Do a "git export" (like "svn export")?

Greg Hewgill picture Greg Hewgill · Oct 2, 2008 · Viewed 654.8k times · Source

I've been wondering whether there is a good "git export" solution that creates a copy of a tree without the .git repository directory. There are at least three methods I know of:

  1. git clone followed by removing the .git repository directory.
  2. git checkout-index alludes to this functionality but starts with "Just read the desired tree into the index..." which I'm not entirely sure how to do.
  3. git-export is a third party script that essentially does a git clone into a temporary location followed by rsync --exclude='.git' into the final destination.

None of these solutions really strike me as being satisfactory. The closest one to svn export might be option 1, because both those require the target directory to be empty first. But option 2 seems even better, assuming I can figure out what it means to read a tree into the index.

Answer

CB Bailey picture CB Bailey · Oct 2, 2008

Probably the simplest way to achieve this is with git archive. If you really need just the expanded tree you can do something like this.

git archive master | tar -x -C /somewhere/else

Most of the time that I need to 'export' something from git, I want a compressed archive in any case so I do something like this.

git archive master | bzip2 >source-tree.tar.bz2

ZIP archive:

git archive --format zip --output /full/path/to/zipfile.zip master 

git help archive for more details, it's quite flexible.


Be aware that even though the archive will not contain the .git directory, it will, however, contain other hidden git-specific files like .gitignore, .gitattributes, etc. If you don't want them in the archive, make sure you use the export-ignore attribute in a .gitattributes file and commit this before doing your archive. Read more...


Note: If you are interested in exporting the index, the command is

git checkout-index -a -f --prefix=/destination/path/

(See Greg's answer for more details)