我正在尝试使用 PyParsing 删除前导或尾随空白字符来清理一些代码。删除前导空格非常容易，因为我可以使用 FollowedBy 子类来匹配字符串但不包含它。现在我需要同样的东西来处理我的识别字符串后面的内容。
这里是一个小例子:
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)
作为输出，得到:
>>> print out
annotation (Documentation(info="<html>
<b>FOO</b>
</html>
"));
并且应该得到:
>>> print out2
annotation (Documentation(info="<html>
<b>FOO</b>
</html>"));
我查看了documentation但找不到与 FollowedBy 等效的“LeadBy”或实现该目标的方法。
最佳答案
您要求的是类似“lookbehind”的内容，即仅当某些内容前面有特定模式时才匹配。目前我还没有一个明确的类，但是对于你想做的事情，你仍然可以从左到右转换，只保留前导部分，而不是抑制它，只是抑制空格.
以下是解决您的问题的几种方法:
# 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)
关于python - 如何在 PyParsing 中构造一个相当于 FollowedBy 子类的 "leading"，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/11774840/