使用正则表达式在Java中提取字符串的特定部分

Question

I need to extract a certain part of string in Java using regex.

For example, I have a string completei4e10 , and I need to extract the value that is between the i and e - in this case, the result would be 4 : completei 4 e10 .

How can I do this?

Answer 1

import java.util.regex.Matcher;
import java.util.regex.Pattern;

public class Test {
    public static void main(String[] args) {
        Pattern p = Pattern.compile( "^[a-zA-Z]+([0-9]+).*" );
        Matcher m = p.matcher( "completei4e10" );

        if ( m.find() ) {
            System.out.println( m.group( 1 ) );
        }

    }
}

Answer 2

There are several ways to do this, but you can do:

    String str = "completei4e10";

    str = str.replaceAll("completei(\\d+)e.*", "$1");

    System.out.println(str); // 4

Or maybe the pattern is [^i]*i([^e]*)e.* , depending on what can be around the i and e .

    System.out.println(
        "here comes the i%@#$%@$#e there you go i000e"
            .replaceAll("[^i]*i([^e]*)e.*", "$1")
    );
    // %@#$%@$#

The […] is a character class . Something like [aeiou] matches one of any of the lowercase vowels. [^…] is a negated character class. [^aeiou] matches one of anything but the lowercase vowels.

The (…) is a capturing group . The * and + are repetition specifier in this context.