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我有一个跨模型平台项目,我正在尝试在单独的存储库中具有平台特异性的Android和BlackBerry Eclipse项目,请使用包含J2Me Platform-generic-generic-generic-generic-generic src和Test Code的git子模块。
这是预期的布局:
team-lib-j2me
| - src (platform generic source code)
| | - ...
| - test
| - src (platform generic source code)
| - ...
team-lib-android
| - team-lib-j2me (git submodule)
| - lib (Eclipse Android project)
| | - AndroidManifest.xml
| | - src (platform specific)
| | - ...
| - test (Eclipse Android Project)
| - AndroidManifest.xml
| - src (platform specific)
| - ...
team-lib-blackberry
| - team-lib-j2me (git submodule)
| - lib (Eclipse BB Project)
| | - BlackBerry_App_Descriptor.xml
| | - src (platform specific)
| | - ...
| - test (Eclipse BB Project)
| - BlackBerry_App_Descriptor.xml
| - src (platform specific)
| - ...
我似乎找不到一种干净,简单的方法:
- 获取任何Eclipse" Lib"项目以供链接到相对路径的链接。
- 获取任何一个日食"测试"项目以源链接到相对路径。
关于如何执行此操作的任何建议?
谢谢!
我刚刚创建了一个新的路径变量" team_lib_j2me" =" $ {project_loc} .. team-lib-j2me",然后创建了"链接" s to:
- " team-lib-j2me-src" = team_lib_j2me src
- " team-lib-j2me-test-src" = team_lib_j2me test src