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

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
    }

}

其他回答

Method method = someVariable.class.getMethod(SomeClass);
String status = (String) method.invoke(method);

SomeClass是类,somvariable是变量。

请参考以下代码,可能对您有帮助。

public static Method method[];
public static MethodClass obj;
public static String testMethod="A";

public static void main(String args[]) 
{
    obj=new MethodClass();
    method=obj.getClass().getMethods();
    try
    {
        for(int i=0;i<method.length;i++)
        {
            String name=method[i].getName();
            if(name==testMethod)
            {   
                method[i].invoke(name,"Test Parameters of A");
            }
        }
    }
    catch(Exception ex)
    {
        System.out.println(ex.getMessage());
    }
}

谢谢……

这里是准备使用的方法:

调用一个不带参数的方法:

public static void callMethodByName(Object object, String methodName) throws IllegalAccessException, InvocationTargetException, NoSuchMethodException {
    object.getClass().getDeclaredMethod(methodName).invoke(object);
}

使用Arguments调用一个方法:

    public static void callMethodByName(Object object, String methodName, int i, String s) throws IllegalAccessException, InvocationTargetException, NoSuchMethodException {
        object.getClass().getDeclaredMethod(methodName, int.class, String.class).invoke(object, i, s);
    }

使用上述方法如下:

package practice;

import java.io.IOException;
import java.lang.reflect.InvocationTargetException;

public class MethodInvoke {

    public static void main(String[] args) throws ClassNotFoundException, NoSuchMethodException, SecurityException, IllegalAccessException, IllegalArgumentException, InvocationTargetException, IOException {
        String methodName1 = "methodA";
        String methodName2 = "methodB";
        MethodInvoke object = new MethodInvoke();
        callMethodByName(object, methodName1);
        callMethodByName(object, methodName2, 1, "Test");
    }

    public static void callMethodByName(Object object, String methodName) throws IllegalAccessException, InvocationTargetException, NoSuchMethodException {
        object.getClass().getDeclaredMethod(methodName).invoke(object);
    }

    public static void callMethodByName(Object object, String methodName, int i, String s) throws IllegalAccessException, InvocationTargetException, NoSuchMethodException {
        object.getClass().getDeclaredMethod(methodName, int.class, String.class).invoke(object, i, s);
    }

    void methodA() {
        System.out.println("Method A");
    }

    void methodB(int i, String s) {
        System.out.println("Method B: "+"\n\tParam1 - "+i+"\n\tParam 2 - "+s);
    }
}

输出:

Method A  
Method B:  
	Param1 - 1  
	Param 2 - Test

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

索引(快)

您可以使用FunctionalInterface将方法保存在容器中以索引它们。您可以使用数组容器通过数字调用它们,或者使用hashmap通过字符串调用它们。通过这个技巧,可以为方法建立索引,从而更快地动态调用它们。

@FunctionalInterface
public interface Method {
    double execute(int number);
}

public class ShapeArea {
    private final static double PI = 3.14;

    private Method[] methods = {
        this::square,
        this::circle
    };

    private double square(int number) {
        return number * number;
    }

    private double circle(int number) {
        return PI * number * number;
    }

    public double run(int methodIndex, int number) {
        return methods[methodIndex].execute(number);
    }
}

λ语法

你也可以使用lambda语法:

public class ShapeArea {
    private final static double PI = 3.14;

    private Method[] methods = {
        number -> {
            return number * number;
        },
        number -> {
            return PI * number * number;
        },
    };

    public double run(int methodIndex, int number) {
        return methods[methodIndex].execute(number);
    }
}

编辑2022

刚才我在想为你提供一个通用的解决方案,与所有可能的方法与变量数:

@FunctionalInterface
public interface Method {
    Object execute(Object ...args);
}

public class Methods {
    private Method[] methods = {
        this::square,
        this::rectangle
    };

    private double square(int number) {
        return number * number;
    }

    private double rectangle(int width, int height) {
        return width * height;
    }

    public Method run(int methodIndex) {
        return methods[methodIndex];
    }
}

用法:

methods.run(1).execute(width, height);