Get specific java version with powershell

Question

I have some issues with getting the java version out as a string. In a batch script I have done it like this:

for /f tokens^=2-5^ delims^=.-_^" %%j in ('%EXTRACTPATH%\Java\jdk_extract\bin\java -fullversion 2^>^&1') do set "JAVAVER=%%j.%%k.%%l_%%m"

The output is: 1.8.0_121

Now I want to do this for PowerShell, but my output is: 1.8.0_12, I miss one "1" in the end Now I have tried it with trim and split but nothing gives me the right output can someone help me out? This is what I've got so var with PowerShell

$javaVersion = (& $extractPath\Java\jdk_extract\bin\java.exe -fullversion 2>&1)
$javaVersion = "$javaVersion".Trim("java full version """).TrimEnd("-b13")

The full output is: java full version "1.8.0_121-b13"

Please edit the question and add the original version string that is to be parsed. — vonPryz
– vonPryz, Commented Dec 4, 2018 at 8:59

Manuel Batsching · Accepted Answer · 2018-12-04 09:13:09Z

1

TrimEnd() works a little different, than you might expect:

'1.8.0_191-b12'.TrimEnd('-b12')

results in: 1.8.0_19 and so does:

'1.8.0_191-b12'.TrimEnd('1-b2')

The reason is, that TrimEnd() removes a trailing set of characters, not a substring. So .TrimEnd('-b12') means: remove all occurrences of any character of the set '-b12' from the end of the string. And that includes the last '1' before the '-'.

A better solution in your case would be -replace:

'java full version "1.8.0_191-b12"' -replace 'java full version "(.+)-b\d+"','$1'

answered Dec 4, 2018 at 9:13

Manuel Batsching

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2 Comments

Glen Hams Over a year ago

Can you maybe explain what the "(.+)-b\d+"','$1' code does? so i can use it in the future?

Manuel Batsching Over a year ago

It is called a regular expression. For details on this particular pattern, see here: regex101.com/r/Z3UZqT/1 .

Ansgar Wiechers · Accepted Answer · 2018-12-04 09:20:32Z

0

Use a regular expression for matching and extracting the version number:

$javaVersion = if (& java -fullversion 2>&1) -match '\d+\.\d+\.\d+_\d+') {
    $matches[0]
}

or

$javaVersion = (& java -fullversion 2>&1 | Select-String '\d+\.\d+\.\d+_\d+').Matches[0].Groups[0].Value

answered Dec 4, 2018 at 9:20

Ansgar Wiechers

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1 Comment

Glen Hams Over a year ago

Thanks for your comment, this worked to ;) i have used the -replace for now

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