我不认为你可以，Java在编译时使用类型擦除，这样你的代码就可以与在泛型之前创建的应用程序和库兼容。

来自Oracle文档:

Type Erasure Generics were introduced to the Java language to provide tighter type checks at compile time and to support generic programming. To implement generics, the Java compiler applies type erasure to: Replace all type parameters in generic types with their bounds or Object if the type parameters are unbounded. The produced bytecode, therefore, contains only ordinary classes, interfaces, and methods. Insert type casts if necessary to preserve type safety. Generate bridge methods to preserve polymorphism in extended generic types. Type erasure ensures that no new classes are created for parameterized types; consequently, generics incur no runtime overhead.

http://docs.oracle.com/javase/tutorial/java/generics/erasure.html

我见过类似的东西

private Class<T> persistentClass;

public Constructor() {
    this.persistentClass = (Class<T>) ((ParameterizedType) getClass()
                            .getGenericSuperclass()).getActualTypeArguments()[0];
 }

GenericDataAccessObjects实例

当然，你可以。

出于向后兼容性的考虑，Java在运行时不使用这些信息。但是信息实际上是以元数据的形式呈现的，并且可以通过反射访问(但是它仍然不用于类型检查)。

来自官方API:

http://download.oracle.com/javase/6/docs/api/java/lang/reflect/ParameterizedType.html#getActualTypeArguments%28%29

但是，对于您的场景，我不会使用反射。我个人更倾向于将其用于框架代码。在你的例子中，我只是将类型作为构造函数参数添加。

这是我的解决方案:

import java.lang.reflect.Type;
import java.lang.reflect.TypeVariable;

public class GenericClass<T extends String> {

  public static void main(String[] args) {
     for (TypeVariable typeParam : GenericClass.class.getTypeParameters()) {
      System.out.println(typeParam.getName());
      for (Type bound : typeParam.getBounds()) {
         System.out.println(bound);
      }
    }
  }
}

为了完成这里的一些答案，我必须得到MyGenericClass的paramtrizedtype，不管层次结构有多高，在递归的帮助下:

private Class<T> getGenericTypeClass() {
        return (Class<T>) (getParametrizedType(getClass())).getActualTypeArguments()[0];
}

private static ParameterizedType getParametrizedType(Class clazz){
    if(clazz.getSuperclass().equals(MyGenericClass.class)){ // check that we are at the top of the hierarchy
        return (ParameterizedType) clazz.getGenericSuperclass();
    } else {
        return getParametrizedType(clazz.getSuperclass());
    }
}

我做了相同的@Moesio上面，但在Kotlin可以这样做:

class A<T : SomeClass>() {

    var someClassType : T

    init(){
    this.someClassType = (javaClass.genericSuperclass as ParameterizedType).actualTypeArguments[0] as Class<T>
    }

}