Python Stripped CSS Background Image Calls

[JenAsh]

In my python file I have several css calls--they all works except for
the background image CSS call.

..mystyle { background-image:
url(http://www.mysite.com/images/background.gif); }

When I try to view it in the browser the code is somehow stripped. I
only get the following:

..mystyle { }

How can I get Python to ignore the codes? Sort of like cdata in XML.
Sorry I am very unfamilar with Python.

Can anyone explain?

[Marc 'BlackJack' Rintsch]

Re: Python Stripped CSS Background Image Calls

In <1153800112.281 524.171410@b28g 2000cwb.googleg roups.com>, JenAsh wrote:

In my python file I have several css calls--they all works except for
the background image CSS call.
>
.mystyle { background-image:
url(http://www.mysite.com/images/background.gif); }
>
When I try to view it in the browser the code is somehow stripped. I
only get the following:
>
.mystyle { }
>
How can I get Python to ignore the codes? Sort of like cdata in XML.
Sorry I am very unfamilar with Python.
>
Can anyone explain?

Can you explain in a little more detail what you are doing? Python
doesn't know about CSS and it's illegal syntax to just write CSS into
Python programs. I guess there's a web framework involved? Or CGI?

Ciao,
Marc 'BlackJack' Rintsch

[JenAsh]

Re: Python Stripped CSS Background Image Calls

Thanks Marc,

Yes, there is a web framework, basically the CSS is used as the layout
wrapper of the Python script. It is a include call within the Python
script.

[Bruno Desthuilliers]

Re: Python Stripped CSS Background Image Calls

JenAsh wrote:

Thanks Marc,
>
Yes, there is a web framework, basically the CSS is used as the layout
wrapper of the Python script. It is a include call within the Python
script.
>

Then first post to this framework's mailing list, or to the author.
FWIW, there are few chances you get any useful answer on a unspecified
web framework here.

--
bruno desthuilliers
python -c "print '@'.join(['.'.join([w[::-1] for w in p.split('.')]) for
p in 'onurb@xiludom. gro'.split('@')])"