cherrypy URL问题

Question

我有一个类似于stackoverflow的网址 像：

http://sample.com/post/12345/hello-world

但是我想让以下网址在cherrypy中运行：

http://sample.com/post/12345/hello-world?from=something&else=123

应该更正：

http://sample.com/post/12345/hello-world

我该怎么做？

我使用popargs和_cp_dispatch但没有成功。任何建议将不胜感激。

感谢

修改

我根据saaj的回答让它工作，但我想将代码移动到index（），并且url都返回404.

import cherrypy

class App:

@cherrypy.expose
def index(self, id, name = None, **kwargs):
  if kwargs:
    # do your querystring processing
    raise cherrypy.HTTPRedirect(cherrypy.url())

  if not name:
    # get name part for canonical url
    name = '{0}/hello-world'.format(id)
    raise cherrypy.HTTPRedirect(cherrypy.url(name))

    return '{0} {1} {2}'.format(id, name, kwargs)

if __name__ == '__main__':
  cherrypy.quickstart(App(), '/post', config)

再也非常感谢帮助了。我仍然是一个快乐的初学者。

Answer 1

#!/usr/bin/env python
# -*- coding: utf-8 -*-


import cherrypy


config = {
  'global' : {
    'server.socket_host' : '127.0.0.1',
    'server.socket_port' : 8080,
    'server.thread_pool' : 8
  },
}


class App:

  @cherrypy.expose
  def index(self):
    return '''
      <ul>
        <li>
          <a href='/post/12345/hello-world'>/post/12345/hello-world</a>
        </li>
        <li>
          <a href='/post/12345/hello-world?from=something&else=123'>
            /post/12345/hello-world?from=something&else=123</a>
        </li>
        <li><a href='/post/12345'>/post/12345 (more like SO)</a></li>        
      </ul>
    '''

  @cherrypy.expose
  def post(self, id, name = None, **kwargs):       
    if kwargs:
      # do your querystring processing
      raise cherrypy.HTTPRedirect(cherrypy.url())

    if not name:
      # get name part for canonical url
      name = '{0}/hello-world'.format(id)
      raise cherrypy.HTTPRedirect(cherrypy.url(name))

    return '{0} {1} {2}'.format(id, name, kwargs)


if __name__ == '__main__':
  cherrypy.quickstart(App(), '/', config)