How can I determine who is calling my Python script?

Question

I am writing a Python script, and I'd like its behaviour to depend on who's calling that script:

• if it is called from within a batchfile, I want the Python script to write its output to a logfile.
• if it is called manually from a command prompt, I want the Python script to write its output on screen.

My batchfile looks as follows:

  python py_script.py

Is there any way to get this done? I had a look at os.environ.has_key() but I don't know how to use this.

Answer 1

Just pass an argument to your script to tell him where to write... command line arguments are passed in sys.argv . There are a couple packages in the stdlib to deal with them but you might not have need for here that's the only option for your script.

Answer 2

The easiest way to do this is (as tobias_k remarks in the comments) merely to pipe to a log file when called from a script

REM This is your batch file
python py_script.py >> your_log_file.txt

Bruno's suggestion in the other answer to pull a flag from sys.argv is certainly valid, but you're going to end up doing a lot of duplicate code. Your codebase will then look like:

if LOGGING:
    with open(PATH_TO_LOGFILE, 'a') as logf:
        logf.write(some_txt)
else:
    print(some_txt)

anytime you have a print call. Alternatively you could do some INCREDIBLY ugly stuff like

if __name__ == "__main__":
    if LOGGING:
        old_stdout = sys.sdout
        with open(PATH_TO_LOGFILE, 'a') as logf:
            sys.stdout = logf
            main()
        sys.stdout = old_stdout

Honestly that's ugly and hackish ( and might not even work, I've never tried to reassign sys.stdout ). Just use shell redirection.