Filter a string in Groovy and do something with the matches

Question

 def aString = '[Something-26607] Updating another.txt [Something2] Updating something.txt [Something2] Updating something.txt fsfsfs'

Is there a way in groovy to get only unique parts between the square brackets and store them in a variable so after I can call them in a for loop. So I will iterate over them

Something-26607 Something2

I have tried:

def matcher = aString =~ (/\[(?<issue>[^]]+)]/) def result = matcher.find() ? matcher.group('issue') : "" print result

but I get only get the first match:

Something-26607

You can run while (matcher.find()) { and use a pattern like String regex = "\\[(?<issue>[^]]+)](?!.*\\1)"; to find the unique matches in the string, or use your original pattern and use a Set for example. The values are in group 1 of the pattern. — The fourth bird
– The fourth bird, Commented Mar 31, 2021 at 10:15
Can you give an example. I can't fully understand what you are saying — WhoAmI
– WhoAmI, Commented Mar 31, 2021 at 10:25
An example of the pattern regex101.com/r/6mqnnR/1 as an example ideone.com/3yeGUi — The fourth bird
– The fourth bird, Commented Mar 31, 2021 at 10:30

The fourth bird · Accepted Answer · 2021-03-31 10:52:51Z

I have no experience in Groovy, but reading a little bit about the syntax to get an example, you might use a Set and loop the results using a while loop:

def myList = [] def aString = '[Something-26607] Updating another.txt [Something2] Updating something.txt [Something2] Updating something.txt fsfsfs' def pattern = Pattern.compile("\\[(?<issue>[^]]+)]") def matcher = pattern.matcher(aString) while (matcher.find()) { myList << matcher.group(1) } println myList.toSet()

Output

[Something2, Something-26607]

Groovy demo

Without using a Set, you can also use a negative lookahead to check to get the unique values in the string asserting that what is captured in the group does not occur anymore at the right side.

\\[(?<issue>[^]]+)](?!.*\\1)

Regex demo

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