write server itself not very hard, i simply don't want patch
ZServer/components.xml
is there any way don't touch Zserver?
probably not.
jens
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is there any way create server for new protocol without changing ZServer
module?
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Nikolay Kim wrote:
is there any way create server for new protocol without changing ZServer
module?
Nope, have you looked at Twisted?
cheers,
Chris
--
Simplistix - Content Management, Zope & Python Consulting
- http://www.simplistix.co.uk
_
I have a product that uses a component/interface model
internally. I wrote this against the "scarecrow" Interface
module in Zope 2.5.1.
However that module has limitations. One particular one is
what I call a "SubInterface" -- which is a special case of
Interface.Attribute (a subclass of it) t
Hi
I have these function
def Propiedad(self,
Name):
"""Propiedad""" Obj =
self Maestro = getattr(Obj,
'Maestro', None) Paso =
1 while Maestro is not
None: Obj
=
self.restrictedTraverse(Maestro.contenido).__of__(self)
Maestro = g
Hi,
When we do a search on a ZCatalog we have the message below. We work with Zope 2.6.2
and our Catalog have more than 200 000 objects referenced.
We think our catalog is corrupted but we would correct it.
Anybody have the same problem ?
An idea for fix it ?
Thank
Eric Brun
Type d'erre
Nikolay Kim wrote at 2004-2-26 17:50 +0600:
>is there any way create server for new protocol without changing ZServer
>module?
You must implement the server for the new protocol
and you will need some integration code.
You have lots of examples at hand.
--
Dieter
__
> You must implement the server for the new protocol
> and you will need some integration code.
>
> You have lots of examples at hand.
write server itself not very hard, i simply don't want patch
ZServer/components.xml
is there any way don't touch Zserver?
_
