import urllib html=urllib.urlopen(URL(.....)).read()
On Sep 14, 2:47 pm, "V. K" <[email protected]> wrote: > Hi, > > I want to call a web2py web service from the shell and obtain the HTML > response (not using a web browser). > From example: > > For the service:http://www....../a/c/f/x/y?z=t > > Construct the URL object: > > URL('a','c','f',args=['x','y'],vars = dict(z='t')) > > and then how do I call the controller using this object to obtain the > response HTML? > > Thanks in advance! --~--~---------~--~----~------------~-------~--~----~ You received this message because you are subscribed to the Google Groups "web2py-users" 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/web2py?hl=en -~----------~----~----~----~------~----~------~--~---
