我正在做一个项目，一个要求是如果 main 方法的第二个参数以/"(对于 linux)开头，它应该将其视为绝对路径(不是问题)，但如果它不以/"开头，它应该获取类的当前工作路径，并将给定的参数附加到它上面.
I'm working on a project and one requirement is if the 2nd argument for the main method starts with "/" (for linux) it should consider it as an absolute path (not a problem), but if it doesn't start with "/", it should get the current working path of the class and append to it the given argument.
我可以通过以下几种方式获取类名:System.getProperty("java.class.path")、new File(".") 和 >getCanonicalPath() 等等……
I can get the class name in several ways: System.getProperty("java.class.path"), new File(".") and getCanonicalPath(), and so on...
问题是，这只给了我存储包的目录 - 即，如果我有一个存储在.../project/this/is/package/name"中的类，它只会给我 "/project/" 并忽略实际 .class 文件 所在的包名.
The problem is, this only gives me the directory in which the packages are stored - i.e. if I have a class stored in ".../project/this/is/package/name", it would only give me "/project/" and ignores the package name where the actual .class files lives.
有什么建议吗?
这是从练习描述中摘录的解释
Here's the explanation, taken from the exercise description
sourcedir 可以是绝对的(以/"开头)或相对于我们运行程序的位置
sourcedir can be either absolute (starting with "/") or relative to where we run the program from
sourcedir 是 main 方法的给定参数.我怎样才能找到那条路?
sourcedir is a given argument for the main method. how can I find that path?
使用 this.getClass().getCanonicalName() 获取完整的类名.
Use this.getClass().getCanonicalName() to get the full class name.
请注意，包/类名 ("abC") 与 .class 文件 (a/b/C.class) 的路径不同，使用包名/类名派生路径通常是不好的做法.类文件/包的集合可以在多个不同的类路径中，可以是目录，也可以是jar文件.
Note that a package / class name ("a.b.C") is different from the path of the .class files (a/b/C.class), and that using the package name / class name to derive a path is typically bad practice. Sets of class files / packages can be in multiple different class paths, which can be directories or jar files.
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