python 正则提取一个字符串时如何只返回字符串而不返回边界？

加括号，比如\<(.*)\>

p = re.compile(r'<(.*)>)
m = p.search(string)
if m:
s = m.group(1)
return s

@banbanchs 试了一下不管用啊

哈，这个问题好熟悉，以前刚用re，很纠结匹配之后还要去切片甚至多次re多次切片，直到我知道了()之后眼泪掉下来

@banbanchs 不好意思，我自己弄错了，刚才用这个网站http://tool.oschina.net/regex试验结果不对，用python重新做了一下是对了

@vulgur 如果是用sub替换字符串的时候这样写规则无效哦，还有什么办法吗

@decken
RTFM
https://docs.python.org/3/library/re.html#re.sub
> Backreferences, such as \6, are replaced with the substring matched by group 6 in the pattern.
> In string-type repl arguments, in addition to the character escapes and backreferences described above, \g<name> will use the substring matched by the group named name, as defined by the (?P<name>...) syntax. \g<number> uses the corresponding group number; \g<2> is therefore equivalent to \2, but isn’t ambiguous in a replacement such as \g<2>0. \20 would be interpreted as a reference to group 20, not a reference to group 2 followed by the literal character '0'. The backreference \g<0> substitutes in the entire substring matched by the RE.

a="aaaa<ab>xxx"
re.findall("<(.+)>",a)

regex look ahead/behind

这题这么简单还用不上capture

前后环视

顺带一问，如果是在js里面要达到同样效果，有没有办法呀

正则，你知道捕获括号和非捕获括号么？不想匹配，你知道有四种环视么？
知道一个会有问题么？

啥都不知道，就别说会正则！

————没错，我是愤青！