从Powershell脚本调用可执行文件（带有参数） [英] Invoke executable (w/parameters) from powershell script

问题描述

我正在从Powershell调用zip实用程序，并且很难直接获取其参数。代码如下：

I'm calling a zip utility from powershell and having a difficult time getting its parameters straight. Here's the code:

    if (-not (test-path "C:\Program Files (x86)\7-Zip\7z.exe")) {throw "C:\Program Files (x86)\7-Zip\7z.exe needed"} 
    set-alias sz "C:\Program Files (x86)\7-Zip\7z.exe" 

    $argument_1 = "c:\temp\DeployTemp\"
    $argument_0 = "c:\temp\Release\Web_Feature_2012R10_1_1112.prod.com.zip"

    sz x $argument_0 -o$argument_1

问题是7zip可执行调用实际上是提取到名为$ argument_1的目录中，而不是提取存储在字符串中的实际值。我尝试了几种方法来逃避价值，但是没有运气。不幸的是，7zip -o标志和输出目录之间不能有空格...

The problem is the 7zip executable call literally extracts to a directory named $argument_1, instead of the actual value stored in the string. I've tried escaping the value in a few ways but without luck. Unfortunately the 7zip "-o" flag can't have a space between it and the output directory...

推荐答案

尝试一些操作像这样：

& "$sz" x $argument_0 "-o$argument_1"

&符告诉PowerShell处理表达式就像CMD.exe一样，但仍然允许变量扩展（以$开头的令牌）。

The ampersand tells PowerShell to treat the expression more like CMD.exe would, but still allow for the variable expansion (tokens that start with a $).

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