Question

这是我的main.py，我想在等号后提取所有内容。

示例网址将是 /loggedin?frob=72157645687641769-37c9ec9b8fb35d48-125787757 我想提取72157645687641769-37c9ec9b8fb35d48-125787757

import webapp2
from flickr.views import Flickr, FlickrAuthorized


class MainHandler(webapp2.RequestHandler):
    def get(self):
    self.response.write('Hello world!')

app = webapp2.WSGIApplication([
    ('/', MainHandler),
    ('/index', Flickr),
    (r'/loggedin?frob=<:((?:[a-z][a-z]*[0-9]+[a-z0-9]*))>', FlickrAuthorized)
], debug=True)

我的处理程序有以下类：

class FlickrAuthorized(webapp2.RequestHandler):
     def get(self, frob):
     print frob    
     //code

Answer 1

您应该从处理程序中访问查询字符串部分，而不是在URL中匹配它们：

class FlickrAuthorized(webapp2.RequestHandler):
    def get(self):
      frob = self.request.get('frob')
      # code...


app = webapp2.WSGIApplication([
    ('/', MainHandler),
    ('/index', Flickr),
    (r'/loggedin', FlickrAuthorized)
], debug=True)

GAE错误的URI路由

1 个答案: