-----Original Message-----
From: Dave Angel [mailto:da...@ieee.org]
Sent: Sunday, August 02, 2009 12:36
To: Barak, Ron
Cc: 'python-list@python.org'
Subject: Re: Run pyc file without specifying python path ?
Barak, Ron wrote:
Hi Dave,
It seems like I don't understand your solution.
I use the (attached) soapAPI.py as the wrapper to parsing.pyc.
However, if I do (for instance):
$ python -u parsing.pyc -U aaa
The last line of the output is (as expected):
return_code: 12 ; params: {'username': 'aaa'}
But, if I try the following:
$ soapAPI.py -U aaa
I don't get this line. Only the output to stderr gets
printed to the screen.
Bye,
Ron.
Hi Ron,
To make it easier for anybody following this thread, let me
post the minimum equivalent source files, inline.
parsing.py:
------------------------------
#!/usr/bin/env python
import sys
def main():
print >> sys.stderr, "This is stderr output"
return 5, sys.argv
if __name__ == "__main__":
return_code, params = main()
print "return_code:",return_code,"; params:",params
sys.exit(return_code)
-------------------------------
soapapi.py:
-------------------------------
#!/usr/bin/env python
import sys
import parsing
parsing.main()
------------------------------
When I run soapapi.;py, it indeed prints only the stderr output.
The solution is to move (most or all) of the top-level code
of parsing.py into a main() function. Since you already have
a main(), I'll rename that, and make a new one that calls it.
new parsing.py:
-------------------------------
#!/usr/bin/env python
import sys
def innermain():
print >> sys.stderr, "This is stderr output"
return 5, sys.argv
def main():
return_code, params = innermain()
print "return_code:",return_code,"; params:",params
sys.exit(return_code)
if __name__ == "__main__":
main()
-------------------------------
The output is now two lines, one from innermain(), and one
from main().
And it's the same whether the user runs parsing.py or soapAPI.py
To clarify what happened, realize that when the user invokes
parsing.py, the module is considered a script, and gets a
pseudo-name of
"__main__" When that same module is imported by another one, it is
considered a library module, and gets its own name "parsing"
So any logic that explicitly checks for "__main__" has to
change, because we want identical behavior in the two cases.
DaveA
