Sorry I hadn't thought about the fact that those urls would be activated and cut off. The point is that the URL created by the createLoginUrl call is: has "https : //www.foo.com/ _ah/login %253Fcontinue..." as the "continue" URL (hopefully adding a few spaces will foil the URL activator...) Causing the problem explained above with HTTPS and custom domains.
Thanks. On May 18, 8:59 pm, gmiller74 <[email protected]> wrote: > This appears to be a fairly basic question, but I couldn't find any > answer after looking through this group. > > I'm using a custom google Apps Domain "foo.com". When I browse to my > app at "www.foo.com" the front page comes up fine. On that page, > there is a link generated using UserService.createLoginUrl() that > looks something like this: > > https://www.google.com/a/foo.com/ServiceLogin?service=ah&passive=true...<... > additional continue url omitted for brevity> > > You may notice the problem right away, that the "continue" url is an > HTTPS url that points to my custom domain. Now as we all know you > can't serve HTTPS off of your custom domain. > (http://code.google.com/appengine/kb/general.html#httpsapps) So the > "continue" redirect breaks once the Google SSO sends me back. > > Given that I don't appear to have any control over how that URL gets > generated, how do I do user authentication using my custom domain? > > Thanks in advance for any help. > > -- > You received this message because you are subscribed to the Google Groups > "Google App Engine" group. > To post to this group, send email to [email protected]. > To unsubscribe from this group, send email to > [email protected]. > For more options, visit this group > athttp://groups.google.com/group/google-appengine?hl=en. -- You received this message because you are subscribed to the Google Groups "Google App Engine" group. To post to this group, send email to [email protected]. To unsubscribe from this group, send email to [email protected]. For more options, visit this group at http://groups.google.com/group/google-appengine?hl=en.
