G
GMTaglia
Hi guys,
I was wondering why optparse accept an option if in the command line is not
*exactly* the one present in the source code....may be an example will
explain better....
#!/usr/bin/env python
import optparse as opt
parser = opt.OptionParser()
parser.add_option("-f", "--filein", dest="fileinput", help="filename to
read", metavar="FILE")
(options, args) = parser.parse_args()
count = 0
filename = open(options.fileinput, 'r')
for lines in filename.xreadlines():
count += 1
print count
if you run this dumb example in this way:
liquid@jupiter liquid $ ./testi.py --fil elenco_tm.txt
367
it goes on, but the option must be --filein not --fil
Is true that if you have another option like
parser.add_option("-f", "--fileout", dest="fileoutput", help="filename to
write", metavar="FILE")
the program will stop raising a conflict option, but why accept only the
beginning of the long option instead of the exact one?
Have a nice sunday, and as usual sorry for the english...
GMario
I was wondering why optparse accept an option if in the command line is not
*exactly* the one present in the source code....may be an example will
explain better....
#!/usr/bin/env python
import optparse as opt
parser = opt.OptionParser()
parser.add_option("-f", "--filein", dest="fileinput", help="filename to
read", metavar="FILE")
(options, args) = parser.parse_args()
count = 0
filename = open(options.fileinput, 'r')
for lines in filename.xreadlines():
count += 1
print count
if you run this dumb example in this way:
liquid@jupiter liquid $ ./testi.py --fil elenco_tm.txt
367
it goes on, but the option must be --filein not --fil
Is true that if you have another option like
parser.add_option("-f", "--fileout", dest="fileoutput", help="filename to
write", metavar="FILE")
the program will stop raising a conflict option, but why accept only the
beginning of the long option instead of the exact one?
Have a nice sunday, and as usual sorry for the english...
GMario