Need help in regex matching

Question

It may be very simple, but I am extremely new to regex and have a requirement where I need to do some regex matches in a string and extract the number in it. Below is my code with sample i/p and required o/p. I tried to construct the Pattern by referring to https://www.freeformatter.com/java-regex-tester.html , but my regex match itself is returning false.

Pattern pattern = Pattern.compile(".*/(a-b|c-d|e-f)/([0-9])+(#[0-9]?)");
String str = "foo/bar/Samsung-Galaxy/a-b/1"; // need to extract 1.
String str1 = "foo/bar/Samsung-Galaxy/c-d/1#P2";// need to extract 2.
String str2 = "foo.com/Samsung-Galaxy/9090/c-d/69"; // need to extract 69

System.out.println("result " + pattern.matcher(str).matches());
System.out.println("result " + pattern.matcher(str1).matches());
System.out.println("result " + pattern.matcher(str1).matches());

All of above SOPs are returning false. I am using java 8, is there is any way by which in a single statement I can match the pattern and then extract the digit from the string.

I would be great if somebody can point me on how to debug/develop the regex.Please feel free to let me know if something is not clear in my question.

Answer 1

You may use

Pattern pattern = Pattern.compile(".*/(?:a-b|c-d|e-f)/[^/]*?([0-9]+)");

See the regex demo

When used with matches() , the pattern above does not require explicit anchors, ^ and $ .

Details

• .* - any 0+ chars other than line break chars, as many as possible
• / - the rightmost / that is followed with the subsequent subpatterns
• (?:ab|cd|ef) - a non-capturing group matching any of the alternatives inside: ab , cd or ef
• / - a / char
• [^/]*? - any chars other than / , as few as possible
• ([0-9]+) - Group 1: one or more digits.

Java demo :

List<String> strs = Arrays.asList("foo/bar/Samsung-Galaxy/a-b/1","foo/bar/Samsung-Galaxy/c-d/1#P2","foo.com/Samsung-Galaxy/9090/c-d/69");
Pattern pattern = Pattern.compile(".*/(?:a-b|c-d|e-f)/[^/]*?([0-9]+)");
for (String s : strs) {
    Matcher m = pattern.matcher(s);
    if (m.matches()) {
        System.out.println(s + ": \"" + m.group(1) + "\"");
    }
}

A replacing approach using the same regex with anchors added:

List<String> strs = Arrays.asList("foo/bar/Samsung-Galaxy/a-b/1","foo/bar/Samsung-Galaxy/c-d/1#P2","foo.com/Samsung-Galaxy/9090/c-d/69");
String pattern = "^.*/(?:a-b|c-d|e-f)/[^/]*?([0-9]+)$";
for (String s : strs) {
    System.out.println(s + ": \"" + s.replaceFirst(pattern, "$1") + "\"");
}

See another Java demo .

Output:

foo/bar/Samsung-Galaxy/a-b/1: "1"
foo/bar/Samsung-Galaxy/c-d/1#P2: "2"
foo.com/Samsung-Galaxy/9090/c-d/69: "69"

Answer 2

Because you match always the last number in your regex, I would Like to just use replaceAll with this regex .*?(\\d+)$ :

String regex = ".*?(\\d+)$";

String strResult1 = str.replaceAll(regex, "$1");
System.out.println(!strResult1.isEmpty() ? "result " + strResult1 : "no result");

String strResult2 = str1.replaceAll(regex, "$1");
System.out.println(!strResult2.isEmpty() ? "result " + strResult2 : "no result");

String strResult3 = str2.replaceAll(regex, "$1");
System.out.println(!strResult3.isEmpty() ? "result " + strResult3 : "no result");

If the result is empty then you don't have any number.

Outputs

result 1
result 2
result 69

Answer 3

Here is a one-liner using String#replaceAll :

public String getDigits(String input) {
    String number = input.replaceAll(".*/(?:a-b|c-d|e-f)/[^/]*?(\\d+)$", "$1");
    return number.matches("\\d+") ? number : "no match";
}

System.out.println(getDigits("foo.com/Samsung-Galaxy/9090/c-d/69"));
System.out.println(getDigits("foo/bar/Samsung-Galaxy/a-b/some other text/1"));
System.out.println(getDigits("foo/bar/Samsung-Galaxy/9090/a-b/69ace"));

69
no match
no match

This works on the sample inputs you provided. Note that I added logic which will display no match for the case where ending digits could not be matched fitting your pattern. In the case of a non-match, we would typically be left with the original input string, which would not be all digits.