getting from command line options to file

[CSUIDL PROGRAMMEr]

hi folks
I am new to python. I have a module does call a os.command(cmd) where
cmd is a rpm command.
Instead of using os.command and getting the results on command line ,
i would like to dump the output in a file. Is os.command(cmd >
filename) the most efficient command??


thanks

 

[Michael Hoffman]

hi folks
I am new to python. I have a module does call a os.command(cmd) where
cmd is a rpm command.
Instead of using os.command and getting the results on command line ,
i would like to dump the output in a file. Is os.command(cmd >
filename) the most efficient command??

I think the best thing to do would be something like this (Python 2.5):

from __future__ import with_statement
import subprocess

with file("test.out", "w") as outfile:
subprocess.check_call(["ls", "/etc"], stdout=outfile)

 

[CSUIDL PROGRAMMEr]

hi folks
I am new to python. I have a module does call a os.command(cmd) where
cmd is a rpm command.
Instead of using os.command and getting the results on command line ,
i would like to dump the output in a file. Is os.command(cmd >
filename) the most efficient command??

I think the best thing to do would be something like this (Python 2.5):

from __future__ import with_statement
import subprocess

with file("test.out", "w") as outfile:
subprocess.check_call(["ls", "/etc"], stdout=outfile)

but what if i have python which is 2.4??

 

[Gabriel Genellina]

En Fri, 20 Apr 2007 17:48:10 -0300, CSUIDL PROGRAMMEr

hi folks
I am new to python. I have a module does call a os.command(cmd) where
cmd is a rpm command.
Instead of using os.command and getting the results on command line ,
i would like to dump the output in a file. Is os.command(cmd >
filename) the most efficient command??

I think the best thing to do would be something like this (Python 2.5):

from __future__ import with_statement
import subprocess

with file("test.out", "w") as outfile:
subprocess.check_call(["ls", "/etc"], stdout=outfile)

but what if i have python which is 2.4??

Omit the with statement:

import subprocess

outfile = open("test.out", "w")
try:
subprocess.check_call(["ls", "/etc"], stdout=outfile)
finally:
outfile.close()