Using 7-Zip from command line - name zip file with directory name

Question

I know this has to be so easy, but I do not know how to do it. Any help would be appreciated.

I want to name zip files with the directory name.

Say I have a directory "Zip1" with files "Doc1.txt" and "Doc2.txt"

I understand everything about zipping the folder with the exception of naming it. I can find NO examples of how to name a file through the command line using directory name information.

How can I name this zip file as "Zip1.7z" without actually typing the directory name but having the command line pull this information from the directory the files reside in?

Can anyone help?

Answer 1

I assume that you are in a directory where all subdirectories should be zipped and that the 7-Zip binaries are on your PATH.

You can then use the following on Windows (directory is %i):

for /d %i in (*) do 7z ... %i.7z ...

On Linux (directory is $i):

for i in $(find -mindepth 1 -maxdepth 1 -type d) ; do 7zr ... $i.7z ... ; done

Shorter but less robust:

for i in */ ; do 7zr ... ${i%/}.7z ... ; done

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Example to zip all .txt files in all directories which start with backup-:

Windows:

for /d %i in (backup-*) do 7z ... %i.7z %i\*.txt

Linux:

for i in backup-*/ ; do 7zr ... ${i%/}.7z $i/*.txt ; done