Question

我有一个应用程序，它生成一个带有HTML文档的ZIP文件。但是，提取的HTML文件的名称与zip文件的名称不同（出于某些我无法控制的奇怪原因）。

我需要解压缩zip文件，并将提取的文件重命名为与zip文件名相同但是我不确定这会在我的脚本中插入一些帮助，请参阅下面的脚本：

#Files Location
$ZipFilesPath = "C:\Test\"
#Unzip To Same Location
$UnzipPath = "C:\Test\"

$Shell = New-Object -com Shell.Application
$Location = $Shell.NameSpace($UnzipPath)
$ZipFiles = Get-Childitem $ZipFilesPath -Recurse -Include *.ZIP
$FileCounter = 1

#Clear Initilisation Vars from Console
clear

foreach ($ZipFile in $ZipFiles) {
    #Get The Base Filename without the Filepath
    $ZipFileActualName = [io.path]::GetFileNameWithoutExtension($ZipFile.FullName)

    write-host "File: "  $ZipFileActualName

    $ZipFolder = $Shell.NameSpace($ZipFile.fullname)
    $Location.Copyhere($ZipFolder.items(), 1040)

    #Move Along to Next File
    $FileCounter++
}

最终解决方案（下面添加来自Nick Cox的作品）

#Location Of ZIP Files Defined Here
$ZipFilesPath = "C:\Test\";
#Unzip To Temp Folder Defined Here
$TempPath = "C:\Test\Temp"

#Create Our Temp File If It Doesnt Exist (Will Be Deleted Again)
If(!(test-path $TempPath))
{
      New-Item -ItemType Directory -Force -Path $TempPath
}

$Shell = New-Object -com Shell.Application
$Location = $Shell.NameSpace($TempPath)
$ZipFiles = Get-Childitem $ZipFilesPath -Recurse -Include *.ZIP
$FileCounter = 1

#Clear Initilisation Vars from Console
clear

foreach ($ZipFile in $ZipFiles) {
    #Get The Base Filename without the Filepath
    $ZipFileActualName = [io.path]::GetFileNameWithoutExtension($ZipFile.FullName)
    write-host "File: "  $ZipFileActualName
    $ZipFolder = $Shell.NameSpace($ZipFile.fullname)
    $Location.Copyhere($ZipFolder.items(), 1040)
    #Move Our Temp Files
    $HtmlFiles = Get-ChildItem $TempPath *.html
    $HtmlFiles |% {Move-Item  $_.Fullname "$UnzipPath/$ZipFileActualName.html"}
    #Move Along to Next File
    $FileCounter++
}

#Remove Our .ZIP Folders
Get-ChildItem -Path $ZipFilesPath -Include *.zip* -File -Recurse | foreach { $_.Delete()}

#Remove Our Temp Folder
Remove-Item –path $TempPath –recurse

Answer 1

为什么不提取到临时位置并从那里复制到目的地？

#Files Location
$ZipFilesPath = "C:\Test\"
#Unzip To Same Location
$UnzipPath = "C:\Test\"
$TempPath = "C:\Test\Temp"

$Shell = New-Object -com Shell.Application
$Location = $Shell.NameSpace($TempPath)
$ZipFiles = Get-Childitem $ZipFilesPath -Recurse -Include *.ZIP
$FileCounter = 1

#Clear Initialisation Vars from Console
clear

foreach ($ZipFile in $ZipFiles) {
    #Get The Base Filename without the Filepath
    $ZipFileActualName = [io.path]::GetFileNameWithoutExtension($ZipFile.FullName)

    write-host "File: "  $ZipFileActualName

    $ZipFolder = $Shell.NameSpace($ZipFile.fullname)
    $Location.Copyhere($ZipFolder.items(), 1040)

    $HtmlFiles = Get-ChildItem $TempPath *.html
    $HtmlFiles |% {Move-Item  $_.Fullname "$UnzipPath/$ZipFileActualName.html"}
    #Move Along to Next File
    $FileCounter++
}

顺便说一句，如果您使用的是PowerShell＆gt; = 5.0，则可以使用Expand-Archive进行本地解压缩。

Answer 2

    foreach ($ZipFile in $ZipFiles) {
    #Get The Base Filename without the Filepath
    $ZipFileActualName = [io.path]::GetFileNameWithoutExtension($ZipFile.FullName)

    write-host "File: "  $ZipFileActualName

    $ZipFolder = $Shell.NameSpace($ZipFile.fullname)
    $Location.Copyhere($ZipFolder.items(), 1041)

    $HtmlFiles = Get-ChildItem $TempPath *.html
    $HtmlFiles |% {Move-Item  $_.Fullname "$UnzipPath/$ZipFileActualName.html"}
    #Move Along to Next File
    $FileCounter++
}

Powershell提取并重命名为ZIP文件名

2 个答案: