2009/6/8 Aaron Brady <castiro...@gmail.com>
> On Jun 7, 6:13 pm, Paul Rubin <http://phr...@nospam.invalid> wrote:
> > Aaron Brady <castiro...@gmail.com> writes:
> > > url+= { '/': '' }.get( url[ -1 ], '/' )
> >
> > > Shorter is always better.
> >
> > url = url.rstrip('/') + '/'
>
> I was joking. Sheesh.
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>
my two cents: a solution based on regex: the pattern replaces all slashes at
the end of the string with a single one. The * usage matches also the empty
group of slashes (example s2)
>>> print s1
aaaaa/
>>> print s2
bbbb
>>> re.sub('[/]*$','/', s1)
'aaaaa/'
>>> re.sub('[/]*$','/', s2)
'bbbb/'
-Roland
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