Powershell start-job -scriptblock 无法识别同一文件中定义的函数？

Question

我有以下代码。

function createZip
{
Param ([String]$source, [String]$zipfile)
Process { echo "zip: $source`n     --> $zipfile" }
}

try {
    Start-Job -ScriptBlock { createZip "abd" "acd" } 
}
catch {
    $_ | fl * -force
}
Get-Job | Wait-Job 
Get-Job | receive-job 
Get-Job | Remove-Job 

但是，该脚本返回以下错误。

Id              Name            State      HasMoreData     Location             Command                  
--              ----            -----      -----------     --------             -------                  
309             Job309          Running    True            localhost            createZip "a...
309             Job309          Failed     False           localhost            createZip "a...
Receive-Job : The term 'createZip' is not recognized as the name of a cmdlet, function, script file, or operable program. Check the spelling of the name, or if a path was included, verify that the path is correct and try again.
At line:17 char:22
+ Get-Job | receive-job <<<<  
    + CategoryInfo          : ObjectNotFound: (function:createZip:String) [Receive-Job], CommandNotFoundException
    + FullyQualifiedErrorId : CommandNotFoundException

在start-job的脚本 block 内似乎无法识别函数名称。我也尝试过function:createZip。

Answer 1

Start-Job 实际上会启动另一个没有 createZip 函数的 PowerShell.exe 实例。您需要将其全部包含在脚本 block 中:

$createZip = {
    param ([String]$source, [String]$zipfile)
    Process { echo "zip: $source`n     --> $zipfile" }
}

Start-Job -ScriptBlock $createZip  -ArgumentList "abd", "acd"

从后台作业返回错误消息的示例:

$createZip = {
    param ([String] $source, [String] $zipfile)

    $output = & zip.exe $source $zipfile 2>&1
    if ($LASTEXITCODE -ne 0) {
        throw $output
    }
}

$job = Start-Job -ScriptBlock $createZip -ArgumentList "abd", "acd"
$job | Wait-Job | Receive-Job

另请注意，通过使用抛出，作业对象State将为“失败”，因此您只能获取失败的作业:Get-Job -状态失败。