I need to render the page differently if it's acessed by an iphone/ipod touch. I suppose the information is in the request object, but what would be the syntax?
views:
1145answers:
7
+2
A:
This article outlines a few ways of detecting an iPhone through by checking the HTTP_USER_AGENT agent variable. Depending on where you want to do the check at (HTML level, Javascript, CSS, etc.), I'm sure you can extrapolate this into your Python app. Sorry, I'm not a python guy. 8^D
Dillie-O
2009-03-05 23:53:04
A:
Check the user agent. It will be
Mozilla/5.0 (iPhone; U; CPU like Mac OS X; en) AppleWebKit/420+ (KHTML, like Gecko) Version/3.0 Mobile/1A543a Safari/419.3
I'm not sure how to do this with appengine, but the equivalent PHP code can be found here: http://www.mattcutts.com/blog/iphone-user-agent/
Keltex
2009-03-05 23:53:29
Well, it will change. The key bits an app should check for are 'iPhone' and 'AppleWebKit' - using that exact string will bite you in the ass the next time Safari gets a point release.
ceejayoz
2009-03-05 23:58:31
No, the key bits are Mobile/(something) and Safari/(something). "iPhone" won't be there on an iPod touch.
Brent Royal-Gordon
2009-03-06 01:11:09
What does an android show up as? Its based on webkit, and I wouldn't be surprised if it puts Safari in its UA (chrome does this, iirc).
Richard Levasseur
2009-03-06 06:47:53
A:
Here's how to do implement it as middleware in Django, assuming that's what you're using on appengine.
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
Basically look for 'iPhone' in the HTTP_USER_AGENT. Note that iPod Touch has a slightly different signature than the iPhone, hence the broad 'iPhone' search instead of a more restrictive search.
Parand
2009-03-06 08:17:57
A:
if you're using the standard webapp framework the user agent will be in the request instance. This should be good enough:
if "iPhone" in request.headers["User-Agent"]:
# do iPhone logic
Patrick
2009-03-06 08:25:39
I think your code should say self.request.headers["User-Agent"] This works great for me. Thanks
mcotton
2009-12-22 06:20:14
A:
The Using the Safari on iPhone User Agent String article on the apple website indicate the different user agents for iPhone and iPod touch.
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
VirtualBlackFox
2009-03-06 08:31:02
A:
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...')
+2
A:
This is the syntax I was looking for, works with iphone and ipod touch:
uastring = self.request.headers.get('user_agent')
if "Mobile" in uastring and "Safari" in uastring:
# do iphone / ipod stuff
Steph Thirion
2009-04-10 20:27:02