[Twisted-web] Getting URL
Alberto Trujillo
alberto.trujillo at ucd.ie
Fri Aug 19 05:14:07 MDT 2005
Could anybody help me with this piece of code. I would like to add to
the url the information that is placed in the line "50", but looks not
possible. The url shoul be like the url in the line "51".
def data_header(self, ctx, data):
"""return the content for the head of the template"""
http_equiv = ''
content = ''
self.loginOK = False
myform = ctx.arg('form')
if (myform == 'login') and (self.executeLogin(ctx)):
self.loginOk = True
http_equiv = 'refresh'
href = url.here.curdir()
href = href.add('content', 'home')
50 #content = '3;URL=' + href
51 content = '2;URL=http://localhost:8088/'
return {'http_equiv':http_equiv, 'content':content}
Another question.
While I'm processing the call from an url, is possible to change the
values from the arguments? I mean, in this example I have a template
that calls to different fragments (header, navigation, content, footer),
the data_header method inside the template is the first one in the
calls. Could I change here the value of one argumentin such a way that
when the data_content method inside the template is called, this one
receives the new value.
Thank you very much
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