Martin, web.found should be raised as exception, not returned. And I would use raise web.seeother for this purpose.
http://www.w3.org/Protocols/rfc2616/rfc2616-sec10.html#sec10.3.3 On Jul 22, 6:30 pm, "W. Martin Borgert" <[email protected]> wrote: > Quoting digistam <[email protected]>: > > > I do not want to run the POST > > action again when I hit the webbrowser refresh button. I only want to > > run the POST action again when I turn back to the form page. I hope I > > make myself clear. What will be the solution for my problem? > > You need to redirect the POSTs directly to the GETs. This way the > last method is a GET and a reload in the browser will perform the > GET. You want to change your POST into: > > def POST(self): > form = myform() > session.count += 1 > a = [[],[]] > ... > pages = dict([('val1',a[0]),('val2',a[1])]) > return web.found( > web.ctx.env.get( > u'HTTP_REFERER', u'/a-default-page/')) > > Cheers -- You received this message because you are subscribed to the Google Groups "web.py" 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/webpy?hl=en.
