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I want to call another powershell script with external argument. I try this but return error. anyone can help please

$direct = "D:\Learn"
Start-Process powershell.exe -WindowStyle Minimized ".\Testing.exe" -Path $direct 


Testing.exe
Param(
  [parameter(mandatory=$true)][string]$Loc
)

Get-Content $Loc\API.txt
Pause
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  • hey SBR, let me know if the below post helps Commented Mar 4, 2020 at 8:52

3 Answers 3

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The Start-Process cmdlet has a -AgumentList parameter:

$direct = "D:\Learn"
Start-Process powershell.exe -WindowStyle Minimized ".\Testing.exe" -ArgumentList "-Path $direct"
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6 Comments

it return this eror Start-Process : A positional parameter cannot be found that accepts argument '.\Testing.exe'.@Martin
Please clarify which process you want to start with which arguments
I want to start the process of Testing.exe with argument $direct
I try to change Start-Process with & the argument can work, but got problem, the main script can not continue to running before Testing.exe finish
Why you start powershell first? Just try this: Start-Process".\Testing.exe" -WindowStyle Minimized -ArgumentList "-Path $direct"
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1

If you want to just run a File with some arguments

$filepath = ".\Testing.exe"
$direct = "D:\Learn"
Start-Process -FilePath $filepath -ArgumentList $direct -Wait -NoNewWindow

1 Comment

hi @Clint if you don't mind please check out my other question. Thank you so much stackoverflow.com/q/60598457/11076819
1

I think they all were arguments of powershell.exe. The whole argument can be wrapped in one double quote in argumentlist.

Start-Process "powershell.exe" -ArgumentList "-windowstyle minimized '.\testing.exe' -path $direct"

Or even it can be done without start-process:

& "powershell.exe -windowstyle minimized '.\testing.exe' -path $direct"

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