Python 如何匹配从一个字符开始并以相同字符结束但不'；不包括匹配中的最后一个字符？

如果你觉得标题有点晦涩，我的意思是：我在寻找以散列符号（

#

）开头的每个模式，然后匹配该符号之后的所有内容，直到找到另一个散列或另一个定义的条目，但最后一个散列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>

我尝试过这种模式，但它没有带来我想要的：

---

\w.+\s*=（\s*\s*+）\w.+\n*\s*+（？=#））

这可能会让您开始…您可能会想要去掉找到的

\n

值和不需要的空白

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')

---

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')

---

你所说的定义条目是什么意思？只有以LB开头的sth:？@anubhava:在末尾添加了模式。@mrxra:是的，当找到

LB

时，应该停止匹配。与查找符号类似。定义的条目是什么意思？只有以LB开头的sth:？@anubhava:在末尾添加了模式。@mrxra:是的，当找到

LB

时，应该停止匹配。类似于查找

#

符号。