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，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 的脚本块中似乎无法识别函数名称。我也试过function:createZip。

Answer 1

Start-Job 实际上启动了另一个没有 createZip 函数的 PowerShell.exe 实例。您需要将其全部包含在脚本块中：

$createZip = {
    param ([String]$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 throw，作业对象 State 将“失败”，因此您只能获取失败的作业：Get-Job -State Failed。

Answer 2

如果您对使用 start-job 和 receive-job 还很陌生，并且想更轻松地调试您的功能，请尝试以下表单：

   $createZip = { 
   function createzipFunc {
     param ([String]$source, [String]$zipfile)
     Process { echo "zip: $source`n     --> $zipfile" }
     }
     #other funcs and constants here if wanted...
   }
   # some secret sauce, this defines the function(s) for us as locals
   invoke-expression $createzip

   #now test it out without any job plumbing to confuse you
   createzipFunc "abd" "acd"

   # once debugged, unfortunately this makes calling the function from the job  
   # slightly harder, but here goes...
   Start-Job -initializationScript $createZip -scriptblock {param($a,$b) `
createzipFunc $a $b } -ArgumentList "abc","def"

我没有像你那样将我的函数定义为一个简单的过滤器，这并没有让一切变得更简单，但我之所以这样做是因为我想最终将一些函数传递给我的 Job。

很抱歉挖掘了这个线程，但它也解决了我的问题，而且非常优雅。所以我只需要添加一点点我在调试我的 powershell 工作时写的酱汁。