more python fun!!

bulliver · 05-16-2003, 12:04 AM

Hello.

Having some difficulties with a script. Here's the relevant part:

Code:

if sys.argv[1] == '-dir':  
    convertDirectory(sys.argv[2])  
elif sys.argv[1] == '--help':     
    showUsage()  
elif string.count(sys.argv[1], '.mp3') == 1:
    convertSong(sys.argv[1])  
else:     
    showUsage()

What I want is if no command line arguments are given, the usage details are shown. When I test it with no arguments however, I get an error telling me that sys.argv[1] is out of range, which is true: it doesn't exist, which is kinda the point.

The traceback shows the error at the 'if sys.argv[1] == '-dir': ' line, but everything else works as it should.

I tried testing for sys.argv[1] ie: elif sys.argv[1] == 0 but always get the out of range error.

What am I doing wrong here?

jfabiani · 05-16-2003, 02:04 AM

if len(args) < 1:
raise Exception, "Wrong number of arguments"

bulliver · 05-16-2003, 12:50 PM

I'm afraid that didn't work. Here's the traceback:

'Traceback (most recent call last):
File "./dir2ogg.py", line 82, in ?
if sys.argv[1] == '-dir':
IndexError: list index out of range'

As you can see it is completely ignoring my else: clause, which is supposed to be the catch all right? It seems to check the first test: (if sys.argv[1] == '-dir'

then croak because the variable being checked doesn't exist. I thought it was just supposed to move down to the next test if the first doesn't apply.

Anybody have ideas?

bulliver · 05-16-2003, 12:58 PM

Ok I got it:

Code:

if len(sys.argv) < 2:
    showUsage()

elif sys.argv[1] == '-dir':
	convertDirectory(sys.argv[2])

elif sys.argv[1] == '--help':
    showUsage()

elif string.count(sys.argv[1], '.mp3') == 1:
    convertSong(sys.argv[1])

I just had to test for the missing variable first...

Thanks for the help jfabiani