Question

我想启动后台作业并将其进程ID捕获到.pid文件中。我能够使用Start-Process执行以下操作：

Start-Process C:\process.bat -passthru | foreach { $_.Id } > start.pid

现在，我想用Start-Job包装Start-Process，在后台运行它，如下所示：

$command = "Start-Process C:\process.bat -passthru | foreach { $_.Id }"
$scriptblock = [Scriptblock]::Create($command)
Start-Job -ScriptBlock $scriptblock

不幸的是，这不起作用，Receive-Job给了我以下错误：

The term '.Id' 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.
+ CategoryInfo          : ObjectNotFound: (.Id:String) [], CommandNotFoundException
+ FullyQualifiedErrorId : CommandNotFoundException
+ PSComputerName        : localhost

看起来$ _变量有问题。也许它会被Start-Job覆盖。

任何线索都非常欢迎！

Answer 1

这是因为在使用双引号时正在扩展变量。如果你想保留$ _，那么你需要使用单引号。

$command = 'Start-Process C:\process.bat -passthru | foreach { $_.Id }'
$scriptblock = [Scriptblock]::Create($command)
Start-Job -ScriptBlock $scriptblock

Powershell：捕获后台作业的进程ID

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