How to get a list of directories in a zip?

Question

I am looking for a way to list the directories in a zip file in bash under Linux. I found out there is an application called zipinfo, which can list the paths in the zip (without any extra noise to have to parse through (zipinfo -1 foo.zip), as opposed to unzip -l foo.zip). However, this isn't good enough and I am wondering if there is a better way.

Tova

why zipinfo -1 foo.zip is not good enough? – Hudnut

Answer 1

To list only directories:

unzip -l foo.zip "*/"

Output (e.g.):

Archive:  foo.zip
  Length      Date    Time    Name
---------  ---------- -----   ----
        0  2015-09-10 20:10   work/
        0  2015-08-31 10:45   work/test1/
        0  2015-08-31 10:50   work/test1/cc/
        0  2015-08-31 10:45   work/test1/dd/
        0  2015-08-31 10:45   work/test1/aa/
        0  2015-08-31 10:45   work/test1/bb/
        0  2015-09-09 21:17   work/tmp/
        0  2015-08-23 18:49   work/tmp/work/
        0  2015-09-08 19:33   work/tmp/work/loop/
        0  2015-08-15 16:00   work/tmp/work/1/
        0  2015-08-15 16:00   work/1/
        0  2015-08-24 18:40   work/dir/
        0  2015-09-05 18:07   work/rename/
---------                     -------
        0                     13 files

or use

zipinfo -1 foo.zip "*/"

Output (e.g.):

work/
work/test1/
work/test1/cc/
work/test1/dd/
work/test1/aa/
work/test1/bb/
work/tmp/
work/tmp/work/
work/tmp/work/loop/
work/tmp/work/1/
work/1/
work/dir/
work/rename/

Supinate

zipinfo -1 foo.zip "*/" worked best, thank you! :) – Tova

I get caution: filename not matched: */ although there are plenty of directories in my zip file. EDIT: Is a zip file expected to contain dedicated directory entries? My file only contains file entries... – Maziemazlack

@moi: I recommend to start a new question. – Supinate

Neither unzip -l foo.zip "*/" nor zipinfo -1 foo.zip "*/"are reliable solutions: the Zip file may have been created without dedicated directory entries. For instance the zip tool has the option -D do not add directory entries. – Imamate

Answer 2

You can try piping unzip -l to awk like this:

unzip -l foo.zip | awk '/\/$/ { print $NF }'

All the directories in the unzip -l output end with a slash and $NF prints just the directory path.

Detestation

Answer 3

On Cygwin

I coudn't get either of these to work on my Cygwin. For my immediate use, where I only needed to get one directory deep:

zipinfo -1 foo.zip | grep -o "^[^/]\+[/]" | sort -u

For other directories, you could use xargs in combination with some other parsing thing, something like:

## PSEUDOCODE
$ zipinfo -1 foo.zip | \
# from the pipe, run 
xargs \
# do something like the following while loop, going one file at a time
while there is still a '/' in the line; do
  parse off "[^/]+([/]|)$" from the line
  print out the new line
done | \
sort -u

---

System Info

$ uname -a
CYGWIN_NT-10.0 C-D-ENG-E-INT3 2.11.2(0.329/5/3) 2018-11-08 14:34 x86_64 Cygwin

$ bash --version
GNU bash, version 4.4.12(3)-release (x86_64-unknown-cygwin)
Copyright (C) 2016 Free Software Foundation, Inc.
License GPLv3+: GNU GPL version 3 or later <http://gnu.org/licenses/gpl.html>

This is free software; you are free to change and redistribute it.
There is NO WARRANTY, to the extent permitted by law.

$ systeminfo | sed -n 's/^OS\ *//p'
Name:                   Microsoft Windows 10 Enterprise
Version:                10.0.17134 N/A Build 17134
Manufacturer:           Microsoft Corporation
Configuration:          Member Workstation
Build Type:             Multiprocessor Free

$

Revitalize

Answer 4

The open-source Zip_Dir_List tool will list all directories for you, even for Zip archives that don't have explicit directory entries. From the link, go up to the repository and follow the build instructions in readme.txt .

Imamate