如何使用＆QUOT;回购＆QUOT;克隆最小的Android源？ [英] how to use "repo" to clone minimal android source?
问题描述
我能够通过使用回购工具克隆Android源$ C $ C。然而，我想要做的是克隆源$ C $ C比有一个11GB的足迹更最小的方式。这似乎下载相关的每一个Android设备和每一个先前版本的东西。我试着想我可以检查出具体的分支像这样减少这种
I am able to clone the Android source code by using the "repo" tool. However, what I want to do is clone the source code in a more minimal way than having an 11GB footprint. It seems to download things related to every Android device and every prior release. I tried thought I could reduce this by checking out a specific branch like this:
repo init -u https://android.googlesource.com/platform/manifest -b android-4.0.1_r1
不过，最终发生的是，我仍然可以参与一切，只是在一个特定的快照（可以理解）。但是，有没有什么办法来限制克隆了多少？
However, what ends up happening is that I still get everything involved, just at a specific snapshot (understandable). But is there any way to limit the amount that is cloned?
推荐答案
Android的源代码树是由许多独立的Git仓库，里面全是由回购管理的。你不能真正降低的下载一个给定的Git仓库中的数据量。
The android source tree is made up of many separate git repositories, which are all managed by repo. You can't really reduce the amount of data that's downloaded for a given git repository.
但是，你只能下载git的回购协议可用，使用回购同步℃的子集;项目＆GT; 。即如果你只想要的框架/基础包，你应该能够做到回购同步框架/基，做最初的回购初始化后，。
However, you can only download a subset of the git repos that are available, using repo sync <project>. I.e. if you only wanted the frameworks/base package, you should be able to do repo sync frameworks/base, after doing the initial repo init.
如果你实际上是想打造虽然源，你可能要完整的东西。
If you are actually wanting to build the source though, you probably want the full thing.
您可能能够通过删除你不需要的设备库，以节省一两演出。您可以通过编辑＆LT做到这一点;源方式＆gt;。/回购/ manifest.xml文件和删除您不想要的设备的仓库
You might be able to save a gig or two by removing the device repositories that you don't need. You can do this by editing <source>/.repo/manifest.xml and removing the repositories for the devices you don't want.
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