Question

我正在尝试通过使用PyParsing删除前导或尾随空白字符来清除一些代码。删除前导空格很容易，因为我可以使用匹配字符串但不包含它的FollowedBy子类。现在我需要跟我识别的字符串一样的东西。

这是一个小例子：

from pyparsing import *

insource = """
annotation (Documentation(info="  
  <html>  
<b>FOO</b>
</html>  
 "));
"""
# Working replacement:
HTMLStartref = OneOrMore(White(' \t\n')) + (FollowedBy(CaselessLiteral('<html>')))

## Not working because of non-existing "LeadBy" 
# HTMLEndref = LeadBy(CaselessLiteral('</html>')) + OneOrMore(White(' \t\n')) + FollowedBy('"')

out = Suppress(HTMLStartref).transformString(insource)
out2 = Suppress(HTMLEndref).transformString(out)

输出结果为：

>>> print out
annotation (Documentation(info="<html>
<b>FOO</b>
</html>
 "));

和应该获取：

>>> print out2
annotation (Documentation(info="<html>
<b>FOO</b>
</html>"));

我查看了documentation，但未找到与LeadBy等效的“FollowedBy”或如何实现该目标。

Answer 1

你所要求的是＆＃34; lookbehind＆＃34;，也就是说，只有在特定模式之前有事物时才匹配。我现在还没有明确的课程，但是对于你想做的事情，你仍然可以从左到右转换，只留下领先部分，而不是抑制它，只是压制空白。

以下是解决问题的几种方法：

# define expressions to match leading and trailing
# html tags, and just suppress the leading or trailing whitespace
opener = White().suppress() + Literal("<html>")
closer = Literal("</html>") + White().suppress()

# define a single expression to match either opener
# or closer - have to add leaveWhitespace() call so that
# we catch the leading whitespace in opener
either = opener|closer
either.leaveWhitespace()

print either.transformString(insource) 


# alternative, if you know what the tag will look like:
# match 'info=<some double quoted string>', and use a parse
# action to extract the contents within the quoted string,
# call strip() to remove leading and trailing whitespace,
# and then restore the original '"' characters (which are
# auto-stripped by the QuotedString class by default)
infovalue = QuotedString('"', multiline=True)
infovalue.setParseAction(lambda t: '"' + t[0].strip() + '"')
infoattr = "info=" + infovalue

print infoattr.transformString(insource)

如何在PyParsing中构造一个与FollowedBy子类相当的“前导”

1 个答案: