几年前我就这样做了。最可靠的方法(即使用官方Java api，没有外部依赖)是编写一个自定义doclet来生成一个可以在运行时读取的列表。

你可以像这样从命令行运行它:

javadoc -d build -doclet com.example.ObjectListDoclet -sourcepath java/src -subpackages com.example

或者像这样从ant运行它:

<javadoc sourcepath="${src}" packagenames="*" >
  <doclet name="com.example.ObjectListDoclet" path="${build}"/>
</javadoc>

下面是基本代码:

public final class ObjectListDoclet {
    public static final String TOP_CLASS_NAME =  "com.example.MyClass";        

    /** Doclet entry point. */
    public static boolean start(RootDoc root) throws Exception {
        try {
            ClassDoc topClassDoc = root.classNamed(TOP_CLASS_NAME);
            for (ClassDoc classDoc : root.classes()) {
                if (classDoc.subclassOf(topClassDoc)) {
                    System.out.println(classDoc);
                }
            }
            return true;
        }
        catch (Exception ex) {
            ex.printStackTrace();
            return false;
        }
    }
}

为了简单起见，我删除了命令行参数解析，并将其写入System。Out而不是file。

如果你想加载给定类在同一个包中的所有子类，你可以这样做:

public static List<Class> loadAllSubClasses(Class pClazz) throws IOException, ClassNotFoundException {
    ClassLoader classLoader = pClazz.getClassLoader();
    assert classLoader != null;
    String packageName = pClazz.getPackage().getName();
    String dirPath = packageName.replace(".", "/");
    Enumeration<URL> srcList = classLoader.getResources(dirPath);

    List<Class> subClassList = new ArrayList<>();
    while (srcList.hasMoreElements()) {
        File dirFile = new File(srcList.nextElement().getFile());
        File[] files = dirFile.listFiles();
        if (files != null) {
            for (File file : files) {
                String subClassName = packageName + '.' + file.getName().substring(0, file.getName().length() - 6);
                if (! subClassName.equals(pClazz.getName())) {
                    subClassList.add(Class.forName(subClassName));
                }
            }
        }
    }

    return subClassList;
}

我需要将此作为一个测试用例，以查看是否向代码中添加了新的类。这就是我所做的

final static File rootFolder = new File(SuperClass.class.getProtectionDomain().getCodeSource().getLocation().getPath());
private static ArrayList<String> files = new ArrayList<String>();
listFilesForFolder(rootFolder); 

@Test(timeout = 1000)
public void testNumberOfSubclasses(){
    ArrayList<String> listSubclasses = new ArrayList<>(files);
    listSubclasses.removeIf(s -> !s.contains("Superclass.class"));
    for(String subclass : listSubclasses){
        System.out.println(subclass);
    }
    assertTrue("You did not create a new subclass!", listSubclasses.size() >1);     
}

public static void listFilesForFolder(final File folder) {
    for (final File fileEntry : folder.listFiles()) {
        if (fileEntry.isDirectory()) {
            listFilesForFolder(fileEntry);
        } else {
            files.add(fileEntry.getName().toString());
        }
    }
}

在java中输入链接描述hereService Manager将获得J中接口的所有实现类

还要注意的是，这当然只能找到当前类路径中存在的所有子类。想必这对于你目前所看到的是可以的，而且你也有可能考虑过这一点，但如果你在任何时候发布了一个非最终类(对于不同程度的“狂野”)，那么其他人已经编写了他们自己的子类，而你不知道这是完全可行的。

Thus if you happened to be wanting to see all subclasses because you want to make a change and are going to see how it affects subclasses' behaviour - then bear in mind the subclasses that you can't see. Ideally all of your non-private methods, and the class itself should be well-documented; make changes according to this documentation without changing the semantics of methods/non-private fields and your changes should be backwards-compatible, for any subclass that followed your definition of the superclass at least.

将它们添加到父类构造函数(this. getclass (). getname())内部的静态映射(或创建一个默认映射)，但这将在运行时更新。如果可以选择延迟初始化，可以尝试这种方法。