在Powershell脚本中运行7-Zip

Question

我正在尝试使用7-Zip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;

但是,当我运行这个时,我得到:

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


Error:
Incorrect command line

把它写到屏幕我得到:

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的路径上放置引号,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;     

但后来我收到以下错误:

    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

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时,它按预期工作.我一直试图弄清楚这一点,但我想我错过了一些东西(可能很明显).任何人都可以看到我需要做什么才能让这次运行？

Answer 1

找到此脚本并根据您的需要进行调整.你能试试吗:

$7zipPath = "$env:ProgramFiles\7-Zip\7z.exe"

if (-not (Test-Path -Path $7zipPath -PathType Leaf)) {
    throw "7 zip file '$7zipPath' not found"
}

Set-Alias 7zip $7zipPath

$Source = "c:\BackupFrom\backMeUp.txt"
$Target = "c:\BackupFolder\backup.zip"

7zip a -mx=9 $Target $Source

Answer 2

在7z命令之前加上"&"特殊字符.示例:&7z ...

Answer 3

只需在命令前加上 & 号

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