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这是一个创建于 3977 天前的主题，其中的信息可能已经有所发展或是发生改变。
例如匹配字符串“aaaa<ab>xxx"中尖括号中的ab，用‘\<.*\>'匹配后，返回的结果是<ab>,如何实现返回的结果只有ab呢？
一种简单的方法是对返回的结果做切片，请问怎么用python正则表达可以做到？
一种简单的方法是对返回的结果做切片，请问怎么用python正则表达可以做到？
 | 1 banbanchs 2014-11-22 15:43:24 +08:00  1 加括号，比如\<(.*)\> |
 | 2 vulgur 2014-11-22 15:46:39 +08:00 p = re.compile(r'<(.*)>) m = p.search(string) if m: s = m.group(1) return s |
 | 4 14 2014-11-22 15:49:55 +08:00 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 |
 | 5 tommark OP @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 |
 | 7 aaaa007cn 2014-11-22 17:55:29 +08:00 @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. |
 | 8 xz 2014-11-22 18:04:34 +08:00 a="aaaa<ab>xxx" re.findall("<(.+)>",a) |
 | 9 ryd994 2014-11-23 07:30:57 +08:00 via Android regex look ahead/behind |
| 10 ryd994 2014-11-23 07:31:50 +08:00 via Android 这题这么简单还用不上capture |
 | 11 leoleozhu 2014-11-23 07:51:14 +08:00 via Android 前后环视 |
 | 12 cdxem713 2014-11-23 08:30:59 +08:00 via iPhone 顺带一问，如果是在js里面要达到同样效果，有没有办法呀 |
 | 13 mhohai 2014-11-24 08:51:55 +08:00 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！ |
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