![]() ![]() faultRemote=origin to always checkout remote branches from there if is ambiguous but exists on the origin remote. You can look around, make experimental changes and commit them, and you can discard any commits you make in this state without impacting any branches by switching back to a branch. ![]() ![]() If the branch exists in multiple remotes and one of them is named by the faultRemote configuration variable, weâll use that one for the purposes of disambiguation, even if the isnât unique across all remotes. In each case, assuming you dont already have a local branch called release/BranchName, git will work out that what you want is a new local branch which 'tracks' (is associated with for push and pull commands) the remote branch of the same name. I wanted to do the Git equivalent of svn up. I didn't want them anymore so I deleted the file, thinking I can just checkout a fresh copy. If is not found but there does exist a tracking branch in exactly one remote (call it ) with a matching name, treat as equivalent to: $ git switch -c -track / How do I fix a Git detached head Ask Question Asked 11 years, 3 months ago Modified 20 days ago Viewed 2.3m times 2027 I was doing some work in my repository and noticed a file had local changes. Generally speaking: git checkout will get you out of that.master) you could simply git checkout master to get out of detached HEAD state. If that branch matches a remote tracking one, it will create a local branch, and automatically track the remote one! 9 Answers Sorted by: 519 If you remember which branch was checked out before (e.g. Actually, you don't even (always) need the -create option when creating a new branch with git switch: ![]()
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