I have 100s of directories and within those I have a few zip files. Now there are images named abc.jpg in those zip files. The zip files may be in any folder or in any subfolder so its difficult to extract them all in one place.

I just want to collect those image files. Is this possible?

  • 1
    You can use zip -sf foo.zip | grep abc.jpg to determine if an archive has abc.jpg; that should help. I don't have time to figure out the complete command now, but I'll try later if nobody else has answered – Michael Mrozek Apr 26 '11 at 13:46

I once needed something similar to find class files in a bunch of zip files. Here it is:


function process() {
while read line; do
    if [[ "$line" =~ ^Archive:\s*(.*) ]] ; then
        #echo "$ar"
        if [[ "$line" =~ \s*([^ ]*abc\.jpg)$ ]] ; then
            echo "${ar}: ${BASH_REMATCH[1]}"

find . -iname '*.zip' -exec unzip -l '{}' \; | process

Now you only need to add one line to extract the files and maybe move them. I'm not sure exactly what you want to do, so I'll leave that to you.


If your unix variant supports FUSE (Linux, *BSD, OSX, Solaris all do), mount AVFS to access archives transparently. The command mountavfs creates a view of the whole filesystem, rooted at ~/.avfs, in which archive files have an associated directory that contains the directories and files in the archive. For example, if you have foo.zip in the current directory, then the following command is roughly equivalent to unzip -l foo.zip:

mountavfs    # needs to be done once and for all
find ~/.avfs$PWD/foo.zip\# -ls

So, to loop over all images contained in a zip file under the current directory and copy them to /destination/directory (with a prompt in case of clash):

find ~/.avfs"$PWD" -name '*.zip' -exec sh -c '
    find "${0}#" -name "*.jpg" -exec cp -ip {} "$1" \;
' {} /destination/directory \;

In zsh:

cp -ip ~/.avfs$PWD/**/*.zip(e\''REPLY=($REPLY\#/**/*.jpg(N))'\') /destination/directory

Deconstruction: ~/.avfs$PWD/**/*.zip expands to the AVFS view of the zip files under the current directory. The glob qualifier e is used to modify the output of the glob: …/*.zip(e\''REPLY=$REPLY\#'\') would just append a # to each match. REPLY=($REPLY\#/**/*.jpg(N)) transforms each match into the array of .jpg files in the .zip# directory.

  • nice solution, but FUSE? that's a little bit overkill, isn't it? – Kim Apr 26 '11 at 14:38
  • 2
    @Kim: Why should I cripple myself not to use FUSE if it's available? – Gilles Apr 26 '11 at 14:53
  • some reasons not to use FUSE if you don't have to: portability (some OSs don't have FUSE), maintainability (not everybody knows FUSE) – Kim Apr 26 '11 at 14:59

I assume you have a new version of Bash, so you should be able to use this:

shopt -s globstar
for path in topdir/**/*.zip
    unzip "$path" '.*abc.jpg'

Similar to Kims answer but slightly modified. Just use sed:

find . -name *.zip -exec unzip -l '{}' \; | sed -n -e '/^Archive/ {h}' -e '/abc.jpg$/ {x;p;x;}'

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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