python and sys.argv


Question

if len(sys.argv) < 2:
    sys.stderr.write('Usage: sys.argv[0] ')
    sys.exit(1)


if not os.path.exists(sys.argv[1]):
    sys.stderr.write('ERROR: Database sys.argv[1] was not found!')
    sys.exit(1)

This is a portion of code I'm working on. The first part I'm trying to say if the user doesn't type python programname something then it will exit.

The second part I'm trying to see if the database exists. On both places I'm unsure if I have the correct way to write out the sys.argv's by stderr or not.

1
58
7/31/2016 10:25:05 AM

Accepted Answer

BTW you can pass the error message directly to sys.exit:

if len(sys.argv) < 2:
    sys.exit('Usage: %s database-name' % sys.argv[0])

if not os.path.exists(sys.argv[1]):
    sys.exit('ERROR: Database %s was not found!' % sys.argv[1])
96
6/11/2009 10:41:16 PM

In Python, you can't just embed arbitrary Python expressions into literal strings and have it substitute the value of the string. You need to either:

sys.stderr.write("Usage: " + sys.argv[0])

or

sys.stderr.write("Usage: %s" % sys.argv[0])

Also, you may want to consider using the following syntax of print (for Python earlier than 3.x):

print >>sys.stderr, "Usage:", sys.argv[0]

Using print arguably makes the code easier to read. Python automatically adds a space between arguments to the print statement, so there will be one space after the colon in the above example.

In Python 3.x, you would use the print function:

print("Usage:", sys.argv[0], file=sys.stderr)

Finally, in Python 2.6 and later you can use .format:

print >>sys.stderr, "Usage: {0}".format(sys.argv[0])

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