[lang-ref] ソースファイル名(パス)を取得 ( powershell )

It 'should get source file path' {
    # Get-Command .. ScriptBlock.File

    $output = powershell -ExecutionPolicy Bypass -NoProfile -File .\example\GetSourceFilePath.ps1

    $output.Trim() | Should -Be (Join-Path (Get-Location) 'example\GetSourceFilePath.ps1')
}