extract an certain part of string using java

Question

hello, i need to extract an certain part from an string using java

ie i have a string completei4e10 i need to extract the value which is inside i and e - i.e. "4" completei 4 e10

using regex....

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.

Answer 3

The pattern would look like this: .+i([0-9])+e.+

So this outputs 4:

String str = "completei4e10";
Pattern p = Pattern.compile(".+i([0-9])+e.+");
Matcher m = p.matcher(str);
if (m.find()) {
    System.out.println(m.group(1));
}