Powershell - 如果一个进程没有运行，启动它 [英] Powershell - if a process is not running, start it

问题描述

请菜鸟帮忙.我正在尝试编写一个脚本来检查进程是否正在运行，如果没有，则启动它.如果进程正在运行，它应该什么都不做.到目前为止，我已经提出了以下内容，但它正在启动该过程的一个新实例，无论它是否已经在运行.任何帮助表示赞赏.

Noob help please. I'm trying to write a script that will check if a process is running, and if not, start it. If the process is running, it should do nothing. I've come up with the following so far but it is starting a new instance of the process regardless of whether it was running already. Any help is appreciated.

$Prog = "C:\utilities\prog.exe"
$Running = Get-Process prog -ErrorAction SilentlyContinue
$Start = ([wmiclass]"win32_process").Create($Prog)
if($Running -eq $null)
{$Start}
else
{}

推荐答案

首先，这是您的代码中的错误.在你的代码中，进程是在你评估你的程序是否已经运行之前创建的

First of all, here's is what is wrong in your code. In your code, the process is created before you evaluate whether your program is already running

$Prog = "C:\utilities\prog.exe"
$Running = Get-Process prog -ErrorAction SilentlyContinue
$Start = ([wmiclass]"win32_process").Create($Prog) # the process is created on this line
if($Running -eq $null) # evaluating if the program is running
{$Start}

可以通过将其包装在 {}(脚本块)中来创建应在代码中进一步评估的代码块:

It is possible to create a block of code that should be evaluated further in your code by wrapping it in {} (a scriptblock):

$Prog = "C:\utilities\prog.exe"
$Running = Get-Process prog -ErrorAction SilentlyContinue
$Start = {([wmiclass]"win32_process").Create($Prog)} 
if($Running -eq $null) # evaluating if the program is running
{& $Start} # the process is created on this line

但是，如果您正在寻找一种短的单线来解决您的问题:

However, if you're looking for a short one-liner to solve your problem:

if (! (ps | ? {$_.path -eq $prog})) {& $prog}

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