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问题描述

我知道进行解析时,我应该理想地删除所有空格和换行符,但是我只是作为对我正在尝试的操作的快速解决方案,因此我不知道为什么它不起作用.带有"#### 1"之类的包装器的文档中的文本,并且正在尝试基于此进行解析,但是无论我如何尝试,它都无法正常工作,我认为我正确地使用了多行. >

这根本不返回任何结果:

string='
####1
ttteest
####1
ttttteeeestt

####2

ttest
####2'

import re
pattern = '.*?####(.*?)####'
returnmatch = re.compile(pattern, re.MULTILINE).findall(string)
return returnmatch
解决方案

尝试re.findall(r"####(.*?)\s(.*?)\s####", string, re.DOTALL)(当然也可以与re.compile一起使用).

此正则表达式将返回包含节号和节内容的元组.

对于您的示例,这将返回[('1', 'ttteest'), ('2', ' \n\nttest')].

(顺便说一句:您的示例将无法运行,对于多行字符串,请使用'''""")

I know that for parsing I should ideally remove all spaces and linebreaks but I was just doing this as a quick fix for something I was trying and I can't figure out why its not working.. I have wrapped different areas of text in my document with the wrappers like "####1" and am trying to parse based on this but its just not working no matter what I try, I think I am using multiline correctly.. any advice is appreciated

This returns no results at all:

string='
####1
ttteest
####1
ttttteeeestt

####2

ttest
####2'

import re
pattern = '.*?####(.*?)####'
returnmatch = re.compile(pattern, re.MULTILINE).findall(string)
return returnmatch
解决方案

Try re.findall(r"####(.*?)\s(.*?)\s####", string, re.DOTALL) (works with re.compile too, of course).

This regexp will return tuples containing the number of the section and the section content.

For your example, this will return [('1', 'ttteest'), ('2', ' \n\nttest')].

(BTW: your example won't run, for multiline strings, use ''' or """)

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08-15 01:18