How to extract string between two strings using java Pattern

Question

I have a string /subscription/ffcc218c-985c-4ec8-82d7-751fdcac93f0/subscribe from which I want to extract the middle string /subscription/<....>/subscribe. I have written the below code to get the string

String subscriber = subscriberDestination.substring(1);
int startPos = subscriber.indexOf("/") + 2;
int destPos = startPos + subscriber.substring(startPos + 2).indexOf("/");
return subscriberDestination.substring(startPos, destPos + 2);

Gives back ffcc218c-985c-4ec8-82d7-751fdcac93f0

How can I use java Pattern library to write better code?

No it doesn't. It prints ffcc218c-985c-4ec8-82d7-751fdcac93f. Without the last character. — Gonen I
– Gonen I, Commented Feb 15, 2018 at 11:53
Why would you want to use java Pattern library? Do you expect any performance gain? I doubt you'll get some by using java Pattern library. But I recommend to profile it to be absolute sure about it. — MrSmith42
– MrSmith42, Commented Feb 15, 2018 at 12:36

assylias · Accepted Answer · 2018-02-15 11:48:49Z

3

If you want to use a regular expression, a simple way would be:

return subscriber.replaceAll("/.*/([^/]*)/.*", "$1");

/.*/ is for the /subscription/ bit
([^/]*) a capturing group that matches all characters until the next /
/.* is for the /subscribe bit

And the second argument of replaceAll says that we want to keep the first group.

You can use a Pattern to improve efficiency by compiling the expression:

Pattern p = Pattern.compile("/.*/([^/]*)/.*"); ///store it outside the method to reuse it

Matcher m = p.matcher(subscriber);
if (m.find()) return m.group(1);
else return "not found";

edited Feb 15, 2018 at 11:48

answered Feb 15, 2018 at 11:43

assylias

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2 Comments

AxelH Over a year ago

Carefull that regex .* is greedy, so the first .* will consume as many "/" as possible. So it will basicly take the value just before the last "/". The initial code was usin the value after the first "/" instead. No difference here, since only 2 "/" are used. (after he consume the first one)

assylias Over a year ago

@AxelH Agreed - I had originally put a ? but decided to remove it to stick to the question and make the answer a bit simpler.

Oleg Cherednik · Accepted Answer · 2018-02-15 12:45:51Z

1

5c from me. I recommend to use Pattern for extracting substring with known format:

public final class Foo {
    private static final Pattern PATTERN = Pattern.compile(".*subscription\\/(?<uuid>[\\w-]+)\\/subscribe");

    public static String getUuid(String url) {
        Matcher matcher = PATTERN.matcher(url);
        return matcher.matches() ? matcher.group("uuid") : null;
    }
}

RegEx Demo

edited Feb 15, 2018 at 12:45

answered Feb 15, 2018 at 12:40

Oleg Cherednik

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Comments

MrSmith42 · Accepted Answer · 2018-02-15 12:39:05Z

0

Performance can be improved by:

not creating a substrings.

Also indexOf(..) with a char should be faster than with String

final int startPos = subscriberDestination.indexOf('/',1) + 1 ;
final int destPos = subscriberDestination.indexOf('/',startPos+1);
return subscriberDestination.substring(startPos, destPos );

About useing the java Pattern library:
Do you expect any performance gain? I doubt you'll get some by using java Pattern library. But I recommend to profile it to be absolute sure about it.

edited Feb 15, 2018 at 12:39

answered Feb 15, 2018 at 11:50

MrSmith42

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4 Comments

Gonen I Over a year ago

The original question has a wrong character count. You must have relied on it, because you transferred the mistake to your answer.

MrSmith42 Over a year ago

@Gonen I: I didn't check that because the question was not about an error in the result. I think I fixed that now (but feel free to edit my answer if it is still wrong.

AxelH Over a year ago

"How can I use java Pattern library to write better code?"

MrSmith42 Over a year ago

@AxelH: Since better is not yet defined in the question, I wrote this answer about how to improve performance in case of object-creation and speed.

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