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

为了完成我同事的回答,你可能需要密切关注以下问题:

static or instance calls (in one case, you do not need an instance of the class, in the other, you might need to rely on an existing default constructor that may or may not be there) public or non-public method call (for the latter,you need to call setAccessible on the method within an doPrivileged block, other findbugs won't be happy) encapsulating into one more manageable applicative exception if you want to throw back the numerous java system exceptions (hence the CCException in the code below)

下面是一个旧的java1.4代码,它考虑了这些要点:

/**
 * Allow for instance call, avoiding certain class circular dependencies. <br />
 * Calls even private method if java Security allows it.
 * @param aninstance instance on which method is invoked (if null, static call)
 * @param classname name of the class containing the method 
 * (can be null - ignored, actually - if instance if provided, must be provided if static call)
 * @param amethodname name of the method to invoke
 * @param parameterTypes array of Classes
 * @param parameters array of Object
 * @return resulting Object
 * @throws CCException if any problem
 */
public static Object reflectionCall(final Object aninstance, final String classname, final String amethodname, final Class[] parameterTypes, final Object[] parameters) throws CCException
{
    Object res;// = null;
    try {
        Class aclass;// = null;
        if(aninstance == null)
        {
            aclass = Class.forName(classname);
        }
        else
        {
            aclass = aninstance.getClass();
        }
        //Class[] parameterTypes = new Class[]{String[].class};
    final Method amethod = aclass.getDeclaredMethod(amethodname, parameterTypes);
        AccessController.doPrivileged(new PrivilegedAction() {
    public Object run() {
                amethod.setAccessible(true);
                return null; // nothing to return
            }
        });
        res = amethod.invoke(aninstance, parameters);
    } catch (final ClassNotFoundException e) {
        throw new CCException.Error(PROBLEM_TO_ACCESS+classname+CLASS, e);
    } catch (final SecurityException e) {
        throw new CCException.Error(PROBLEM_TO_ACCESS+classname+GenericConstants.HASH_DIESE+ amethodname + METHOD_SECURITY_ISSUE, e);
    } catch (final NoSuchMethodException e) {
        throw new CCException.Error(PROBLEM_TO_ACCESS+classname+GenericConstants.HASH_DIESE+ amethodname + METHOD_NOT_FOUND, e);
    } catch (final IllegalArgumentException e) {
        throw new CCException.Error(PROBLEM_TO_ACCESS+classname+GenericConstants.HASH_DIESE+ amethodname + METHOD_ILLEGAL_ARGUMENTS+String.valueOf(parameters)+GenericConstants.CLOSING_ROUND_BRACKET, e);
    } catch (final IllegalAccessException e) {
        throw new CCException.Error(PROBLEM_TO_ACCESS+classname+GenericConstants.HASH_DIESE+ amethodname + METHOD_ACCESS_RESTRICTION, e);
    } catch (final InvocationTargetException e) {
    throw new CCException.Error(PROBLEM_TO_ACCESS+classname+GenericConstants.HASH_DIESE+ amethodname + METHOD_INVOCATION_ISSUE, e);
    } 
    return res;
}

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

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]

对我来说,一个非常简单和愚蠢的方法是简单地创建一个方法调用者,就像这样:

public static object methodCaller(String methodName)
{
    if(methodName.equals("getName"))
        return className.getName();
}

然后当你需要调用这个方法时,简单地输入如下内容

//calling a toString method is unnessary here, but i use it to have my programs to both rigid and self-explanitory 
System.out.println(methodCaller(methodName).toString()); 
//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");

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);
     }
  }
}