我试图更好地了解optparse
,但仍在努力理解以下代码为何以这种方式运行。我在做蠢事吗?
import optparse
def store_test(option, opt_str, value, parser, args=None, kwargs=None):
print 'opt_str:', opt_str
print 'value:', value
op = optparse.OptionParser()
op.add_option('-t', '--test', action='callback', callback=store_test, default='test',
dest='test', help='test!')
(opts, args) = op.parse_args(['test.py', '-t', 'foo'])
print
print 'opts:'
print opts
print 'args:'
print args
输出:
opt_str: -t value: None opts: {'test': 'test'} args: ['foo']
Why is 'foo'
not being passed to store_test()
and instead being interpreted as an extra argument? Is there something wrong with op.parse_args(['-t', 'foo'])
?
↓
http://codepad.org/vq3cvE13
Edit:
Here's the example from the docs:
def store_value(option, opt_str, value, parser):
setattr(parser.values, option.dest, value)
[...]
parser.add_option("--foo",
action="callback", callback=store_value,
type="int", nargs=3, dest="foo")
最佳答案
您缺少“type”或“nargs”选项属性:
op.add_option('-t', '--test', action='callback', callback=store_test, default='test',
dest='test', help='test!', type='str')
此选项将导致它消耗下一个参数。
引用:
http://docs.python.org/library/optparse.html#optparse-option-callbacks
这似乎是
optparse.py
的相关代码:def takes_value(self):
return self.type is not None
def _process_short_opts(self, rargs, values):
[...]
if option.takes_value():
# Any characters left in arg? Pretend they're the
# next arg, and stop consuming characters of arg.
if i < len(arg):
rargs.insert(0, arg[i:])
stop = True
nargs = option.nargs
if len(rargs) < nargs:
if nargs == 1:
self.error(_("%s option requires an argument") % opt)
else:
self.error(_("%s option requires %d arguments")
% (opt, nargs))
elif nargs == 1:
value = rargs.pop(0)
else:
value = tuple(rargs[0:nargs])
del rargs[0:nargs]
else: # option doesn't take a value
value = None
option.process(opt, value, values, self)
关于python - Optparse回调不消耗参数,我们在Stack Overflow上找到一个类似的问题:https://stackoverflow.com/questions/6583963/