如何通过没有辅助文件的批处理文件运行Powershell脚本

Question

I have two files, one is called Startmenu.bat and the second one is test.ps1

Startmenu.bat:

@echo ON
PowerShell -NoProfile -ExecutionPolicy Bypass -Command "& {Start-Process PowerShell -ArgumentList '-NoProfile -ExecutionPolicy Bypass -File "C:\test.ps1"' -Verb RunAs}";

test.ps1:

Get-AppXPackage -AllUsers | Foreach {Add-AppxPackage -DisableDevelopmentMode -Register "$($_.InstallLocation)\AppXManifest.xml"}

These two files are working properly if I run Startmenu.bat and test.ps1 is placed to C:\\

But my question if can I make just one file? Can I somehow include the Powershell script inside first file, I mean Startmenu.bat ? I want to make just a single file and I really don't know how.

My Start menu in Windows 10 is broken and I have to run Powershell command to access Start Menu (at every boot).

Answer 1

The answer is in the question! You're running PowerShell first with the -Command parameter, which runs PowerShell commands directly; as it happens you're using this to open PowerShell again (with the RunAs verb so that it's elevated). You could simply use -Command again instead of using -File .

Another way is to use the -EncodedCommand parameter and Base64 encode your PowerShell commands.

See the very end of the help when you run powershell.exe -help :

# To use the -EncodedCommand parameter:
$command = 'dir "c:\program files" '
$bytes = [System.Text.Encoding]::Unicode.GetBytes($command)
$encodedCommand = [Convert]::ToBase64String($bytes)
powershell.exe -encodedCommand $encodedCommand