pyparsing Optional()& Optional()允许重复

时间:2015-10-31 04:17:25

标签: python pyparsing

我有这个简单的语法:

word = Word(alphanums + '_')
with_stmt = Suppress('with') + OneOrMore(Group(word('key') + Suppress('=') + word('value')))('overrides')
using_stmt = Suppress('using') + Regex('id-[0-9a-f]{8}')('id')
modifiers = Optional(with_stmt('with_stmt')) & Optional(using_stmt('using_stmt'))
pattern = StringStart() + modifiers + StringEnd()

似乎Optional() & Optional()错误地允许多次重复modifier,并且只标记最后一个:

>>> print dict(pattern.parseString('with foo=bar bing=baz using id-deadbeef using id-feedfeed'))
{
  'with_stmt': (
    [
      (['foo', 'bar'], {'value': [('bar', 1)], 'key': [('foo', 0)]}), 
      (['bing', 'baz'], {'value': [('baz', 1)], 'key': [('bing', 0)]})
    ],
    {'overrides': 
      [(([
        (['foo', 'bar'], {'value': [('bar', 1)], 'key': [('foo', 0)]}),
        (['bing', 'baz'], {'value': [('baz', 1)], 'key': [('bing', 0)]})
      ], {}), 0)]
    }
  ), 
  'overrides': 
    (
      [(['foo', 'bar'], {'value': [('bar', 1)], 'key': [('foo', 0)]}),
      (['bing', 'baz'], {'value': [('baz', 1)], 'key': [('bing', 0)]})], {}
    ),
  'id': (['id-deadbeef', 'id-feedfeed'], {}),
  'using_stmt': (['id-deadbeef', 'id-feedfeed'], {'id': [('id-deadbeef', 0), ('id-feedfeed', 1)]})
}

using_stmtid-deadbeefid-feedfeed匹配,而不是在using id-feedfeed处抛出错误。

奇怪的是,如果使modifiers非可选,那么重复问题就会消失,解析会按预期失败:

>>> dict(pattern.parseString('with foo=bar bing=baz using id-deadbeef using id-feedfeed'))
Traceback (most recent call last):
  File "parse.py", line 10, in <module>
    print dict(pattern.parseString('with foo=bar bing=baz using id-deadbeef using id-feedfeed'))
  File "/path/to/lib/python2.7/site-packages/pyparsing.py", line 1139, in parseString
    raise exc
pyparsing.ParseException: Expected end of text (at char 40), (line:1, col:41)

切换到+而不是&也会导致其按预期失败。 with_stmt表现出同样的问题,并且使其成为非可选项也会修复它。

将模式标记为可选的允许在Each()内重复的内容是什么?

1 个答案:

答案 0 :(得分:1)

这是pyparsing的每个类的错误 - 将在2.0.6中修复。