How can I use a variable filled with a path to navigate or execute in PowerShell?

Question

I'm currently writing a PowerShell script for the first time. I just want to find out which Java versions I have running on my Windows machines. It is searching for all java.exe instants, writes it in a file and then it should execute each line and write the output in a file. But I can't use my variable $command to execute anything. So the for loop is just running one time, because of an error and then quits.

#declaration
$file = ".\text.txt"
$i = 3
$out = ".\output.txt"

Get-ChildItem -Path C:\ -Filter java.exe -Recurse -ErrorAction SilentlyContinue -Force |
    Select-Object Directory >> $file
$count = Get-Content $file | Measure-Object -Line

#remove spaces and append "java.exe"
(Get-Content $file | Foreach {$_.TrimEnd()}) | Set-Content $file
(Get-Content $file | foreach {$_ + "\java.exe' -version"}) | Set-Content $file
(Get-Content $file).Replace("C:", "'C:") | Set-Content $file

#remove last 2 lines of the file
$count = $count.Lines - 2

#execute the stored paths

$i = 3
for ($i=3; $i -le $count; $i++) {
    $command = Get-Content $file | Select -Index $i;
    $command >> $out;
    & $command 2> $out;   # <----------------This line wont work
    echo "_____________________" >> $out;
}

So basically you want to run C:\paht\java.exe -version on each installation you have to find the version of java installed? — DarkLite1
– DarkLite1, Commented Sep 13, 2019 at 13:18
Using your code, you would need to run & exe argument on a line not & somestring. If you are adding the single quotes to make this part of the code work, then that is not necessary. For yours to work as is, you would need to do something like this & ($command -split " (-version)" -replace "'")[0] ($command -split " (-version)" -replace "'")[1]. The reason is because the first set of double quotes after & is interpreted as the command rather than the command and arguments. — AdminOfThings
– AdminOfThings, Commented Sep 13, 2019 at 13:21

Theo · Accepted Answer · 2019-09-13 13:29:57Z

1

Although I'm not sure what your desired output should look like, but below creates an array of PSObjects that you can output on screen or write to CSV file.

$result = Get-ChildItem -Path C:\ -Filter java.exe -File -Recurse -ErrorAction SilentlyContinue -Force | ForEach-Object {
    $pinfo = New-Object System.Diagnostics.ProcessStartInfo
    $pinfo.FileName = $_.FullName
    $pinfo.Arguments = "-version"
    $pinfo.RedirectStandardError = $true
    $pinfo.RedirectStandardOutput = $true
    $pinfo.UseShellExecute = $false
    $process = New-Object System.Diagnostics.Process
    $process.StartInfo = $pinfo
    $process.Start() | Out-Null
    $process.WaitForExit()
    $output = $process.StandardOutput.ReadToEnd()
    if (!$output) { $output = $process.StandardError.ReadToEnd() }
    [PsCustomObject]@{
        'FileName'    = $_.FullName
        'VersionInfo' = $output
    }
}
# output on screen
$result | Format-List

# output to CSV file (VersionInfo field uses multiple lines)
$result | Export-Csv -Path 'D:\javaversions.csv' -UseCulture -NoTypeInformation

Result on screen wil look something like:

FileName    : C:\Program Files\Java\jdk1.8.0_31\bin\java.exe
VersionInfo : java version "1.8.0_31"
              Java(TM) SE Runtime Environment (build 1.8.0_31-b13)
              Java HotSpot(TM) 64-Bit Server VM (build 25.31-b07, mixed mode)


FileName    : C:\Program Files\Java\jdk1.8.0_31\jre\bin\java.exe
VersionInfo : java version "1.8.0_31"
              Java(TM) SE Runtime Environment (build 1.8.0_31-b13)
              Java HotSpot(TM) 64-Bit Server VM (build 25.31-b07, mixed mode)


FileName    : C:\Program Files\Java\jre1.8.0_221\bin\java.exe
VersionInfo : java version "1.8.0_221"
              Java(TM) SE Runtime Environment (build 1.8.0_221-b11)
              Java HotSpot(TM) 64-Bit Server VM (build 25.221-b11, mixed mode)

answered Sep 13, 2019 at 13:29

Theo

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3 Comments

Bastian Jaeger Over a year ago

Thank you for your help. But there is a problem. Your script just writes the java-version into my file. But not the further information like "Java(TM) SE Runtime Environment (build 1.8.0_221-b11)". Fixed it with removing the parameters for Export-Csv

Theo Over a year ago

@BastianJaeger Nope, it's all there. As i said, the result is multi-lined. When opened in Excel you'll have to go into the cell properties and turn TextWrapping on.

Bastian Jaeger Over a year ago

Ah, my bad. Thank you!

Ansgar Wiechers · Accepted Answer · 2019-09-16 13:17:23Z

0

Your code doesn't work because you're writing entire commandlines to the intermediate file and then try to invoke the command strings from this file via the call operator. That doesn't work because a command "'C:\path\to\java.exe' -version" doesn't exist. You should be getting an error like this:

PS C:\> $command = "'C:\path\to\java.exe' -version"
PS C:\> & $command
& : The term ''C:\path\to\java.exe' -version' is not recognized as the name of
a cmdlet, function, script file, or operable program. Check the spelling of the
name, or if a path was included, verify that the path is correct and try again.
At line:1 char:3
+ & $command
+   ~~~~~~~~
    + CategoryInfo          : ObjectNotFound: ('C:\path\to\java.exe' -version:String) [], CommandNotFoundException
    + FullyQualifiedErrorId : CommandNotFoundException

The simplest way to make your code work is to just invoke the executables directly:

Get-ChildItem -Path 'C:\' -Filter 'java.exe' -Recurse -EA SilentlyContinue -Force |
    ForEach-Object {
        $_.FullName + ' -version'
        & $_.FullName -version 2>&1
        '__________________'
    } |
    Set-Content $out

If you want to stick with the approach using an intermediate file you need to prepend the commandline with the call operator, convert it to a scriptblock, and then invoke that.

Get-Content $file | ForEach-Object {
    $command
    & ([Scriptblock]::Create("& $command")) 2>&1
    '__________________'
} | Out-File $out

edited Sep 16, 2019 at 13:17

answered Sep 13, 2019 at 13:47

Ansgar Wiechers

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4 Comments

Bastian Jaeger Over a year ago

I like your first code pretty much. But it doesnt work the way i want. It just writes the path in the $out-file with -version sticked to it. But not the actual output of the command.

Ansgar Wiechers Over a year ago

@BastianJaeger What is the output of the Get-ChildItem? Also, you shouldn't accept an answer when your question is not actually answered yet.

Bastian Jaeger Over a year ago

It answered my question and its all i need, the output of the console ist enough for me. I just noticed it today that the file is not the output of the console. The Ouput gives me this: Mode LastWriteTime Length Name ---- ------------- ---- -a---- 21.08.2017 15:51 191040 java.exe

Ansgar Wiechers Over a year ago

@BastianJaeger Ah, the stupid Java executable writes the version information to STDERR. <_< See updated answer.

DarkLite1 · Accepted Answer · 2019-09-13 13:44:06Z

An alternative solution to @Theo is using a temporary file like so:

$GetParams = @{
    Path        = 'C:\'
    Filter      = 'java.exe'
    Recurse     = $true
    Force       = $true
    File        = $true
    ErrorAction = 'SilentlyContinue'
}
$JavaExeFiles = Get-ChildItem @GetParams | select -First 1

$StartParams = @{
     ArgumentList          = '-version'
     NoNewWindow           = $true
     RedirectStandardError = '.\javaver.txt'
}

$JaveExeDetails = foreach ($JaveExe in $JavaExeFiles) {
    Remove-Item -Path $StartParams.RedirectStandardError -EA Ignore

    Start-Process @StartParams -FilePath $JaveExe.FullName -Wait

    [PSCustomObject]@{
        File    = $JaveExe.FullName
        Version = Get-Content -Path $StartParams.RedirectStandardError
    }
}

# For viewing in the console
$JaveExeDetails | fl *

# For export to a file
$JaveExeDetails | fl * | Out-File -FilePath ./JavaVersions.txt

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