Avoid unwanted path in Zip file

I'm making a shell script to package some files. I'm zipping a directory like this:

zip -r /Users/me/development/something/out.zip /Users/me/development/something/folder/

The problem is that the resultant out.zip archive has the entire file path in it. That is, when unzipped, it will have the whole "/Users/me/development/anotherthing/" path in it. Is it possible to avoid these deep paths when putting a directory into an archive?

When I run zip from inside the target directory, I don't have this problem.

zip -r out.zip ./folder/

In this case, I don't get all the junk. However, the script in question will be called from wherever.

FWIW, I'm using bash on Mac OS X 10.6.

jerwood

Posted 2010-03-14T01:30:53.160

Reputation: 1 139

Answers

Your script should use cd or pushd and popd to move into the directory that will be the root of the archive before issuing the zip command. How you do this exactly will depend on how the script knows what to zip up. But, if you want /Users/me/development/something/folder zipped with internal paths of just ./folder, you'd need to do this:

pushd /Users/me/development/something
zip -r /path/to/out.zip ./folder/
popd

That will result in your out.zip containing the relative paths you want.

If you need assistance with scripting that, you'll need to show us your script.

quack quixote

Posted 2010-03-14T01:30:53.160

Reputation: 37 382

2Why are you playing games with the path? Why not use zip -j? – jww – 2015-01-04T01:15:38.683

Perfect. Thanks for the shell command schooling. I like the idea of pushing and popping paths onto the shell "stack". – jerwood – 2010-03-14T02:16:08.847

8no problem. i do this in one-liners on the commandline all the time, eg: $ pushd /some/path ; do-something ; popd ... or even with subshells: $ ( cd /some/path ; do-something ) – quack quixote – 2010-03-14T02:25:27.630

1@~quack: +1 especially for the sub-shell technique in the comment. – Jonathan Leffler – 2010-03-14T04:39:21.200

1Though using && instead of ; is a good idea so that the other command does not run if the cd failed (typo, or other problem): (cd /some/path && do-something) – Chris Johnsen – 2010-03-14T06:16:23.087

@Chris Johnsen: that's a good tip. i don't use && and || as often in my one-liners as i do in scripting, but that's part personal style and part personal failing. in this case it is the more suitable grammar. – quack quixote – 2010-03-14T10:11:17.593

Life-saving answer. I was looking for option to only zip relative paths while running the script from root. Thanks a lot! – Moseleyi – 2016-02-27T20:35:43.727

@jww i can verify that using the junk-paths argument didn't work for my senario. i needed to cd to the directory. – Andy – 2016-11-23T15:46:19.560

assuming, /path/to was your previous working directory, you could use zip -r ~1/out.zip ... (directory stack shortcuts) – phil294 – 2017-06-13T15:47:50.777

zip -j works for me and thus, it zips the files inside that folder without including the folder name. – Jun – 2018-09-14T00:11:41.500

3Is there not a way to do this without specifying the absolute path for the zip file? – orange80 – 2013-06-28T07:18:09.800

@orange80 - I was reading over this Q&A and wondered the same thing. I did a little research and found what seems to be a solution, but I am definitely a bash novice, so take it with a grain of salt. If you want the zip file to be in the directory you are navigating from, you can use dirs and awk to get that directory. I assign that to a variable and use it in my script: LASTDIR=\dirs | awk '{print $2}'``, then instead of zip -r /path/to/out.zip ./folder/, I do zip -r $LASTDIR/bundle.zip – ajh158 – 2014-01-10T16:01:29.883

The problem is that the resultant out.zip archive has the entire file path in it.
...
Is it possible to avoid these deep paths when putting a directory into an archive?

Yes. Use the -j option with zip. -j is "junk the path". According to the man page on zip:

Store just the name of a saved file (junk the path), and do not store directory names. By default, zip will store the full path (relative to the current directory).

Using -j means the following command:

zip -j myarchive.zip file1.txt dir1/file2.txt dir2/dir3/file3.txt ../file4.txt

Will create an archive that looks like:

myarchive.zip
    |
    +-- file1.txt
    +-- file2.txt
    +-- file3.txt
    +-- file4.txt

jww

Posted 2010-03-14T01:30:53.160

Reputation: 1

6Doesn't this also junk all paths, even those in the 'folder' directory? I think the OP only wants to get rid of the path on the command line. – Peter Jaric – 2015-09-14T07:37:54.867

1This might not be the answer for this question, but it's what I need. – Aaron McMillin – 2017-03-16T00:50:01.763

1This answer incorrectly assumes user wants a flat zip. This is a bad assumption and can cause the script to fail if two files in different sub-folders have the same name. – Andrew Schwartz – 2019-02-12T19:45:18.217

There is pushd and popd and $OLDPWD. Assuming the $PWD is /Users/me/development do:

pushd something/folder
zip -r $OLDPWD/something/out.zip *
popd

Now the $PWD is back to /Users/me/development

Igor Ostaptchenko

Posted 2010-03-14T01:30:53.160

Reputation: 21

Zip without including paths, but include parent directory, and without having to hardcode its name

A slightly improved version of @quack's accepted answer.

If you are doing this in a script and don't want to have to hardcode the parent directory, you can do this:

pushd /Users/me/development/something/folder/;
zip -r ../out.zip ../$(basename $PWD)
popd;

The ../$(basename $PWD) will ensure that the files are retained in the parent directory when extracted.

So now unzip my.zip will give a folder containing all your files:

folder
├── file1
├── file2
├── dir1
│   ├── file3
│   ├── file4

Instead of littering the current directory with the unzipped files:

file1
file2
dir1
├── file3
├── file4

AndrewD

Posted 2010-03-14T01:30:53.160

Reputation: 363