如果有两个变量:

Object obj;
String methodName = "getName";

在不知道obj的类的情况下,我如何调用由methodName识别的方法?

被调用的方法没有参数,只有一个String返回值。它是Java bean的getter。


当前回答

如果多次调用,则可以使用Java 7中引入的新方法句柄。现在我们开始你的方法返回一个字符串:

Object obj = new Point( 100, 200 );
String methodName = "toString";  
Class<String> resultType = String.class;

MethodType mt = MethodType.methodType( resultType );
MethodHandle methodHandle = MethodHandles.lookup().findVirtual( obj.getClass(), methodName, mt );
String result = resultType.cast( methodHandle.invoke( obj ) );

System.out.println( result );  // java.awt.Point[x=100,y=200]

其他回答

对于jooR,它仅仅是:

on(obj).call(methodName /*params*/).get()

这里有一个更详细的例子:

public class TestClass {

    public int add(int a, int b) { return a + b; }
    private int mul(int a, int b) { return a * b; }
    static int sub(int a, int b) { return a - b; }

}

import static org.joor.Reflect.*;

public class JoorTest {

    public static void main(String[] args) {
        int add = on(new TestClass()).call("add", 1, 2).get(); // public
        int mul = on(new TestClass()).call("mul", 3, 4).get(); // private
        int sub = on(TestClass.class).call("sub", 6, 5).get(); // static
        System.out.println(add + ", " + mul + ", " + sub);
    }
}

这个打印:

3, 12, 1

对于那些想要Java 7中简单的代码示例的人:

狗类:

package com.mypackage.bean;

public class Dog {
    private String name;
    private int age;

    public Dog() {
        // empty constructor
    }

    public Dog(String name, int age) {
        this.name = name;
        this.age = age;
    }

    public String getName() {
        return name;
    }

    public void setName(String name) {
        this.name = name;
    }

    public int getAge() {
        return age;
    }

    public void setAge(int age) {
        this.age = age;
    }

    public void printDog(String name, int age) {
        System.out.println(name + " is " + age + " year(s) old.");
    }
}

ReflectionDemo类:

package com.mypackage.demo;

import java.lang.reflect.*;

public class ReflectionDemo {

    public static void main(String[] args) throws Exception {
        String dogClassName = "com.mypackage.bean.Dog";
        Class<?> dogClass = Class.forName(dogClassName); // convert string classname to class
        Object dog = dogClass.newInstance(); // invoke empty constructor

        String methodName = "";

        // with single parameter, return void
        methodName = "setName";
        Method setNameMethod = dog.getClass().getMethod(methodName, String.class);
        setNameMethod.invoke(dog, "Mishka"); // pass arg

        // without parameters, return string
        methodName = "getName";
        Method getNameMethod = dog.getClass().getMethod(methodName);
        String name = (String) getNameMethod.invoke(dog); // explicit cast

        // with multiple parameters
        methodName = "printDog";
        Class<?>[] paramTypes = {String.class, int.class};
        Method printDogMethod = dog.getClass().getMethod(methodName, paramTypes);
        printDogMethod.invoke(dog, name, 3); // pass args
    }
}

输出: 米什卡3岁了。


你可以用这种方法调用带形参的构造函数:

Constructor<?> dogConstructor = dogClass.getConstructor(String.class, int.class);
Object dog = dogConstructor.newInstance("Hachiko", 10);

或者,您可以删除

String dogClassName = "com.mypackage.bean.Dog";
Class<?> dogClass = Class.forName(dogClassName);
Object dog = dogClass.newInstance();

,做

Dog dog = new Dog();

Method method = Dog.class.getMethod(methodName, ...);
method.invoke(dog, ...);

推荐阅读:创建新类实例

首先,不要。避免使用这类代码。它往往是非常糟糕的代码,也不安全(请参阅安全编码指南的第6节) Java编程语言,版本2.0)。

如果你必须这样做,你更喜欢java。豆子要反射。Beans包装反射,允许相对安全和常规的访问。

对于那些在同一个类中调用非静态方法的方法,请参见下面的代码:

class Person {
    public void method1() {
        try {
            Method m2 = this.getClass().getDeclaredMethod("method2");
            m1.invoke(this);
        } catch (NoSuchMethodException e) {
            e.printStackTrace();
        } catch (IllegalAccessException e) {
            e.printStackTrace();
        } catch (InvocationTargetException e) {
            e.printStackTrace();
        }
    }

    public void method2() {
        // Do something
    }

}
//Step1 - Using string funClass to convert to class
String funClass = "package.myclass";
Class c = Class.forName(funClass);

//Step2 - instantiate an object of the class abov
Object o = c.newInstance();
//Prepare array of the arguments that your function accepts, lets say only one string here
Class[] paramTypes = new Class[1];
paramTypes[0]=String.class;
String methodName = "mymethod";
//Instantiate an object of type method that returns you method name
 Method m = c.getDeclaredMethod(methodName, paramTypes);
//invoke method with actual params
m.invoke(o, "testparam");