在scriptblock中更改输入

Question

我有一个scriptblock脚本：

$RoboArgs = @{
  Source = '01'
  Target = '02'
  ExtraArgs = '/e', '/purge'
}
Write-Host @RoboArgs
Start-ThreadJob -InputObject $RoboArgs -ScriptBlock {
  . robocopy_invoke.ps1
  Write-Host @input
} | Receive-Job -Wait -AutoRemoveJob

我想使用input参数(Invoke-Robocopy @RoboArgs)调用robocopy_invoke.ps1模块中定义的函数，但是输入参数的内容一旦进入scriptblock就会改变。这是输出：

-Target: 02 -ExtraArgs: /e /purge -Source: 01
System.Management.Automation.Runspaces.PipelineReader`1+<GetReadEnumerator>d__20[System.Object]

为什么两个Write-Host调用的输出不同？我怎样才能把第二个和第一个一样？

Answer 1

您可以使用'-InputObject‘将对象输送到作业中(使用$input检索)。所以Start-ThreadJob -InputObject $RoboArgs -ScriptBlock {}等同于$RoboArgs | Start-ThreadJob -ScriptBlock {}

相反，您需要的是'-ArgumentList‘(使用$args检索)：

$RoboArgs = @{
  Source = '01'
  Target = '02'
  ExtraArgs = '/e', '/purge'
}
$RoboArgs

Start-ThreadJob -ArgumentList $RoboArgs -ScriptBlock {
  $args
} | Receive-Job -Wait -AutoRemoveJob

示例2(解包$args数组)

$RoboArgs = @{
  Source = '01'
  Target = '02'
  ExtraArgs = '/e', '/purge'
}
Write-Host @RoboArgs

Start-ThreadJob -ArgumentList $RoboArgs -ScriptBlock {
  $robo = $args[0]
  Write-Host @robo
} | Receive-Job -Wait -AutoRemoveJob

示例3(显式定义参数而不是使用$args)

$RoboArgs = @{
    Source    = '01'
    Target    = '02'
    ExtraArgs = '/e', '/purge'
}
Write-Host @RoboArgs


Start-ThreadJob -ArgumentList $RoboArgs -ScriptBlock {
    param ( $myRoboArgs )
    Write-Host @myRoboArgs
} | Receive-Job -Wait -AutoRemoveJob