On Fri, 11 Aug 2006 01:23:14 +0800
Angelo Zhou <[EMAIL PROTECTED]> wrote:
#> Slawomir Nowaczyk wrote:
#> > Hello,
#> >
#> > Let's say I have a module "emacs", defining function eexecfile(file):
#> >
#> > def eexecfile(file):
#> > # do other stuff
#> > execfile(file,globals())
#> > # do other stuff
#> >
#> > Now, assume I have file test.py containing an assignment "x=1"
#> >
#> > If I run python and do:
#> >
#> > import emacs
#> > emacs.eexecfile("test.py")
#> > print emacs.x # works, x was put in module namespace
#> > print x # doesn't work, x is not defined in main script namespace
#> >
#> > What is the best way to make "print x" work? Using the following:
#> >
#> > import __main__
#> > def eexecfile(file):
#> > # do other stuff
#> > execfile(file, __main__.__dict__)
#> > # do other stuff
#> >
#> > seems to work, but it gives me a slightly uneasy feeling. Is this the
#> > right way?
#> >
#>
#> "from emacs import x" will expose x to the current namespace
True... but I do not know in advance what is the contents of test.py
file -- it could be "a=1" :) Sure, I could go over emacs.__dict__ and
expose everything except eexecfile, but that's even less satisfying
than the solution above.
Anyway, thanks for the suggestion.
--
Best wishes,
Slawomir Nowaczyk
( [EMAIL PROTECTED] )
Zawinski's Law: "Every program attempts to expand until it can read mail.
Those programs which cannot so expand are replaced by ones which can."
--
http://mail.python.org/mailman/listinfo/python-list
