如果python中没有选项，则选择工作

Question

I need to load the argument even if there is a option and if not a option.

#!/usr/bin/python
import optparse

parser = optparse.OptionParser()
parser.add_option('-i', dest='name', help='some')
parser.add_option('-c', dest='name', help='some')
parser.add_option('-p', action='store', help='password')

print parser.parse_args()

[root@server tmp]# ./test -i abc
(<Values at 0x4011368: {'p': None, 'name': 'abc'}>, [])
[root@server tmp]# ./test  abc
(<Values at 0x5855368: {'p': None, 'name': None}>, ['abc'])

Now I need have the value "abc" even if I am not using any option. So please let know how can I access that value.

Answer 1

Based solely on your output, you should be able to see that parse_args returns a tuple. The first element of that tuple is an object containing values for defined options and the second element is a list of arguments leftover after parsing options. You can read more about it in the official tutorial .

Having this in mind, you can simply write

options, arguments = parser.parse_args()

and use arguments to do whatever you want with that list.

However, your problem seems to be that when you supply an option, argument is parsed as an option's value. This is caused by your way of defining options because options -i and -c need values.

If you want those options to be boolean, you need to define that manually. Example for one option code would be something like

# This defines an option which set name to True if option is provided, otherwise
# name is set to False
parser.add_option('-i', dest='name', help='some', action="store_true", default=False)

This would also mean that you don't need to provide value for that option, so argument won't be consumed when the parser reads options.

If you want your options to be non-boolean, but don't want to proved values for them, then I'm not sure I get what you're trying to do.

Answer 2

See this maybe : 15.5.2.3. Handling boolean (flag) options