Reputation: 2630
I am trying to have a custom URL which looks like this:
example.com/site/yahoo.com
which would hit this script like this=
example.com/details?domain=yahoo.com
can this be done using app.yaml?
the basic idea is to call "details" with the input "yahoo.com"
Upvotes: 3
Views: 595
Reputation: 20930
You can't really rewrite the URLs per se, but you can use regular expression groups to perform a similar kind of thing.
In your app.yaml file, try something like:
handlers:
- url: /site/(.+)
script: site.py
And in your site.py:
SiteHandler(webapp.RequestHandler):
def get(self, site):
# the site parameter will be what was passed in the URL!
pass
def main():
application = webapp.WSGIApplication([('/site/(.+)', SiteHandler)], debug=True)
util.run_wsgi_app(application)
What happens is, whatever you have after /site/
in the request URL will be passed to SiteHandler
's get()
method in the site
parameter. From there you can do whatever it is you wanted to do at /details?domain=yahoo.com, or simply redirect to that URL.
Upvotes: 4