Git - 使用过滤器分支删除带有空变更集的提交
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Git - remove commits with empty changeset using filter-branch
提问by Paul Pladijs
How do I remove commits which have no changeset using git filter-branch?
如何使用 git filter-branch 删除没有变更集的提交？
I rewrote my git history using:
我使用以下方法重写了我的 git 历史记录：
git filter-branch --tree-filter 'rm -r -f my_folder' -f HEAD
this worked out well but now I have lots of commits with empty changesets. I would like to remove those commits. Preferably in msysgit.
这很有效，但现在我有很多提交空变更集。我想删除这些提交。最好在 msysgit 中。
Rebasing is not really an option because I have over 4000 commits and half of them must be removed.
Rebase 并不是一个真正的选择，因为我有超过 4000 次提交，其中一半必须被删除。
采纳答案by Cascabel
Just add on the --prune-emptyoption:
只需添加--prune-empty选项：
git filter-branch --tree-filter 'rm -rf my_folder' --prune-empty -f HEAD
(And of course, if you have other refs, you might want to rewrite everything with -- --allinstead of just HEAD.)
（当然，如果你有其他引用，你可能想用 重写所有内容，-- --all而不仅仅是HEAD。）
Note that this isn't compatible with --commit-filter; in that case, Charles Bailey has your answer.
请注意，这与--commit-filter;不兼容。在这种情况下，查尔斯贝利有你的答案。
回答by CB Bailey
Just looking a the documentation for filter-branch, you should be able to do this:
只需查看 的文档filter-branch，您就应该能够做到这一点：
git filter-branch --commit-filter 'git_commit_non_empty_tree "$@"' HEAD
