how to pass a variable in "get-content command " in powershell

Question

I have to get the particular file from one remote desktop to local machine or another server.How to pass a variable in get-content to fetch the file from remote desktop connection?

I store the file path as variable and try to pass it in get-content.

Invoke-Command -Computername $Server -ScriptBlock{get-content -path $file }
Invoke-Command -Computername $Server -ScriptBlock{get-content -path ${file} }

$file="C:\Users\Documents\new DB_connection\\log2.txt"

 $Server="servername"

 $answer=  Invoke-Command -Computername $Server -ScriptBlock{get-content -path $file } 
write-output $answer

Cannot bind argument to parameter 'Path' because it is null. + CategoryInfo : InvalidData: (:) [Get-Content], ParameterBindingValidationException + FullyQualifiedErrorId : ParameterArgumentValidationErrorNullNotAllowed,Microsoft.PowerShell.Commands.GetContentCommand

i tried this code "Invoke-Command -Computername $Server -ScriptBlock{get-content -path $file } but it again showing a same error — Divya Mani
– Divya Mani, Commented Aug 26, 2019 at 5:27
Once you assign a value to $file, then simply use $using:file inside of the script block. You can’t pass it into the script block until it is defined — AdminOfThings
– AdminOfThings, Commented Aug 26, 2019 at 5:57

codewario · Accepted Answer · 2019-08-26 19:41:51Z

You can copy the file over a PSSession instead of using Invoke-Command. To copy from a remote server to a local path or different remote server:

 $session = New-PSSession -ComputerName server.domain.tld
 Copy-Item -Source $originatingServerFilePath -Destination $localOrUNCFilePath -FromSession $session

If you need to copy a local file to a destination server:

Copy-Item -Source $localFilePath -Destination $destinationServerFilePath -ToSession $session

This has the benefit of not double-hopping, though the server you run the command on will need to have access to any remote file paths. If you need to copy a file from one server to another server, but the destination server doesn't have the filepath exposed as a shared folder (or you don't have access to it) you cannot specify -ToSession and -FromSession at the same time, so you'll have to copy the file locally and use two sessions like so:

$sourceSession = New-PSSession -ComputerName source.domain.tld
$destinationSession = New-PSSession -ComputerName destination.domain.tld

# Copy the remote file(s) from the source session to a local path first
Copy-Item -Source $sourceSessionFilePath -Destination $localTempFilePath -FromSession $sourceSession

# Copy the new local files to the destination session from your local path
Copy-Item -Source $localTempFilePath -Destination $destinationSessionFilePath -ToSession $destinationSession

Mahmood Shehab · Accepted Answer · 2019-08-26 05:37:48Z

0

Try to define the variable inside the ScriptBlock

$Server="servername"
$answer=  Invoke-Command -Computername $Server -ScriptBlock{
$file="C:\Users\Documents\new DB_connection\\log2.txt";
get-content -path $file} 
write-output $answer

answered Aug 26, 2019 at 5:37

Mahmood Shehab

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