如何在 JavaScript 中用正则表达式匹配第 n 个字符之后的所有字符？
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How to match all characters after nth character with regex in JavaScript?
提问by Aleksei Chepovoi
I want to match all characters after 8th character. And not include first 8!
我想匹配第 8 个字符之后的所有字符。并且不包括前8个！
I need exactly a regular expression cause a framework (Ace.js) requires a regexp, not a string. So, this is not an option:
我需要一个正则表达式，因为框架 (Ace.js) 需要一个正则表达式，而不是字符串。所以，这不是一个选择：
var substring = "123456789".substr(5);
Can I match everything after nth character using regex in JavaScript?
我可以在 JavaScript 中使用正则表达式匹配第 n 个字符之后的所有内容吗？
Updates:I can't call replace(), substring()etc because I don't have a string. The string is known at run time and I don't have access to it. As I already said above the framework (Ace.js) asks me for a regex.
更新：我不能叫replace()，substring()等等，因为我没有一个字符串。该字符串在运行时是已知的，我无权访问它。正如我上面已经说过的，框架 (Ace.js) 要求我提供正则表达式。
回答by David Berndt
(?<=^.{8}).*
will match everything after the 7th position. Matches 89in 0123456789
将匹配第 7 个位置之后的所有内容。在 0123456789匹配89
or IJKLMin ABCDEFGHIJKLM
或ABCDEFGHIJKLM 中的IJKLM
etc
等等
回答by Niko Sams
console.log("123456789".match(/^.{8}(.*)/)[1])
