Java Regex以前的一切和包括匹配

Question

I need the regex expression to remove any text before a match and including the match

eg. I want to remove "123S" and everything before it, I know I can do this with

    string.replaceAll("^.*?(?=[123S])","");
    string.replaceAll("123S","");

But really want to do it in a single expression (can't find another example anywhere!)

Answer 1

You can do it with:

string.replaceAll("^.*123S","");

Remove non-greedy ? to match last occurence and .* everything before.

Answer 2

You don't need the look ahead:

"abc123Sdef123Sxyz".replaceAll("^.*?123S",""); 

This replaces the first occurence only, if that is what you need (output is def123Sxyz ).

In case you want to replace up to the last 123S, just remove the ? modifier:

"abc123Sdef123Sxyz".replaceAll("^.*123S","");

Output is xyz .

Answer 3

string.replaceAll("^.*?123S", "");

(?=是你不想要的“if follow by”模式，而[123S]甚至不正确它只会捕获'2'。

Answer 4

string.replaceAll("^.*?123S","");

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