如何检测我的appengine应用程序是否被iphone/ipodtouch访问?

这是我一直在寻找的语法,适用于iphone和ipod touch:

uastring = self.request.headers.get('user_agent')
if "Mobile" in uastring and "Safari" in uastring:
  # do iphone / ipod stuff

This article 概述了通过检查HTTP_USER_AGENT代理变量来检测iPhone的几种方法。根据你想在哪里进行检查(HTML级别、Javascript、CSS等),我相信你可以将其推断到你的Python应用程序中。对不起,我不是蟒蛇。8^D

这 Using the Safari on iPhone User Agent String 苹果网站上的文章指出了iPhone和iPod touch的不同用户代理。

Mozilla/5.0 (iPhone; U; CPU like Mac OS X; en) AppleWebKit/420+ (KHTML, like Gecko) Version/3.0 Mobile/1A543 Safari/419.3
Mozilla/5.0 (iPod; U; CPU like Mac OS X; en) AppleWebKit/420.1 (KHTML, like Gecko) Version/3.0 Mobile/4A93 Safari/419.3
Mozilla/5.0 (iPhone; U; CPU iPhone OS 2_0 like Mac OS X; en-us) AppleWebKit/525.18.1 (KHTML, like Gecko) Version/3.1.1 Mobile/XXXXX Safari/525.20

以下是如何在Django中将其实现为中间件,假设这是您在appengine上使用的。

class DetectiPhone(object):
    def process_request(self, request):
        if 'HTTP_USER_AGENT' in request.META and request.META['HTTP_USER_AGENT'].find('(iPhone') >= 0:
            request.META['iPhone'] = True

基本上在HTTP_USER_AGENT中查找“iPhone”。请注意,iPod Touch的签名与iPhone略有不同,因此可以进行广泛的“iPhone”搜索,而不是更严格的搜索。

  if "iPhone" in request.headers["User-Agent"]:
    # do iPhone logic

Mozilla/5.0(iPhone;U;CPU类似于Mac OS X;en)AppleWebKit/420+(KHTML,类似于Gecko)版本/3.0移动版/1A543a Safari/419.3

http://www.mattcutts.com/blog/iphone-user-agent/

import os

class MainPage(webapp.RequestHandler):
  @login_required
  def get(self):
    userAgent = os.environ['HTTP_USER_AGENT']
    if userAgent.find('iPhone') > 0:
      self.response.out.write('iPhone support is coming soon...')
    else:
      self.response.out.write('Hey... you are not from iPhone...')