RVince wrote:
Ha! You;re right -- but is there a way to get it without the filename
appended at the end?
On May 24, 11:52 am, Tim Golden <m...@timgolden.me.uk> wrote:
On 24/05/2011 16:36, RVince wrote:
s = "C:\AciiCsv\Gravity_Test_data\A.csv"
f = open(s,"r")
How do I obtain the full pathname given the File, f? (which should
equal "C:\AciiCsv\Gravity_Test_data"). I've tried all sorts of stuff
and am just not finding it. Any help greatly appreciated !
You're going to kick yourself:
f.name
TJG
path, fileName = os.path.split(os.path.abspath(f.name))
JM
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