PowerShell - 启动进程和命令行开关 [英] PowerShell - Start-Process and Cmdline Switches

问题描述

我可以正常运行:

$msbuild = "C:WINDOWSMicrosoft.NETFrameworkv3.5MSBuild.exe" 
start-process $msbuild -wait

但是当我运行此代码(如下)时，出现错误:

But when I run this code (below) I get an error:

$msbuild = "C:WINDOWSMicrosoft.NETFrameworkv3.5MSBuild.exe /v:q /nologo" 
start-process $msbuild -wait

有没有一种方法可以使用 start-process 将参数传递给 MSBuild?我愿意不使用 start-process，我使用它的唯一原因是我需要将命令"作为变量.

Is there a way I can pass parameters to MSBuild using start-process? I'm open to not using start-process, the only reason I used it was I needed to have the "command" as a variable.

当我有
C:WINDOWSMicrosoft.NETFrameworkv3.5MSBuild.exe/v:q/nologo
在单独的一行中，如何在 Powershell 中处理它?

When I have
C:WINDOWSMicrosoft.NETFrameworkv3.5MSBuild.exe /v:q /nologo
on a line by itself, how does that get handled in Powershell?

我应该使用某种 eval() 函数吗?

Should I be using some kind of eval() kind of function instead?

推荐答案

你想要将你的参数分成单独的参数

you are going to want to separate your arguments into separate parameter

$msbuild = "C:WINDOWSMicrosoft.NETFrameworkv3.5MSBuild.exe"
$arguments = "/v:q /nologo"
start-process $msbuild $arguments 

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