从 powershell 脚本调用可执行文件(带参数)

问题描述

我正在从 powershell 调用 zip 实用程序，但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-Zip7z.exe")) {throw "C:Program Files (x86)7-Zip7z.exe needed"}
    set-alias sz "C:Program Files (x86)7-Zip7z.exe"

    $argument_1 = "c:	empDeployTemp"
    $argument_0 = "c:	empReleaseWeb_Feature_2012R10_1_1112.prod.com.zip"

    sz x $argument_0 -o$argument_1

问题是 7zip 可执行文件调用从字面上提取到名为 $argument_1 的目录，而不是存储在字符串中的实际值.我试过以几种方式逃避价值，但没有运气.不幸的是，7zip "-o" 标志在它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

试试这个:

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

&符号告诉 PowerShell 更像 CMD.exe 那样处理表达式，但仍然允许变量扩展(以 $ 开头的标记).

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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