如何在Mercurial中克隆存储库的子文件夹? [英] How do I clone a sub-folder of a repository in Mercurial?

问题描述

我有一个Mercurial存储库，其中包含一些相关项目.我只想将这些项目之一分支到其他地方.

I have a Mercurial repository containing a handful of related projects. I want to branch just one of these projects to work on it elsewhere.

克隆仅仅是一部分存储库，这是实现这一目标的正确方法吗?

Is cloning just part of a repository possible, and is that the right way to achieve this?

推荐答案

您想要的是 窄或部分克隆，但是很遗憾，尚不支持此克隆.

What you want is a narrow or partial clone, but this is unfortunately not yet supported.

如果您已经有一个大型存储库，并且意识到将其拆分为几个较小的存储库是有意义的，则可以使用

If you already have a big repository and you realize that it would make sense to split it into several smaller repositories, then you can use the convert extension to do a Mercurial to Mercurial conversion. Note that this creates a new repository foo and you cannot push/pull between your-big-repo and foo.

默认情况下未启用convert extension，因此请将以下内容添加到存储库的hgrc文件或mercurial.ini文件中:

The convert extension is not enabled by default so add the following to your repo's hgrc file or your mercurial.ini file:

[extensions]
hgext.convert=

然后使用

include "libs/foo"
rename "libs/foo" .

(请注意，即使在Windows上也可以使用正斜杠)并运行

(note you can use forward slashes even on Windows) and run

$ hg convert --filemap map.txt your-big-repo foo

这将使foo成为具有your-big-repo中libs/foo文件夹的完整历史记录的存储库.

That will make foo a repository with the full history of the libs/foo folder from your-big-repo.

如果要从your-big-repo中删除所有foo证据，则可以进行另一次转换，在其中使用exclude libs/foo摆脱目录.

If you want to delete all evidence of foo from your-big-repo you can make another conversion where you use exclude libs/foo to get rid of the directory.

当您拥有多个这样的存储库并且想要整体使用它们时，则应查看建议.

When you have several repositories like that and you want to use them as a whole, then you should look at subrepositories. This feature lets you include other repositories in a checkout — similarly to how svn:externals work. Please follow the recommendations on that wiki page.

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