pattern matching using regex in groovy

Question

I have the below groovy code

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

String myInput="License_All (12313)"
String myRegex="\\(\\d+\\)"


String ResultString
Pattern regex
Matcher regexMatcher

regex = Pattern.compile(myRegex, Pattern.DOTALL);

regexMatcher = regex.matcher(myInput);

List<String> matchList = new ArrayList<String>();
while (regexMatcher.find()) {
matchList.add(regexMatcher.group());
}

for(int i=0;i<matchList.size();i++)
{
myInput=myInput.replaceAll( matchList[i], '')
println matchList[i]
}
println myInput

It is supposed to remove (12313) part but instead it gives me the below output

(12313)
License_All ()

How do i remove all the bracket portion (12313)?

Howard · Accepted Answer · 2011-09-28 08:27:33Z

2

The matchList will contain the matched part, i.e. (12313). This one is then again taken as regular expression within replaceAll and therefore interpreted as group. Thus the brackets are not removed.

Instead you can use your regex directly in replaceAll:

myInput.replaceAll(myRegex, "")

answered Sep 28, 2011 at 8:27

Howard

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tim_yates · Accepted Answer · 2011-09-28 08:51:19Z

1

Becaue you are using replaceAll, it is expecting a Pattern as the first element

myInput = myInput.replaceAll( Pattern.compile(myRegex, Pattern.DOTALL), '' )

Should do it...

As an aside, your original code can be made into more idiomatic Groovy (rather than basically Java) like so -- obviously, this has the same issue as in your original question, it's just a more Groovy way of writing code:

import java.util.regex.Pattern

String myInput = "License_All (12313)"

def regex = Pattern.compile( /\(\d+\)/, Pattern.DOTALL);

List<String> matchList = ( myInput =~ regex ) as List

matchList.each { m ->
  myInput = myInput.replaceAll( m, '')
  println m
}
println myInput

edited Sep 28, 2011 at 8:51

answered Sep 28, 2011 at 8:31

tim_yates

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