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How to pass arguments in Invoke-Command in PowerShell?
To pass the argument in the Invoke-command, you need to use -ArgumentList parameter. For example, we need to get the notepad process information on the remote server.
Example
Invoke-Command -ComputerName Test1-Win2k12 -
ScriptBlock{param($proc) Get-Process -Name $proc} -
ArgumentList "Notepad"Output
Handles NPM(K) PM(K) WS(K) CPU(s) Id SI ProcessName PSComputerName ------- ------ ----- ----- ------ -- -- ----------- -------------- 67 8 1348 7488 0.08 104 notepad Test1-Win2k12
In the above example, we are passing "Notepad" name as the argument to the command and the same has been caught by the $proc variable inside Param().
If you have the multiple, check the below command to pass the multiple parameters.
Example
Invoke-Command -ComputerName Test1-Win2k12 -ScriptBlock{param($proc,$proc2) Get-Process -Name $proc,$proc2} -ArgumentList "Notepad","Calc"Output
Handles NPM(K) PM(K) WS(K) CPU(s) Id SI ProcessName PSComputerName ------- ------ ----- ----- ------ -- -- ----------- -------------- 96 20 5980 11392 0.19 288 calc Test1-Win2k12 67 8 1344 7556 0.08 104 notepad Test1-Win2k12
Same can be achieved with the Session variable.
Example
$sess = New-PSSession -ComputerName Test1-win2k12
Invoke-Command -Session $sess -ScriptBlock{param($proc) Get-Process $proc} -ArgumentList "Notepad"Output
Handles NPM(K) PM(K) WS(K) CPU(s) Id SI ProcessName PSComputerName ------- ------ ----- ----- ------ -- -- ----------- -------------- 67 8 1348 7488 0.08 104 notepad Test1-Win2k12
Updated on: 11-Nov-2020
11K+ Views
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