使用Powershell将zip文件解压缩到多个文件夹中

Question

I have 5 zip files and 5 folders respectively. Files are: FranceData.zip, USAdata.zip, GermanData.zip, Italydata.zip and DanemarkData.zip. Folders are: FranceFolder, USAFolder, GermanFolder, ItalyFolder and DanemarkFolder. I would like to know how would I unzip these files into their respective folders using Array in Powershell. I am knew to Powershell and need help. Thank you

Answer 1

This can be done using the Expand-Archive cmdlet, which is available in PowerShell 5 and above. (PowerShell's version can be checked via the $PSVersionTable variable.)

To extract a zip file to a specific folder the following syntax is used:

Expand-Archive -Path 'ZipFile.zip' -DestinationPath 'ZipFolder'

or

Expand-Archive -LiteralPath 'ZipFile.zip' -DestinationPath 'ZipFolder'

If the -Path parameter is used, PowerShell will recognise characters such as * , ? , [ , ] , as wildcards, which can cause unexpected behaviour for file-paths that contain square-brackets.
If the -LiteralPath parameter is used, PowerShell will not treat any characters as wildcard characters.

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Assuming all your zip files and folders follow the same naming pattern you could use an array like so:

$Countries = @(
    'France',
    'USA',
    'German'
    'Italy',
    'Danemark'
)

foreach ($Country in $Countries)
{
    $ZipFilePath = $Country + 'Data.zip'
    $DestinationPath = $Country + 'Folder'

    Expand-Archive -LiteralPath $ZipFilePath -DestinationPath $DestinationPath
}

---

If your files and folders don't follow the same naming pattern you could use a dictionary (or a collection of KeyValuePair s), like so:

$ZipFilesAndFolders = @{
    'FranceData.zip' = 'FranceFolder'
    'USAData.zip' = 'USAFolder'
    'GermanData.zip' = 'GermanFolder'
    'ItalyData.zip' = 'ItalyFolder'
    'DanemarkData.zip' = 'DanemarkFolder'
}

foreach ($KeyAndValue in $ZipFilesAndFolders.GetEnumerator())
{
    $ZipFilePath = $KeyAndValue.Key
    $DestinationPath = $KeyAndValue.Value

    Expand-Archive -LiteralPath $ZipFilePath -DestinationPath $DestinationPath
}

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With PowerShell 4 (and 3)

If you have .Net Framework 4.5 installed, you can use Microsoft.PowerShell.Archive which was created by the PowerShell team.
PowerShell 4 requires .Net Framework 4.5 so this should work without any system changes needed.
The module can be found at https://github.com/PowerShell/Microsoft.PowerShell.Archive

The syntax used is identical, the function is Expand-Archive and the parameters are -Path , -LiteralPath , and -DestinationPath .
Just ensure that you have imported the module before using it, this can be done using the Import-Module cmdlet like so:

Import-Module -Name 'Microsoft.PowerShell.Archive' -Force

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PowerShell 2

A solution for PowerShell 2 can be found here: https://stackoverflow.com/a/37814462/9447234