我有这样的代码:
super dooper pooper but dooper super pooper
我需要匹配从super
到pooper
(可以包含所有字符)的所有内容,但请停留在第一个pooper
。
我该怎么做?
答案 0 :(得分:0)
正如评论中所说的那样:/super.*?pooper/
如果你不关心跨线的匹配,就会有所作为。如果您需要跨行匹配(并且您在评论中说过),那么您需要:
/super.*?pooper/s
这意味着:
/
= delimiter super
=匹配super
.*?
=在"非贪婪"中匹配零个或多个字符时尚(尽快停止)pooper
=匹配pooper
/
= delimiter s
=让.
匹配换行符(\n
)答案 1 :(得分:0)
试试这个正则表达式:
super[\s\S]*?pooper
super ... Match super
[\s\S] ... Match a single character that is a whitespace character or
match a single character that is not a whitespace character (all characters).
*? ... Between zero and unlimmited times as few times as possible.
pooper ... Match pooper