Git: Checkout only files without repository?

git clone --depth 1 does exactly what you want. I have a git repository with 41,000 commits and this is vastly faster. See this answer or this more general response for a more detailed explanation.

If you want to specify the branch to use (instead of the default "master" branch) use this syntax (requires Git 1.9+):

git clone -b <remoteBranch> --single-branch --depth 1 ssh://[email protected]:serverport/PathToProject <FolderName>

The git-archive manpage contains the following example:

git archive --format=tar --prefix=junk/ HEAD | (cd /var/tmp/ && tar xf -)

Create a tar archive that contains the contents of the latest commit on the current branch, and extract it in the '/var/tmp/junk' directory.

Or you can use low level git-checkout-index, which manpage contains the following example:

Using git checkout-index to "export an entire tree"

The prefix ability basically makes it trivial to use 'git checkout-index' as an "export as tree" function. Just read the desired tree into the index, and do

   $ git checkout-index --prefix=git-export-dir/ -a

git checkout-index will "export" the index into the specified directory.

The final "/" is important. The exported name is literally just prefixed with the specified string.

Or you can try to use --work-tree option to git wrapper, or GIT_WORK_TREE environment variable, e.g. by using "git --work-tree=/somwehere/else checkout -- .".

Git is much easier than Subversion for this, as the whole repository is held in a single directory. You only need to delete the hidden ".git" folder in the root of the project to create a production-ready copy of your site.

In Linux/OSX this would be:

mkdir dist && cd dist
git checkout --depth=1 http://path/to/your/project.git
rm -rf .git