30

How to unzip a file (ex: foo.zip) to a folder with the same name (foo/)?

Basically, I want to create an alias of unzip that unzips files into a folder with the same name (instead of the current folder). That's how Mac's unzip utility works and I want to do the same in CLI.

4
  • @Christopher I don't see how it's a duplicate. Can you find an answer to my question in that question? Dec 17, 2018 at 17:09
  • 1
    The accepted anwer answers your question: unzip -d foo foo.zip.
    – Fabby
    Dec 18, 2018 at 0:28
  • 2
    @Fabby, no it doesn't, I need a dynamic solution that won't require adding the name of the target folder. I believe the second paragraph makes this very clear. Dec 18, 2018 at 9:32
  • It seems to me that another answer there, by n.st does what you want?
    – Jeff Schaller
    Mar 6, 2019 at 15:47

3 Answers 3

29

I use unar for this; by default, if an archive contains more than one top-level file or directory, it creates a directory to store the extracted contents, named after the archive in the way you describe:

unar foo.zip

You can force the creation of a directory in all cases with the -d option:

unar -d foo.zip

Alternatively, a function can do this with unzip:

unzd() {
    if [[ $# != 1 ]]; then echo I need a single argument, the name of the archive to extract; return 1; fi
    target="${1%.zip}"
    unzip "$1" -d "${target##*/}"
}

The

target=${1%.zip}

line removes the .zip extension, with no regard for anything else (so foo.zip becomes foo, and ~/foo.zip becomes ~/foo). The

${target##*/}

parameter expansion removes anything up to the last /, so ~/foo becomes foo. This means that the function extracts any .zip file to a directory named after it, in the current directory. Use unzip $1 -d "${target}" if you want to extract the archive to a directory alongside it instead.

2
  • 1
    Smoothing over this behavior for unzip, 7z, unrar, etc is just crazy....glad to see unar can just do it all (I run parallel unar {} ::: *.zip *.rar *.7z)
    – Colin D
    May 13, 2019 at 13:58
  • now that's an answer, complete with recommendations
    – austin
    Nov 15, 2020 at 23:42
17

Use unzip -d exdir zipfile.zip to extract a zipfile into a particular directory. In principle from reading your post literally you could write a function like this:

unzip_d () {
    unzip -d "$1" "$1"
}

Since you want the .zip extension removed though, you can use special variable syntax to do that:

unzip_d () {
    zipfile="$1"
    zipdir=${1%.zip}
    unzip -d "$zipdir" "$zipfile"
}
1
  • 2
    additional tip: use basename to make it more robust. (already upvoted though) **;-)
    – Fabby
    Dec 18, 2018 at 10:49
9

Just one another variant to do that with a 1-liner for multiple zip files in current dir:

for f in *.zip; do unzip "$f" -d "${f%.zip}"; done

For just one file (i.e. to use in a function) it would be:

unzip "$1" -d "${1%.zip}"
0

Not the answer you're looking for? Browse other questions tagged or ask your own question.