Specify NON-hardcoded path and filename in a scriptblock?

Question

Ok I feel spechul for even asking, but I have tried several iterations of this and nothing has worked except hard coding the scriptname in the Scriptblock statement, which is unacceptable.

Here is the code that works, hard coded, unacceptable....

$Scriptblock = { C:\Scripts\Path1\ScriptName.ps1 -arguement0 $args[0] -arguement1 $args[1] }

Start-Job -ScriptBlock $Scriptblock -ArgumentList $argue0, $argue1 | Out-Null

Ive tried this, and it doesn't work...

$loc = (Get-Location).Path

Set-Location -Path  $loc

And this....

$rootpath = $MyInvocation.MyCommand.Path.Substring(0, ($MyInvocation.MyCommand.Path).LastIndexOf("\"))

Set-Location -Path  $rootpath

And this....

$rootpath = $MyInvocation.MyCommand.Path.Substring(0, ($MyInvocation.MyCommand.Path).LastIndexOf("\"))

$scriptFilename = $([string]::Format("{0}\ScriptName.ps1", $rootpath))

$sb = $([string]::Format("{0} -arguement0 $args[0] -arguement1 $args[1]", $scriptFilename))

$Scriptblock = { $sb }

Start-Job -ScriptBlock $Scriptblock -ArgumentList $argue0, $argue1 | Out-Null

Nothing else has worked except the first code above with hardcoded path and script name - I know it has to be something stupid I am missing - help me fix stoopid please! ;-)

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Answer 1

In your last example, this line:

$ScriptBlock = { $sb }

simply creates a scriptblock with a string inside it. Change it to:

$ScriptBlock = [scriptblock]::Create($sb)

Specify NON-hardcoded path and filename in a scriptblock?

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