Java regular expression word match

Question

I have 3 values IU, PRI and RET. if my input string contains any one or more value(s),
the Java regular expression should return true.

Answer 1

Try

String s= "A IU somehting PRI something RET whatever";

Pattern p= Pattern.compile("(IU|PRI|RET)");
Matcher m= p.matcher(s);
while (m.find()) {
    String matched= m.group(1);
    System.out.println(matched);
}

It prints:

IU
PRI
RET

Answer 2

Ok here's a crazy solution with anagrams of each given String, built into a Pattern just for fun:

public static void main(String[] args) {
    try {
        Pattern pattern = makePattern("IU", "PRI", "RET");
        System.out.println(pattern.pattern());
        String test = "put returns UI between paragraphs, also IRP and TER";
        Matcher matcher = pattern.matcher(test);
        while (matcher.find()) {
            System.out.println(matcher.group());
        }
    }
    catch (Exception e) {
        e.printStackTrace();
    }
}
public static Pattern makePattern(String... words) throws Exception {
    if (words == null || words.length == 0) {
        throw new Exception("TODO handle invalid argument");
    }
    StringBuilder patternBuilder = new StringBuilder("(");
    for (String word : words) {
        if (word == null || word.isEmpty()) {
            throw new Exception("TODO invalid word");
        }
        for (String anagram: doAnagrams(word, null)) {
            patternBuilder.append("\\b").append(anagram).append("\\b").append("|");
        }
    }
    patternBuilder.deleteCharAt(patternBuilder.length() - 1);
    patternBuilder.append(")");
    return Pattern.compile(patternBuilder.toString());
}
public static Set<String> doAnagrams(String original, Set<String> processed) {
    if (original == null || original.isEmpty()) {
        return new LinkedHashSet<String>();
    }
    Set<String> result;
    if (processed == null) {
        result = new LinkedHashSet<String>();
        result.add(original);
    } else {
        result = processed;
    }
    if (original.length() <= 1) {
        return result;
    }
    String sub = original.substring(1);
    String subStart = original.substring(0, 1);
    for (String subAnagram : doAnagrams(sub, null)) {
        result.add(subAnagram.concat(subStart));
    }
    if (sub.concat(original.substring(0, 1)).equals(result.iterator().next())) {
        return result;
    } 
    else {
        return doAnagrams(sub.concat(subStart), result);
    }
}

Output:

(\bIU\b|\bUI\b|\bPRI\b|\bRIP\b|\bIRP\b|\bIPR\b|\bPIR\b|\bRPI\b|\bRET\b|\bETR\b|\bTER\b|\bTRE\b|\bRTE\b|\bERT\b)
UI
IRP
TER

Answer 3

You need word boundaries for that:

boolean foundMatch = false;
Pattern regex = Pattern.compile("\\b(?:UI|PRI|RET)\\b");
Matcher regexMatcher = regex.matcher(subjectString);
foundMatch = regexMatcher.find();

Answer 4

I don't know if you are still looking for the solution of this. But here's the code for your question. I assumed that the anagrams you are looking for are separated by spaces and the words appear in Uppercase.

    String text = "put returns UI between IU paragraphs PRI RIP and RET ETR";
    Pattern p = Pattern.compile("([UI]{2}|[PRI]{3}|[RET]{3})");

    Matcher m = p.matcher(text);
    System.out.println(m.find());

If you are try for case insensitive matching, change the pattern to the following;

    (?i)([UI]{2}|[PRI]{3}|[RET]{3})

Answer 5

You could do this in one line and get your boolean value.

boolean matcher = Pattern.matches("[UI]{2}|[PRI]{3}|[RET]{3}", stringToBeMatched);