remove '.part1' from Output directory in 7z

Question

I have 2 files test.part1.rar and test.part2.rar When extracting them using command

7z x -o* test.part1.rar

The output directory is a folder with test.part1 name I want the output directory to be a folder with name test (archive name without '.part1')

    From 7z documantation
    
        -o{dir_path}
        {dir_path}
        This is the destination 
directory path. It's not required to end with a backslash. If you specify * in {dir_path}, 7-Zip substitutes that * character to archive name.

Wouldn't it be less stressful to fix it after? `mv test.part2/* test.part1` / `mv test.part1 test`? — Tim Roberts, Nov 09 '21 at 19:45
Yes but i am writing a code to decompress multiple files not just one — , Nov 10 '21 at 00:23

theherk · Accepted Answer · 2021-11-09T19:43:30.877

0

Maybe using parameter expansion will do what you seek.

f=test.part1.rar; 7z x -o ${f%%.*} $f

You may want to use the 7z e command if you don't want to extract with the full paths within the archive.

The * can be used within the output directory to be replaced by the archive name. It doesn't sound like you need that.

edited Nov 09 '21 at 19:43

answered Nov 09 '21 at 17:23

theherk

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The problem is that * is not replaced by archive name when extracting multipart rar files ,but it is replaced by archive name+part1 – Nov 09 '21 at 18:23
I see what you mean. You could do something like `f=test.part1.rar; 7z x -o ${f%%.*} $f` – theherk Nov 09 '21 at 19:41

remove '.part1' from Output directory in 7z

1 Answers1