regex expression in java using wildcards

You can use the regex, "(?<![a-z])" + Pattern.quote(phrase) + "(?![a-z])"

Demo at regex101 with phrase = "hello".

• (?<![a-z]): Negative lookbehind for [a-z]
• (?![a-z]): Negative lookahead for [a-z]

Java Demo:

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

public class Main {

    public static void main(String[] args) {
        // Test
        String phrase = "hello";
        String regex = "(?<![a-z])" + Pattern.quote(phrase) + "(?![a-z])";
        Pattern pattern = Pattern.compile(regex);
        Stream.of(
                "hi hello world",
                "hihelloworld"
        ).forEach(s -> {
            Matcher matcher = pattern.matcher(s);
            System.out.print(s + " => ");
            if(matcher.find()) {
                System.out.println("Match found");
            }else {
                System.out.println("No match found");
            }
        });
    }
}

Output:

hi hello world => Match found
hihelloworld => No match found

In case you want the full-match, use the regex, .*(?<![a-z]) + Pattern.quote(phrase) +(?![a-z]).* as demonstrated at regex101.com. The pattern, .* means any character any number of times. The rest of the patterns are already explained above. The presence of .* before and after the match will ensure covering the whole string.

Java Demo:

import java.util.regex.Pattern;
import java.util.stream.Stream;

public class Main {

    public static void main(String[] args) {
        // Test
        String phrase = "hello";
        String regex = ".*(?<![a-z])" + Pattern.quote(phrase) + "(?![a-z]).*";
        Stream.of(
                    "hi hello world", 
                    "hihelloworld"
        ).forEach(s -> System.out.println(s + " => " + (s.matches(regex) ? "Match found" : "No match found")));
    }
}

Output:

hi hello world => Match found
hihelloworld => No match found