Is it possible to fake destination directory path with zip

zip

Given a bunch of files in the current directory, I want to be able to zip them such that, when extracted, they'd be placed in some given directories. For example, say I have these in $PWD:

ls
fsck
xz
chroot

And when I run unzip foo.zip, they each get extracted to these directories:

/bin
/sbin
/usr/bin
/usr/sbin

The closest I could find is that these files must already exist in whatever directory paths I want the files to be extracted to (using $PWD as root directory):

$ zip foo.zip bin/ls sbin/fsck usr/bin/xz usr/sbin/chroot

Then, on a system where I want this extraction to happen, I would have to:

$ cd /
$ sudo unzip foo.zip

Best Answer

I don't see a way to do this using the zip command, but it's easy in python.

Note that the zip file format specification, section 4.4.17.1, says that the pathname cannot start with a '/', so I can't help with that part.

The python zipfile module will let you override a file's pathname when you add it to the zip archive; just pass the desired name as the optional 2nd argument to ZipFile.write:

ZipFile.write(filename[, arcname[, compress_type]])
Write the file named filename to the archive, giving it the archive name arcname (by default, this will be the same as filename, but without a drive letter and with leading path separators removed). If given, compress_type overrides the value given for the compression parameter to the constructor for the new entry.
Note: Archive names should be relative to the archive root, that is, they should not start with a path separator.

Here's an example:

$ touch 1 2 3
$ python
Python 2.7.6 (default, Mar 22 2014, 22:59:56) 
>>> import zipfile
>>> with zipfile.ZipFile('bundle.zip', 'w') as bundle:
...   bundle.write('1', '/bin/1')
...   bundle.write('2', '/sbin/2')
...   bundle.write('3', '/usr/bin/3')
... 
>>> 
$ unzip -l bundle
Archive:  bundle.zip
  Length      Date    Time    Name
---------  ---------- -----   ----
        0  2014-08-11 13:00   bin/1
        0  2014-08-11 13:00   sbin/2
        0  2014-08-11 13:00   usr/bin/3
---------                     -------
        0                     3 files

Note that zipfile.write will remove any leading '/' from the pathname, to conform with the standard.

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'|[^/]*/||'

Convert a Windows-created ZIP to Linux (internal paths issue)

I think something went wrong with the creation of the zip file, because when I create a zip file on Windows is has (portable) forward slashes:

zip.exe -r pip pip
updating: pip/ (244 bytes security) (stored 0%)
  adding: pip/pip.log (164 bytes security) (deflated 66%)

But now that you have the files with file names that contain "paths" with backslashes, you can run the following program in unzip_dir:

#! /usr/bin/env python

# already created directories, walk works topdown, so a child dir
# never creates a directory if there is a parent dir with a file.
made_dirs = set()

for root, dir_names, file_names in os.walk('.'):
    for file_name in file_names:
        if '\\' not in file_name:
            continue
        alt_file_name = file_name.replace('\\', '/')
        if alt_file_name.startswith('/'):
            alt_file_name = alt_file_name[1:]  # cut of starting dir separator
        alt_dir_name, alt_base_name = alt_file_name.rsplit('/', 1)
        print 'alt_dir', alt_dir_name
        full_dir_name = os.path.join(root, alt_dir_name)
        if full_dir_name not in made_dirs:
            os.makedirs(full_dir_name)  # only create if not done yet
            made_dirs.add(full_dir_name)
        os.rename(os.path.join(root, file_name),
                  os.path.join(root, alt_file_name))

This handles files in any directory under the directory from where the program is started. Given the problem that you describe, the unzip_dir probably doesn't have any subdirectories to start with, and the program could just walk over the files in the current directory only.

Best Answer

Related Solutions

Unzip file contents, but without creating archive folder

Convert a Windows-created ZIP to Linux (internal paths issue)

Related Question