Compare version strings in groovy

Hey I have created a Groovy script that will extract the version numbers of some folder. I would then like to compare the version numbers and select the highest.

I got my script to run through the dir folder and I then get the versions in this format: 02.2.02.01

So I could get something like this:

02.2.02.01
02.2.02.02
02.2.03.01

I don't have them as a list but like this:

baseDir.listFiles().each { file -> 
  def string = file.getName().substring(5, 15)
  // do stuff
}

Also I have tested that Groovy could compare them with the > operator and it can! But now I need to select the one with the highest version

This appears to work

String mostRecentVersion(List versions) {
  def sorted = versions.sort(false) { a, b -> 

    List verA = a.tokenize('.')
    List verB = b.tokenize('.')
     
    def commonIndices = Math.min(verA.size(), verB.size())
    
    for (int i = 0; i < commonIndices; ++i) {
      def numA = verA[i].toInteger()
      def numB = verB[i].toInteger()
      
      if (numA != numB) {
        return numA <=> numB
      }
    }
    
    // If we got this far then all the common indices are identical, so whichever version is longer must be more recent
    verA.size() <=> verB.size()
  }
  
  println "sorted versions: $sorted"
  sorted[-1]
}

Here is an inadequate set of tests. You should add some more.

assert mostRecentVersion(['02.2.02.01', '02.2.02.02', '02.2.03.01']) == '02.2.03.01' 
assert mostRecentVersion(['4', '2']) == '4'
assert mostRecentVersion(['4.1', '4']) == '4.1'
assert mostRecentVersion(['4.1', '5']) == '5'

Run this code and the tests in the Groovy console to verify that it works

If we're going for the shortest answer, this must come close ;-)

String mostRecentVersion( List versions ) {
  versions.sort( false ) { a, b ->
    [a,b]*.tokenize('.')*.collect { it as int }.with { u, v ->
      [u,v].transpose().findResult{ x,y-> x<=>y ?: null } ?: u.size() <=> v.size()
    }
  }[-1]
}

Mine is the shortest! lol )

versions = versions.sort {a, b ->
  def a1 = a.tokenize('.')*.toInteger(), b1 = b.tokenize('.')*.toInteger()
  for (i in 0..<[a1.size(), b1.size()].min()) 
    if (a1[i] != b1[i]) return a1[i] <=> b1[i]
  0
}

If anyone is using Grails (e.g. Grails 2.2.3), I think VersionComparator already provides exactly what we need.

If you are not using Grails, you can always Google the source code of this class.

Example of working tests:

import org.codehaus.groovy.grails.plugins.VersionComparator

assert ['1.13.4', '1.4.5'].sort( new VersionComparator() ) == ['1.4.5', '1.13.4']
assert ['3.1.20', '3', '3.0.1', '3.1'].sort( new VersionComparator() ) == ['3', '3.0.1', '3.1', '3.1.20']
assert ['02.2.02.02', '02.2.03.01', '02.2.02.01'].sort( new VersionComparator() ) == ['02.2.02.01', '02.2.02.02', '02.2.03.01']
assert ['4', '2'].sort( new VersionComparator() ) == ['2', '4']
assert ['4.1', '4'].sort( new VersionComparator() ) == ['4', '4.1']
assert ['4.1', '5'].sort( new VersionComparator() ) == ['4.1', '5']

assert new VersionComparator().compare( '1.13.4', '1.4.5' ) > 0
assert new VersionComparator().compare( '1.4.5', '1.13.4' ) < 0

Hope this helps.

Compare version strings in groovy

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