javascript - 如何捕捉两个（编号）标记之间的所有内容，包括换行符？ [重复](How to catch everything between two (numbered) markers, including line breaks? [duplicate])

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How do I match any character across multiple lines in a regular expression? (如何在正则表达式中跨多行匹配任何字符？) (23 answers) (（23个答案）)

Closed 8 hours ago . (8小时前关闭。)

I'm trying to capture everything between two markers ( START[number] and END[number] ) where number also needs to be extracted.(我试图捕获两个标记（ START[number]和END[number] ）之间的所有内容，其中数字也需要提取。)

I need to support line breaks.(我需要支持换行符。) For instance the following :(例如以下内容：) START[1] message to capture ...END[1] must return :(必须返回：) group 1 : 1(小组1： 1) group 2 : message to capture ...(第2组： message to capture ...) group 3 : 1(第3组： 1) Here is my attempt ( demo ):(这是我的尝试（ 演示 ）：) START[(d+)]((.| | )*?)END[(d+)] It doesn't work as I obtain a third group made of the last character of the message to capture and I don't know why.(当我获得由要捕获的消息的最后一个字符组成的第三组时，它不起作用，我也不知道为什么。) Can someone may help me with this ?(有人可以帮我吗？) Thanks.(谢谢。)   ask by Malick translate from so

Answer 1

Use [\s\S] instead of (.|\n|\r)(使用[\s\S]代替(.|\n|\r))

START[(d+)]([sS]+?)END[(d+)] Demo(演示版) To be sure to have the same number in START and END, use a backreference to group 1:(为确保在START和END中具有相同的编号，请对组1使用向后引用：) (Credit to Aaron de Windt in comment)(（对Aaron de Windt进行评论）) START[(d+)]([sS]+?)END[(1)]