我正在尝试使用7-Zip来备份PowerShell(V2)脚本中的一些文件。
我有：

$zipPath = "C:\Program Files\7-Zip\7z.exe"
[Array]$zipArgs = "-mx=9 a", "`"c:\BackupFolder\backup.zip`"", "`"c:\BackupFrom\backMeUp.txt`""

&$zipPath $zipArgs;

但当我运行这段代码时，我得到的是：

7-Zip [64] 9.20  Copyright (c) 1999-2010 Igor Pavlov  2010-11-18

Error:
Incorrect command line

把这段文字写到屏幕上，我会得到：

C:\Program Files\7-Zip\7z.exe -mx=9 a "c:\BackupFolder\backup.zip" "c:\BackupFrom\backMeUp.txt"

所以我假设我需要在7z.exe的路径周围加上引号，这给了我：

$zipPath = "C:\Program Files\7-Zip\7z.exe"
$zipPath = " `"$zipPath`" "
[Array]$zipArgs = "-mx=9 a", "`"c:\BackupFolder\backup.zip`"", "`"c:\BackupFrom\backMeUp.txt`""

&$zipPath $zipArgs;

但随后我收到以下错误：

The term '"C:\Program Files\7-Zip\7z.exe"' is not recognized as the name of a cmdlet, function, script file
, or operable program. Check the spelling of the name, or if a path was included, verify that the path is c
orrect and try again.
At C:\BackupScript\Backup.ps1:45 char:22
+                     & <<<< `"$zipPath`" $zipArgs;                    
    + CategoryInfo          : ObjectNotFound: ("C:\Program Files\7-Zip\7z.exe":String) [], CommandNotFound 
   Exception
    + FullyQualifiedErrorId : CommandNotFoundException

把它写出来给了我：

"C:\Program Files\7-Zip\7z.exe" -mx=9 a "c:\BackupFolder\backup.zip" "c:\BackupFrom\backMeUp.txt"

当直接粘贴到命令窗口中时，其工作方式与预期相同。我试图弄清楚这一点已经有一段时间了，但假设我遗漏了什么(可能非常明显)。有没有人知道我需要做些什么才能让这一切顺利进行？