从Powershell脚本使用参数运行可执行文件 [英] Run Executable from Powershell script with parameters

问题描述

所以我有一个powershell脚本，它应该运行一个可执行文件的参数传递设置我想运行的方法，我需要传递一个参数，这是一个目录到一个配置文件。所以这是我有

So I have a powershell script that is supposed to run an executable with an argument to pass to set which method I want to run, and I need to pass a parameter, which is a directory to a config file. So this is what I have

Start-Process -FilePath "C:\Program Files\MSBuild\test.exe" -ArgumentList /genmsi/f $MySourceDirectory\src\Deployment\Installations.xml

f是短名称，文件是我的属性的长名...我在powershell中得到一个错误，告诉我无法找到/ f或/ file的位置参数。

/f is the shortname and file is the long name for my attribute... I get an error in powershell telling me that a positional parameter cannot be found for /f or /file.

任何想法？

推荐答案

我能够使用 Invoke-表达式 cmdlet。

I was able to get this to work by using the Invoke-Expression cmdlet.

Invoke-Expression "& `"$scriptPath`" test -r $number -b $testNumber -f $FileVersion -a $ApplicationID"

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