Question

I have a small problem trying to unzip a file using the 7za command-line utility in Powershell.

I set the $zip_source variable to the zip file s path and the $unzip_destination to the desired output folder.

但是,<条码>7za<>>>的指挥系统使用需要具体表述如下:

7za x -y <zip_file> -o<output_directory>

因此,我目前的呼吁是:

&  7za  x -y "$zip_source" -o$unzip_destination

Due to the fact that there can be no space between -o and the destination it seems that PowerShell will not expand the $unzip_destination variable, whereas $zip_source is expanded.

Currently the program simply extracts all the files into the root of C: in a folder named $unzip_destination. Setting different types of quotes around the variable won t work neither:

-o"$unzip_destination" : still extracts to C:$unzip_destination
-o $unzip_destination  : still extracts to C:$unzip_destination
-o $unzip_destination  : Error: Incorrect command line

是否有办法在指挥之前强行扩大?

Answer 1

努力这样做:

-o $($unzip_destination)

^{Editor s note: This solution only works with a space after -o (in which case just -o $unzip_destination would do) - if you remove it, the command doesn t work as intended.

This approach is therefore not suitable for appending a variable value directly to an option name, as required by the OP.}

Answer 2

引证:

&  7za  x -y "$zip_source" "-o$unzip_destination"

Answer 3

这一工作应当:

&  7za  x -y $zip_source -o${unzip_destination}

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