如何在Java中获取当前类名包括包名？
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How to get current class name including package name in Java?
提问by La bla bla
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 pathof the class and append to it the given argument.
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“ /”开头（对于 linux），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。
I can get the class name in several ways: System.getProperty("java.class.path"), new File(".")and getCanonicalPath(), and so on...
我可以通过多种方式获取类名：System.getProperty("java.class.path")、new File(".")和getCanonicalPath()，等等...
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 fileslives.
问题是，这只会给我存储包的目录 - 即如果我有一个存储在“ .../project/this/is/package/name”中的类，它只会给我“ /project/”并忽略实际所在的包名称.class files。
Any suggestions?
有什么建议？
EDIT: Here's the explanation, taken from the exercise description
编辑：这是来自练习描述的解释
sourcedir can be either absolute (starting with “/”) or relative to where we run the program from
sourcedir 可以是绝对的（以“/”开头）或相对于我们运行程序的位置
sourcedir is a given argument for the main method. how can I find that path?
sourcedir 是 main 方法的给定参数。我怎样才能找到那条路？
采纳答案by The Nail
Use this.getClass().getCanonicalName()to get the full class name.
使用this.getClass().getCanonicalName()来获得完整的类名。
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.
请注意，包/类名（“abC”）与 .class 文件的路径（a/b/C.class）不同，使用包名/类名来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，可以是目录或 jar 文件。
回答by Jon Egeland
There is a class, Class, that can do this:
有一个类，Class可以做到这一点：
Class c = Class.forName("MyClass"); // if you want to specify a class
Class c = this.getClass(); // if you want to use the current class
System.out.println("Package: "+c.getPackage()+"\nClass: "+c.getSimpleName()+"\nFull Identifier: "+c.getName());
If crepresented the class MyClassin the package mypackage, the above code would print:
如果c表示MyClass包中的类mypackage，则上面的代码将打印：
Package: mypackage
Class: MyClass
Full Identifier: mypackage.MyClass
包：mypackage
类：MyClass
完整标识符：mypackage.MyClass
You can take this information and modify it for whatever you need, or go check the APIfor more information.
您可以获取此信息并根据需要对其进行修改，或者查看 API以获取更多信息。
回答by Bozho
The fully-qualified name is opbtained as follows:
完全限定名称的获取方式如下：
String fqn = YourClass.class.getName();
But you need to read a classpath resource. So use
但是您需要读取类路径资源。所以用
InputStream in = YourClass.getResourceAsStream("resource.txt");
回答by Zahangir Alam
use this.getClass().getName()to get packageName.classNameand use this.getClass().getSimpleName()to get only class name
use this.getClass().getName()to getpackageName.className和 use this.getClass().getSimpleName()to get only class name
