Git Clone: Just the files, please?

Question

I want to clone a GIT repo and NOT end up with a .git directory. In other words I just want the files. Is there a way to do this?

git clone --no-chec

Answer 1

git checkout -f

There's another way to do this by splitting the repo from the working tree.

This method is useful if you need to update these git-less git files on a regular basis. For instance, I use it when I need to check out source files and build an artifact, then copy the artifact into a different repo just for deployment to a server, and I also use it when pushing source code to a server when I want the source code to checkout and build into the www directory.

We'll make two folders, one for the git one for the working files:

mkdir workingfiles
mkdir barerepo.git

initialize a bare git repo:

cd barerepo.git
git --bare init 

Then create a post-receive hook:

touch hooks/post-receive
chmod ug+x hooks/post-receive

Edit post-receive in your favorite editor:

GIT_WORK_TREE=/path/to/workingfiles git checkout -f
# optional stuff:
cd down/to/some/directory
[do some stuff]

Add this as a remote:

git remote add myserver ssh://user@host:/path/to/barerepo.git

Now every time you push to this bare repo it will checkout the working tree to /workingfiles/. But /workingfiles/ itself is not under version control; running git status in /workingfiles/ will give the error fatal: Not a git repository (or any parent up to mount point /data). It's just plain files.

Unlike other solutions rm -r .git command is not needed, so if /workingfiles/ is some other git repo you don't have to worry about the command used removing the other repo's git files.