Powershell start-job -scriptblock 无法识别同一文件中定义的函数? [英] Powershell start-job -scriptblock cannot recognize the function defined in the same file?
问题描述
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.
However, the script returns the following error.
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
.
It seems the function name cannot be recognized inside the script block of start-job
. I tried function:createZip
too.
推荐答案
Start-Job
实际上启动了另一个 PowerShell.exe 实例,它没有您的 createZip 函数.您需要将其全部包含在一个脚本块中:
Start-Job
actually spins up another instance of PowerShell.exe which doesn't have your createZip function. You need to include it all in a script block:
$createZip = {
param ([String]$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:
An example returning an error message from the background job:
$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",因此您只能获得失败的作业:Get-Job -状态失败
.
Also note that by using a throw
the job object State
will be "Failed" so you can get only the jobs which failed: Get-Job -State Failed
.
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