如何匹配从一个字符开始并以相同字符结束的模式，但不包括匹配中的最后一个字符？

Question

如果你觉得标题有点晦涩，我的意思是：我在寻找每一个以哈希符号（#）开头的模式，然后匹配该符号之后的所有内容，直到找到另一个哈希或另一个定义的条目，但最后一个哈希或其他条目不应该是匹配的一部分

举个例子：

#one_liner = some cr4zy && weird stuff h3r3 $% ()
#multi_liner =some other s7uff,
    but put in (other) line
#one_liner_again = again, some stuff here...
LB: this line shouldn't be taken into consideration!
#multi_liner_again=You guessed:
going to another line!
<EOF>

我想以四个匹配结束，例如包含这样一组元组：

("one_liner", "some cr4zy && weird stuff h3r3 \$\% ()")
("multi_liner", "some other s7uff, but put in (other) line")
("one_liner_again", "again, some stuff here...")
("multi_liner_again", "You guessed: going to another line!")

我尝试过这种模式，但它没有带来我想要的：

#\w.+\s*=(\s*|\S*.+)\w.+\n*\s*.+(?=#)

Answer 1

这可能会让您开始…您可能会希望除去找到的\n值和不需要的空格

import re

data = """
#one_liner = some cr4zy && weird stuff h3r3 $% ()
#multi_liner =some other s7uff,
    but put in (other) line
#one_liner_again = again, some stuff here...
   LB: this line shouldn't be taken into consideration!
#multi_liner_again=You guessed: 
going to another line!
"""

for m in re.findall(r"#([^#]+)\s*=\s*((?:[^#](?!LB:))*)", data, re.MULTILINE|re.DOTALL):
    print(m)

印刷品：

('one_liner ', 'some cr4zy && weird stuff h3r3 $% ()\n')
('multi_liner ', 'some other s7uff,\n    but put in (other) line\n')
('one_liner_again ', 'again, some stuff here...\n  ')
('multi_liner_again', 'You guessed: \ngoing to another line!\n')