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Remove checkout-last alias

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Since `git checkout -` does the same thing as `git checkout-last`
there's no need for the alias.
This commit is contained in:
Kenneth Benzie 2018-08-26 16:42:29 +01:00
parent 8e556960fb
commit 1294ea01f2
2 changed files with 0 additions and 3 deletions
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README.md
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@ -17,8 +17,6 @@ rather than remembering the plumbing term of a specific command.
* `delete` deletes an existing branch, shorthand for `git branch -D`.
* `name` the name of the current branch.
* `last` the name of the last branch checked out before the current branch.
* `checkout-last` checkout the last branch, shorthand for
`git checkout $(git last)`.
* `publish` push and set the tracking branch of a local branch to origin,
shorthand for `git push -u origin <branch>`.
* `unpublish` delete a remote branch, shorthand for `git push origin :<branch>`.

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config
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@ -14,7 +14,6 @@
delete = branch -D
name = rev-parse --abbrev-ref HEAD
last = !sh -c 'git reflog | grep \" checkout: moving from\" | head -n 1 | awk \"{ print \\$6 }\"'
checkout-last = !git checkout `git last`
publish = !git push -u origin `git name`
unpublish = !git push -u origin :`git name`