Question

我尝试使用7-Zip备份Powershell（v2）脚本中的某些文件。

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;

但是当我跑步时，我得到了：

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


Error:
Incorrect command line

将此内容写入我得到的屏幕：

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"

直接粘贴到命令窗口时，可以正常工作。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些什么来实现这个目标？

Answer 1

找到this脚本并根据您的需要进行调整。你可以尝试一下：

if (-not (test-path "$env:ProgramFiles\7-Zip\7z.exe")) {throw "$env:ProgramFiles\7-Zip\7z.exe needed"} 
set-alias sz "$env:ProgramFiles\7-Zip\7z.exe"  

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

sz a -mx=9 $Target $Source

Answer 2

put＆＃34;＆amp;＆＃34; 7z命令之前的特殊字符。示例：＆amp; 7z ...

Answer 3

也许更简单的解决方案是通过cmd在您的Powershell上运行7-zip：

cmd /c 7za ...

Answer 4

尝试使用参数-file指定程序或脚本的位置：

-file＆＃34; C：\ Program Files \ someting.exe＆＃34;

Answer 5

简单地使用＆符号后缀命令

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

Answer 6

如果正确调整了它：不要忘记“ $ Target”上的“” 并避免在c：\ programm文件中的$ 7zipPath路径中留有空格

Set-Alias 7zip $7zipPath

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

7zip a -mx=9 "$Target" "$Source"

或

7z a "$ArchiveName" -t7z '@listfile.txt'

在Powershell脚本中运行7-Zip

6 个答案: