On Aug 30, 8:52 am, naugiedoggie wrote:
> On Aug 29, 1:14 pm, MRAB wrote:
>
>
>
>
>
> > On 29/08/2010 15:22, naugiedoggie wrote:
> > > I'm having a problem with using a function as the replacement in
> > > re.sub().
> > > Here is the function:
> > > def normalize(s) :
> > > return
> > > urll
On Aug 29, 1:14 pm, MRAB wrote:
> On 29/08/2010 15:22, naugiedoggie wrote:
> > I'm having a problem with using a function as the replacement in
> > re.sub().
> > Here is the function:
> > def normalize(s) :
> > return
> > urllib.quote(string.capwords(urllib.unquote(s.group('provider'
>
On 29/08/2010 15:22, naugiedoggie wrote:
Hello,
I'm having a problem with using a function as the replacement in
re.sub().
Here is the function:
def normalize(s) :
return
urllib.quote(string.capwords(urllib.unquote(s.group('provider'
This normalises the provider and returns only tha
On 8/29/2010 10:22 AM, naugiedoggie wrote:
Hello,
I'm having a problem with using a function as the replacement in
re.sub().
Here is the function:
def normalize(s) :
return
urllib.quote(string.capwords(urllib.unquote(s.group('provider'
To debug your problem, I would start with print
In article
<9170aad0-478a-4222-b6e2-88d00899d...@t2g2000yqe.googlegroups.com>,
naugiedoggie wrote:
> Hello,
>
> I'm having a problem with using a function as the replacement in
> re.sub().
>
> Here is the function:
>
> def normalize(s) :
> return
> urllib.quote(string.capwords(urllib.unq
Hello,
I'm having a problem with using a function as the replacement in
re.sub().
Here is the function:
def normalize(s) :
return
urllib.quote(string.capwords(urllib.unquote(s.group('provider'
The purpose of this function is to proper-case the words contained in
a URL query string param
