Powershell:捕获后台作业的进程 ID | 捕获后台作业的进程

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问题描述

我想启动一个后台作业并将它的进程 ID 捕获到一个 .pid 文件中.我能够使用 Start-Process 来做到这一点，如下所示:

I want to start a background job and capture it's process id into a .pid file. I was able to do it with the Start-Process as follows:

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

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

Now, I want to wrap Start-Process with Start-Job, to run it in the background, like this:

$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:

Unfortunatelly, this doesn't work and Receive-Job gives me the following error:

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.

Looks like it's something wrong with the $_ variable. Maybe it gets overwritten by the Start-Job.

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推荐答案

那是因为使用双引号时变量被扩展了.如果要保留$_，则需要使用单引号.

That is because the variable is being expanded when using double quotes. If you want to keep the $_, then you need to use single quotes.

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

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