为什么传递给脚本块的参数不起作用？

Question

完全披露：我的问题可能是基于对 Xen Desktop 的 Citrix PowerShell 模块的不完全理解。

我有以下脚本块。它在循环中被调用，对列表中的每个 VM 调用一次。我正在使用 PowerShell Jobs 因为我想保持 UI 线程自由更新 UI 而作业运行.

代码"A"

$j = Start-Job -Name $jobName -ScriptBlock {
    param($url, $uuid, $cred, $snapshotName)
    $ErrorActionPreference = "Stop" 
    try
    {
        $error.clear()
        $xSS = $cred | Connect-XenServer -url $url -NoWarnCertificates -SetDefaultSession -PassThru; 
        $vm = (Get-XenVM -SessionOpaqueRef $xss.opaque_ref -uuid $uuid)
        #Create snapshot
        Invoke-XenVM -Async -SessionOpaqueRef $xss.opaque_ref -VM $vm -XenAction Snapshot -NewName $snapshotName
        return "OK"
    }
    catch
    {
        return ("ERROR: "+$error)
    }
} -ArgumentList $global:configFileVmMetaData.poolUrl, $xenVm.key, $global:cred, $snapshotName

代码 "A" 工作正常，但它需要的时间比必要的时间更长，因为我每次调用脚本块时都在执行 Connect-XenServer cmdlet。

因此，我尝试在循环外调用 Connect-XenServer 一次并传入会话变量，如下面 代码 "B" 中所示。结果是错误 Could not find open sessions to any XenServers 被抛入脚本块内。我假设 $xss 会话变量在传递到脚本块时以某种方式被破坏。

知道为什么 $xss 会话变量会被打吗？

代码"B"

$xSS = $cred | Connect-XenServer -url $global:configFileVmMetaData.poolUrl -NoWarnCertificates -SetDefaultSession -PassThru; 

loop
{

    $j = Start-Job -Name $jobName -ScriptBlock {
        param($xss, $uuid, $snapshotName)
        $ErrorActionPreference = "Stop" 
        try
        {
            $error.clear()
            $vm = (Get-XenVM -SessionOpaqueRef $xss.opaque_ref -uuid $uuid)
            #Create snapshot
            Invoke-XenVM -Async -SessionOpaqueRef $xss.opaque_ref -VM $vm -XenAction Snapshot -NewName $snapshotName
            return "OK"
        }
        catch
        {
            return ("ERROR: "+$error)
        }
    } -ArgumentList $xss, $xenVm.key, $snapshotName

}

Additional examples inspired by the Robert Cotterman answer

In all cases, I still get the Could not find open sessions to any XenServers error

FYI - Using PowerShell 5.1

示例使用 $using。变量内容按预期传入和传出

cls

$aLocal = "AAA"
$bLocal = "BBB"

$j = Start-Job -Name "TestJob" -ScriptBlock {
    return ($using:aLocal + " *** " + $using:bLocal)
    }

while ($true)
{
    $g = get-job -name "TestJob"

    write-host ("get-job " + $g.Name + " is " + $g.State)

    if ($g.State -ne "Running")
    {
        break
    }

    start-sleep -Seconds 1
}

write-host ("receive-Job='" + (receive-Job -Name "TestJob") +"'")

$g = get-Job -Name "TestJob"
Write-Host ("get-Job "+$g.name + " " + $g.state + " " + $g.HasMoreData + " " + $g.id)

if($g)
{
    Remove-Job -Name "TestJob"
}

输出

get-job TestJob is Running
get-job TestJob is Completed
receive-Job='AAA *** BBB'
get-Job TestJob Completed False 45
Remove-Job

使用参数的示例。变量内容按预期传入和传出

cls

$aLocal = "AAA"
$bLocal = "BBB"

$j = Start-Job -Name "TestJob" -ScriptBlock {
    return ($args[0] + " *** " + $args[1])
    } -ArgumentList ($aLocal, $bLocal)

while ($true)
{
    $g = get-job -name "TestJob"

    write-host ("get-job " + $g.Name + " is " + $g.State)

    if ($g.State -ne "Running")
    {
        break
    }

    start-sleep -Seconds 1
}

write-host ("receive-Job='" + (receive-Job -Name "TestJob") +"'")

$g = get-Job -Name "TestJob"
Write-Host ("get-Job "+$g.name + " " + $g.state + " " + $g.HasMoreData + " " + $g.id)

if($g)
{
    Remove-Job -Name "TestJob"
}

输出

get-job TestJob is Running
get-job TestJob is Completed
receive-Job='AAA *** BBB'
get-Job TestJob Completed False 49

使用命名参数的示例。变量内容按预期传入和传出

cls

$aLocal = "AAA"
$bLocal = "BBB"

$j = Start-Job -Name "TestJob" -ScriptBlock {
    param($a, $b)
    return ($a + " *** " + $b)
    } -ArgumentList ($aLocal, $bLocal)

while ($true)
{
    $g = get-job -name "TestJob"

    write-host ("get-job " + $g.Name + " is " + $g.State)

    if ($g.State -ne "Running")
    {
        break
    }

    start-sleep -Seconds 1
}

write-host ("receive-Job='" + (receive-Job -Name "TestJob") +"'")

$g = get-Job -Name "TestJob"
Write-Host ("get-Job "+$g.name + " " + $g.state + " " + $g.HasMoreData + " " + $g.id)

if($g)
{
    Remove-Job -Name "TestJob"
}

输出

get-job TestJob is Running
get-job TestJob is Completed
receive-Job='AAA *** BBB'
get-Job TestJob Completed False 55

Answer 1

作业和调用命令要求您指定您正在使用变量。只需将变量从

$variable

到

$using:variable

内部变量不需要这个。但是从父脚本调用变量就可以了。

或者，由于您将 $xss 作为参数传递，因此您不会使用 $xss 来调用它，而是

$args[0]

因为这是你的第一个论点。第二个是 $args[1] 等等... 原因是因为整个 xss 变量被作为参数打印，并且没有在作业中命名。它被命名为 $args 并在第一个位置 (0) 中占有一席之地。

我更喜欢 $using:variable，因为它可以减少混淆

Answer 2

我只是不认为它是那样工作的。与 Xen 服务器的连接是为您的 PowerShell 会话创建的，有关该连接的信息收集在 $xss 变量中。每个作业都运行自己的 PowerShell 会话。所以只是将$xss传递给作业是不一样的，它只是对连接细节的描述。

为什么传递给脚本块的参数不起作用？

Why does an argument passed into a script block fail to work?

powershell

xen

scope

scriptblock