我如何调用Java方法时，给予方法名作为字符串?

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

这听起来像是Java Reflection包可以做到的事情。

http://java.sun.com/developer/technicalArticles/ALT/Reflection/index.html

特别是在按名称调用方法下面:

进口数组;*;

public class method2 {
  public int add(int a, int b)
  {
     return a + b;
  }

  public static void main(String args[])
  {
     try {
       Class cls = Class.forName("method2");
       Class partypes[] = new Class[2];
        partypes[0] = Integer.TYPE;
        partypes[1] = Integer.TYPE;
        Method meth = cls.getMethod(
          "add", partypes);
        method2 methobj = new method2();
        Object arglist[] = new Object[2];
        arglist[0] = new Integer(37);
        arglist[1] = new Integer(47);
        Object retobj 
          = meth.invoke(methobj, arglist);
        Integer retval = (Integer)retobj;
        System.out.println(retval.intValue());
     }
     catch (Throwable e) {
        System.err.println(e);
     }
  }
}

2008-10-02 05:22:24

其他回答

从反射使用方法调用:

Class<?> c = Class.forName("class name");
Method method = c.getDeclaredMethod("method name", parameterTypes);
method.invoke(objectToInvokeOn, params);

地点:

“class name”是类的名称 objectToInvokeOn类型为Object，是您希望在其上调用方法的对象 "method name"是你想要调用的方法的名称 parameterTypes的类型是Class[]，声明方法所接受的参数 params类型为Object[]，声明要传递给方法的参数

2008-10-02 05:21:20

//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");

2013-04-30 05:46:18

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。

2008-10-02 14:08:40

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

2019-02-26 10:32:47