On Aug 29, 1:14 pm, MRAB pyt...@mrabarnett.plus.com 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
On Aug 30, 8:52 am, naugiedoggie michael.a.p...@gmail.com wrote:
On Aug 29, 1:14 pm, MRAB pyt...@mrabarnett.plus.com 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)
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
In article
9170aad0-478a-4222-b6e2-88d00899d...@t2g2000yqe.googlegroups.com,
naugiedoggie michael.a.p...@gmail.com 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
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
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
