CherryPy和RESTful web api

在CherryPy中创建RESTful Web API的最佳方法是什么？我一直在寻找几天,似乎没什么好看的.对于Django来说,似乎有很多工具可以做到这一点,但不是CherryPy或我不知道它们.

稍后编辑:我应该如何使用Cherrypy将/ getOrders？account = X&type = Y等请求转换为/ orders/account/type之类的内容？

我不知道这是否是"最好的"方式,但这是我的方式:

import cherrypy

class RESTResource(object):
   """
   Base class for providing a RESTful interface to a resource.

   To use this class, simply derive a class from it and implement the methods
   you want to support.  The list of possible methods are:
   handle_GET
   handle_PUT
   handle_POST
   handle_DELETE
   """
   @cherrypy.expose
   def default(self, *vpath, **params):
      method = getattr(self, "handle_" + cherrypy.request.method, None)
      if not method:
         methods = [x.replace("handle_", "")
            for x in dir(self) if x.startswith("handle_")]
         cherrypy.response.headers["Allow"] = ",".join(methods)
         raise cherrypy.HTTPError(405, "Method not implemented.")
      return method(*vpath, **params);

class FooResource(RESTResource):
    def handle_GET(self, *vpath, **params):
        retval = "Path Elements:
" + '
'.join(vpath)
        query = ['%s=>%s' % (k,v) for k,v in params.items()]
        retval += "
Query String Elements:
" + \
            '
'.join(query)
        return retval

class Root(object):
    foo = FooResource()

    @cherrypy.expose
    def index(self):
        return "REST example."

cherrypy.quickstart(Root())

您只需从RESTResource类派生并使用前缀相同名称的方法处理您想要的任何RESTful动词(GET,PUT,POST,DELETE)handle_.如果您不处理特定动词(例如POST),则基类将为您引发405 Method Not Implemented错误.

传入路径项并传入vpaths任何查询字符串params.使用上面的示例代码,如果您要求/foo/bar?woo=hoo,vpath[0]将会是bar,并且params将会{'woo': 'hoo'}.

因为HTTP定义了这些调用方法,所以使用CherryPy实现REST的最直接方法是使用MethodDispatcher而不是默认调度程序.

更多内容可以在CherryPy文档中找到:http: //cherrypy.readthedocs.io/en/latest/tutorials.html#tutorial-7-give-us-a-rest

以下是有关如何使用CherryPy Tools发送和接收JSON的详细说明:http: //tools.cherrypy.org/wiki/JSON