Question

我有两个网络应用程序：parent_app和sub_app

说，http://www.some.com/parent.png将由parent_app处理。

如果它是另一个网站的refererd，则parent_app会获得一个HTTP_REFERER，比如http://www.other.com/path?query=value，

我希望sub_app将此HTTP_REFERER的路径和query_string作为自己的路径和query_string，并将结果返回给parent_app，因此，parent_app的url不会被更改，访问者浏览器也不会获得303跳。 / p>

的 sub.py： 的

import web
class Sub(object):
    def GET(self):
        return web.input().query             # I want it to be 'value', from "query=value"
urls = (r'/path', 'Sub')
sub_app = web.application(urls, locals())

的 parent.py： 的

import web
from sub import sub_app
class Parent(object):
    def GET(self):
        return sub_app.request('/path?query=value').data  #=========(1)
urls = (
    r'/parent.png', 'Parent',
    r'', sub_app
    )
parent_app = web.application(urls, locals())

并运行：

>>>python parent.py

当我访问'http://www.some.com/parent.png'时？

我收到了以下错误：

Traceback (most recent call last):
  File "/home/netroyal/Documents/program/studame/web/wsgiserver/__init__.py", line 1245, in communicate
    req.respond()
  File "/home/netroyal/Documents/program/studame/web/wsgiserver/__init__.py", line 775, in respond
    self.server.gateway(self).respond()
  File "/home/netroyal/Documents/program/studame/web/wsgiserver/__init__.py", line 2018, in respond
    response = self.req.server.wsgi_app(self.env, self.start_response)
  File "/home/netroyal/Documents/program/studame/web/httpserver.py", line 306, in __call__
    return self.app(environ, xstart_response)
  File "/home/netroyal/Documents/program/studame/web/httpserver.py", line 274, in __call__
    return self.app(environ, start_response)
  File "/home/netroyal/Documents/program/studame/web/application.py", line 279, in wsgi
    result = self.handle_with_processors()
  File "/home/netroyal/Documents/program/studame/web/application.py", line 249, in handle_with_processors
    return process(self.processors)
  File "/home/netroyal/Documents/program/studame/web/application.py", line 246, in process
    raise self.internalerror()
  File "/home/netroyal/Documents/program/studame/web/application.py", line 473, in internalerror
    parent = self.get_parent_app()
  File "/home/netroyal/Documents/program/studame/web/application.py", line 458, in get_parent_app
    if self in web.ctx.app_stack:
AttributeError: 'ThreadedDict' object has no attribute 'app_stack'

/ home / netroyal / Documents / program / studame / web /是web.py包路径。

那么，如何让（1）正确运行？

我希望获得相同的结果，例如（1）在shell中运行：

>>> import web
>>> class Sub(object):
...     def GET(self):
...         return web.input().query
>>> urls = ('/path', 'Sub')
>>> sub_app = web.application(urls, locals())
>>> sub_app.request('/path?query=value').data             #=========(1)'
'value'
>>>

我知道，我可以使用

web.seeother('/path?query=value')

让访问者看到结果，但我不希望浏览器跳转到另一个网址。

我认为

urllib2.urlopen('http://www.some.com/path?query=value')

会有效，但是有更好的方法可以在一个请求中完成吗？

感谢您的帮助！----- 也可以阅读此内容！ ==========================================编辑======= ===============================

好的，经过一些代码黑客攻击，我解决了我的一部分问题：

我添加了simulation.py：

import web, re
class Simulation(object):
    def __init__(self, urls, fvars):
        self._urls = urls
        self._fvars = fvars
    def request(self, localpart='/', method='GET', data=None, host='0.0.0.0:8080', headers=None, https=False):   
        #nnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnn    
        # get path and args(parameters from new url)
        try:
            path, query = localpart.split('?', 1)
        except:
            path, query = (localpart, '')
        # get all arguments: args(parameters)
        parts = query.split('&')
        args = {}
        for part in parts:
            try:
                name, value = part.split('=')
            except:
                pass
            else:
                args[name] = value
        #uuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuu
        patterns = [self._urls[2*i] for i in range(0, len(self._urls)/2)]
        for i in range(0, len(patterns)):
            result = re.match(patterns[i], path)
            if result:
                web.input = lambda: web.storage(args)
                Worker = self._fvars[self._urls[2*i + 1]]
                return Worker().GET(*result.groups())
        #----------------------------------------------------------------------------------------
        raise web.notfound()

并且在的 sub.py：

from simulation import *
class Sub(object):
    def GET(self):
        return web.input().query             # I want it to be 'value', from "query=value"
urls = (r'/path', 'Sub')
sub_app = web.application(urls, locals())
sub_sim = Simulation(urls, locals())         # new class to run request in parent_app

加入的 parent.py：

import web
from sub import sub_app, sub_sim               #sub_sim is new
class Parent(object):
    def GET(self):
        return sub_sim.request('/path?query=value')  # sub_app changed to sub_sim, and no (.data)
urls = (
    r'/parent.png', 'Parent',
    r'', sub_app
    )
parent_app = web.application(urls, locals())

我也可以使用'/ sub'来访问sub_app，保持它是一个独立的应用程序。

我解决了我的问题，但并不完美，但有些困难。我想我会用它，当我有更多时间时，我会找到另一种方式。如果您有更好的解决方案，请告诉我，谢谢。

最好的问候。

========================================= 我觉得我在跟自己说话，人们在哪里？

如何在parent_application中创建一个sub_application请求？

0 个答案: