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Home/ Questions/Q 8194635
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Editorial Team
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Editorial Team
Asked: June 7, 20262026-06-07T04:53:20+00:00 2026-06-07T04:53:20+00:00

I want to write a script which checks and opens a settings-file called .mysettings

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I want to write a script which checks and opens a settings-file called “.mysettings” if it exists in the HOME-directory. If the file is not present in the HOME-directory it should try to open one in the current directory if it exists there.

Is there a idiom or a one-liner in python to program something like that?

The best way I can think of right now is to try to open the first file with a try-catch block like explained in this question and then trying the second file.

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  1. Editorial Team
    Editorial Team
    2026-06-07T04:53:21+00:00Added an answer on June 7, 2026 at 4:53 am

    This is the python way to do it. No one liner, but clear, and easy to read.

    try:
        with open("/tmp/foo.txt") as foo:
            print foo.read()
    except:
        try:
            with open("./foo.txt") as foo:
                print foo.read()
        except:
            print "No foo'ing files!"
    

    Of course, you could always do something like this as well:

    for f in ["/tmp/foo.txt", "./foo.txt"]:
        try:
            foo = open(f)
        except:
            pass
        else:
            print foo.read()
    
    • Beautiful is better than ugly.
    • Readability counts.
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