其进行了修改，并使用Neo4j REST Client for Python获取数据库的实例并执行请求，它的工作原理很吸引人.

So I took the example given in the link, modified it and used Neo4j REST Client for Python to get an instance of the db and perform requests, and it worked like a charm.

感谢您的所有答复:)

Thanks for all your responses :)

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