os.path.normpath

[nathanbullock]

I am using a windows box and passing a string like "../foo/../foo2" to
normpath which then returns "..\\foo2". But if this string is going
into a webpage link it should really be "../foo".

Is there any way to tell os.path.normpath to act like we are an a unix
style box?

What about in the new python 2.5 Path class?

Nathan

 

[BartlebyScrivener]

But if this string is going into a webpage link

http://docs.python.org/lib/module-urlparse.html

rd

 

[Gabriel Genellina]

I am using a windows box and passing a string like "../foo/../foo2" to
normpath which then returns "..\\foo2". But if this string is going
into a webpage link it should really be "../foo".

You could just .replace('\\','/') on the resulting string. Or use the
urlparse module.

Is there any way to tell os.path.normpath to act like we are an a unix
style box?

The fact than '/' is used as a path separator both on unix and on
HTTP URLs should be considered as a mere coincidence (in fact it isn't...)
URLs dont necesarily point to a real file on a real file system (Zope
is an example).



Gabriel Genellina
Softlab SRL





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[grahamd]

I am using a windows box and passing a string like "../foo/../foo2" to
normpath which then returns "..\\foo2". But if this string is going
into a webpage link it should really be "../foo".

Is there any way to tell os.path.normpath to act like we are an a unix
style box?

Use posixpath.normpath() instead.

 

[Dennis Lee Bieber]

I am using a windows box and passing a string like "../foo/../foo2" to
normpath which then returns "..\\foo2". But if this string is going
into a webpage link it should really be "../foo".

Somehow I missed this the first time...

If this is for a URL, shouldn't the a URL specific library be
considered?
--
Wulfraed Dennis Lee Bieber KD6MOG
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