java 获取程序路径,如何获取运行java程序的路径

Is there a way to get the path of main class of the running java program.

structure is

D:/

|---Project

|------bin

|------src

I want to get the path as D:\Project\bin\.

I tried System.getProperty("java.class.path"); but the problem is, if I run like

java -classpath D:\Project\bin;D:\Project\src\ Main

Output

Getting : D:\Project\bin;D:\Project\src\

Want : D:\Project\bin

Is there any way to do this?

===== EDIT =====

Got the solution here

package foo;

public class Test

{

public static void main(String[] args)

{

ClassLoader loader = Test.class.getClassLoader();

System.out.println(loader.getResource("foo/Test.class"));

}

}

This printed out:

file:/C:/Users/Jon/Test/foo/Test.class

URL main = Main.class.getResource("Main.class");

if (!"file".equalsIgnoreCase(main.getProtocol()))

throw new IllegalStateException("Main class is not stored in a file.");

File path = new File(main.getPath());

Note that most class files are assembled into JAR files so this won't work in every case (hence the IllegalStateException). However, you can locate the JAR that contains the class with this technique, and you can get the content of the class file by substituting a call to getResourceAsStream() in place of getResource(), and that will work whether the class is on the file system or in a JAR.

解决方案

Try this code:

final File f = new File(MyClass.class.getProtectionDomain().getCodeSource().getLocation().getPath());

replace 'MyClass' with your class containing the main method.

Alternatively you can also use

System.getProperty("java.class.path")

Above mentioned System property provides

Path used to find directories and JAR archives containing class files.

Elements of the class path are separated by a platform-specific

character specified in the path.separator property.