Regex to parse percentage range from given string

Question

I was trying to form regex which should be able to extract percentage range from any String. For example- Adidas 30-60% Off on footwear,

I am able to extract 60% from this string using - \\\\b\\\\d+(?:%|percent\\\\b) , But, this should work for both 56% as well as 30-60% .

Any help will be appriciaed.

Answer 1

You can use

\b(?:\d+[—–-])?\d+(?:%|percent\b)

With fractions support it may look like

\b(?:\d+(?:\.\d+)?[—–-])?\d+(?:\.\d+)?(?:%|percent\b)

See the regex demo . Details :

• \\b - a word boundary
• (?:\\d+(?:\\.\\d+)?[—–-])? - an optional sequence of one or more digits followed with an optional sequence of . and one or more digits and then a dash
• \\d+(?:\\.\\d+)? - one or more digits followed with an optional sequence of . and one or more digits
• (?:%|percent\\b) - % or percent followed with a word boundary.

See a Java demo :

import java.util.regex.*;

class Test
{
    public static void main (String[] args) throws java.lang.Exception
    {
        String s = "both 56% as well as 30-60%.";
        Pattern pattern = Pattern.compile("\\b(?:\\d+[—–-])?\\d+(?:%|percent\\b)");
        Matcher matcher = pattern.matcher(s);
        while (matcher.find()){
            System.out.println(matcher.group()); 
        } 
    }
}

Output:

56%
30-60%

Answer 2

public static void main(String[] args) {
        String line = "Adidas 30-60% Off";
        String pattern = "\\b[\\d|-]+(?:%|percent\\b)";

        Pattern r = Pattern.compile(pattern);

        Matcher m = r.matcher(line);
        if (m.find( )) {
            System.out.println("Found value: " + m.group(0) );
        } else {
            System.out.println("NO MATCH");
        }
    }