Java/Groovy - string: replace characters on matched regex

Question

I have a problem with creating regex of match that will get from string example: NotificationGroup_n+En where n are numbers from 1-4 and when let's say i match desired number from range i will replace or remove it with that specific number.

String BEFORE process: NotificationGroup_4+E3

String AFTER process: NotificationGroup_E3

I removed n (number from 1-4) and leave _E with number

My question is how to write regex in string.replace function to match number and than the plus sign and leave out only the string with _En

def String string = "Notification_Group_4+E3";
    println(removeChar(string));
}
public static def removeChar(String string) { 
    if ((string.contains("1+"))||(string.contains("2+")||(string.contains("3+"))||(string.contains("4+")))) {
        def stringReplaced = string.replace('4+', "");
        return stringReplaced;
    }
}

Matias Bjarland · Accepted Answer · 2019-06-05 08:29:01Z

3

in groovy:

def result = "Notification_Group_4+E3".replaceFirst(/_\d\+(.*)/, '_$1')
println result

output:

~>  groovy solution.groovy
Notification_Group_E3

~>

Try it online!

A visualization of the regex look like this:

Regex explanation:

we use groovy slashy strings /.../ to define the regex. This makes escaping simpler
we first match on underscore _
Then we match on a single digit (0-9) using the predefined character class \d as described in the javadoc for the java Pattern class.
We then match for one + character. We have to escape this with a backslash \ since + without escaping in regular expressions means "one or more" (see greedy quantifiers in the javadocs) . We don't want one or more, we want just a single + character.
We then create a regex capturing group as described in the logical operators part of the java Pattern regex using the parens expression (.*). We do this so that we are not locked into the input string ending with E3. This way the input string can end in an arbitrary string and the pattern will still work. This essentially says "capture a group and include any character (that is the . in regex) any number of times (that is the * in regex)" which translates to "just capture the rest of the line, whatever it is".
Finally we replace with _$1, i.e. just underscore followed by whatever the capturing group captured. The $1 is a "back reference" to the "first captured group" as documented in, for example, the java Matcher javadocs.

edited Jun 5, 2019 at 8:29

answered Jun 5, 2019 at 8:14

Matias Bjarland

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1 Comment

Filip Kostroš Over a year ago

thank you for support, finally understand regex little bit better, appreciated

Akash Shah · Accepted Answer · 2019-06-05 06:31:16Z

1

try this regex (\d.*?\+) here demo

in java :

String string = "Notification_Group_4+E3";
System.out.print(string.replaceAll("\\d.*?\\+", ""));

output :

Notification_Group_E3

answered Jun 5, 2019 at 6:31

Akash Shah

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1 Comment

Filip Kostroš Over a year ago

This solution helped me a lot, thanks, nice site to test regex

injecteer · Accepted Answer · 2019-06-05 10:56:20Z

0

The simple one-liner:

String res = 'Notification_Group_4+E3'.replaceAll( /_\d+\+/, '_' )
assert 'Notification_Group_E3' == res

answered Jun 5, 2019 at 10:56

injecteer

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