Zip the contents of a folder without including the folder itself

command linezip

I have a directory called folder that looks like this:

folder
      -> root_folder
                    -> some files

I want to zip this directory into zipped_dir, I tried:

zip -r zipped_dir.zip folder/*

But this generates a ZIP that looks like this:

zipped_dir
          -> folder
                   -> root_folder
                                 -> some files

in other words, it's including the directory whose contents I want to zip. How can I exclude this parent directory from the ZIP, without moving anything?

IE I would like this end result:

zipped_dir
          -> root_folder
                        -> some files

Best Answer

Try to use this command (you will get the idea)

cd folder; zip -r ../zipped_dir.zip *

Maybe there is other way, but this is fastest and simplest for me :)

Related Solutions

Unzip file contents, but without creating archive folder

My version of unzip has a -j option to not create any directory.

unzip -j /path/to/file.zip

Will extract all the files into the current directory without restoring the directory structure stored in the zip file.

If you want to only remove one level of directories from the archive, (extract myarchive/dir/file as dir/file, not file), you could use bsdtar (which does supports zip files in addition to tar files) instead and its -s option.

bsdtar -xf /path/to/file.zip -s'|[^/]*/||'

Zsh Scripting – Zip Folder Without Including Path and Using One Argument

The argument of the function is $1.

You can use history expansion modifiers to extract the directory part and the last component of the path: if the argument is ~/directory2/directory1 then $1:h is ~/directory2 and $1:t is directory1. (Mnemonic: head and tail.)

Use parentheses instead of braces around the function body. This way the function body runs in a subshell (a separate shell process), and variable assignments, directory changes and so on only affect the subshell.

coolerzip () (
    cd $1:h &&
    zip -r $2:h.zip $2:h
)

Other shells don't have history expansion modifiers, so you'd need something different to parse the argument. For the file name, ${1##*/} strips off all leading directory components; however, this doesn't work if there's a trailing slash in the parameter, whereas "$(basename -- "$1")" works in that case. For the leading directories, use $(dirname -- "$1"). Note that in shells other than zsh, you need to put double quotes around variable substitutions.

coolerzip () (
    name="$(basename -- "$1")"
    cd "$(dirname -- "$1")" &&
    zip -r "$name.zip" "$name"
)

Best Answer

Related Solutions

Unzip file contents, but without creating archive folder

Zsh Scripting – Zip Folder Without Including Path and Using One Argument

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