Python, Keeping a dynamically named file open in a large modular program

Question

I am trying to create a logfile for a large modular program. The name is dynamically produced with a timestamp and I can't think of a good way to pass the open filename other than to make it a global variable:

import time

def CreateLogFile():
    timestr = time.strftime("%Y%m%d-%H%M%S")
    global LogFile
    filenamestring = timestr + 'LogFile.txt'
    LogFile = open('PrintLogs/' + filenamestring, 'w')

CreateLogFile()

This isn't working well and I was wondering if anyone had a better suggestion. Thanks

You could wrap it all into a class to avoid the globals. Then you can pass around an instance of the class to the modules which need to use it. — joeButler, Oct 17 '13 at 23:33

score 0 · Accepted Answer · answered Oct 17 '13 at 23:34

0

Just make your own logging module:

import mylogger

log_file_handle = mylogger.get_log()

The get method can (should) handle all the logic of creating a new logfile if you haven't made one yet, or returning a handle to the already-created-this-session one.

answered Oct 17 '13 at 23:34

roippi

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Thank you, I didn't know the logger module existed – user2883778 Oct 19 '13 at 21:27

Python, Keeping a dynamically named file open in a large modular program

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