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如何在PowerShell中从指定目录运行命令并等待其完成后再继续?

powershell

1

VenerableAgents · 技术社区 · 8 年前

工作:

$job = Start-Job `
    -InitializationScript { Set-Location C:\MyDirectory\ }  `
    -ScriptBlock { C:\MyDirectory\MyCmdLineExecutable.exe }
Wait-Job $job
Receive-Job $job

$Path = "C:\MyDirectory\"
$ExePath = $path+"MyCmdLineExecutable.exe"
$job = Start-Job `
    -InitializationScript { Set-Location $Path }  `
    -ScriptBlock { $ExePath }
Wait-Job $job
Receive-Job $job

以下是错误:

Set-Location : Cannot process argument because the value of argument "path" is null. Change the value of argument "path" to a non-null value.
At line:1 char:2
+  Set-Location $Path
+  ~~~~~~~~~~~~~~~~~~
    + CategoryInfo          : InvalidArgument: (:) [Set-Location], PSArgumentNullException
    + FullyQualifiedErrorId : ArgumentNull,Microsoft.PowerShell.Commands.SetLocationCommand


Id     Name            PSJobTypeName   State         HasMoreData     Location             Command                  
--     ----            -------------   -----         -----------     --------             -------                  
49     Job49           BackgroundJob   Failed        False           localhost             $ExePath                
Running startup script threw an error: Cannot process argument because the value of argument "path" is null. Change the value of argument "path" to a non-null value..
    + CategoryInfo          : OpenError: (localhost:String) [], RemoteException
    + FullyQualifiedErrorId : PSSessionStateBroken

2 回复 | 直到 8 年前

1

2

JosefZ 8 年前

组合来自的信息 Start-Job 带有的文档 About_Scopes 文章,我确信你需要使用 -InputObject 参数:

指定命令的输入。输入包含物体。
在 脚本块 参数,使用 $Input automatic variable 表示输入对象。

$Path = "C:\MyDirectory\"
$ExePath = $path+"MyCmdLineExecutable.exe"

$job = Start-Job -InputObject @( $Path, $ExePath) `
    -InitializationScript { <# $Input variable isn't defined here #> }  `
    -ScriptBlock { 
        $aux = $Input.GetEnumerator()
        Set-Location $aux[0]
        & $aux[1] }
Wait-Job $job
Receive-Job $job

顺便说一句,收件人跑存储在变量中并由字符串表示的命令,使用 & Call operator

$ExePath        ### output only
& $ExePath      ### invocation

2

0

Bill_Stewart 8 年前

我想你想要 Start-Process 和 -Wait -WorkingDirectory 参数指定新进程的工作目录。例子:

Start-Process notepad -WorkingDirectory "C:\Program Files" -Wait
Write-Host "Finished"

运行此脚本时,记事本将打开,但脚本在关闭之前不会继续。当您关闭记事本时 Write-Host 线路运行。