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Editorial Team
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Asked: 2026-06-14T12:50:01+00:00

I am using rsync –list-only command with Django/Python: subprocess.Popen([‘rsync’, ‘–list-only’, source], stdout=subprocess.PIPE, env={‘RSYNC_PASSWORD’: password}).communicate()[0]

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I am using rsync –list-only command with Django/Python:

subprocess.Popen(['rsync', '--list-only', source],
                           stdout=subprocess.PIPE, 
                           env={'RSYNC_PASSWORD': password}).communicate()[0]

It returned result like this:

drwxrwxrwx 4096 2012/11/07 09:56:23 upload

I don’t want all the files information. I only want file name displayed like this:

upload

How can I do that? Thanks

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1 Answer

  1. Editorial Team

    rsync has no option to abbreviate the output, you need to use a regex (or split())

    import re
retval = subprocess.Popen(['rsync', '--list-only', source],
                           stdout=subprocess.PIPE, 
                           env={'RSYNC_PASSWORD': password}).communicate()[0]
retval = re.sub('^.+?\d+:\d+:\d+\s+(\S+.+)', '\g<1>', retval)

    Alternatively (as long as file names have no spaces)…

    retval = subprocess.Popen(['rsync', '--list-only', source],
                           stdout=subprocess.PIPE, 
                           env={'RSYNC_PASSWORD': password}).communicate()[0]
retval = retval.split(' ')[-1]
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