Thanks for the responses!
I'd forgotten about using the sys module:
import sys
filename = sys.argv[0]
Using "__file__" also works.
Thanks
Scott
On Nov 29, 2004, at 9:37 AM, Scott Frankel wrote:
I'm looking for a way to identify a filename remotely. Put
differently,
is there a way a file can get its own name from its globals()?
doit.py calls exec() on a second py script, tpairs.py, to obtain a
dict of the
globals in tpairs.py. How can I add the filename, "tpairs.py," to the
resulting dict?
i.e.:
# - - - - - - - - - - - - - - - -
# -- tpairs.py
# - - - - - - - - - - - - - - - -
this = "this"
that = "that"
gdict = globals()
# - - - - - - - - - - - - - - - -
# -- doit.py
# - - - - - - - - - - - - - - - -
#!/usr/bin/env python
theFile = "./tpairs.py"
theDict = {}
execfile(theFile, theDict)
# somehow add theFile to gdict, i.e.:
# gdict['theFile'] = theFile
Thanks in advance!
Scott
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