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。
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
其他回答
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());
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
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……
这对我来说很有效:
public class MethodInvokerClass {
public static void main(String[] args) throws NoSuchMethodException, SecurityException, IllegalAccessException, IllegalArgumentException, ClassNotFoundException, InvocationTargetException, InstantiationException {
Class c = Class.forName(MethodInvokerClass.class.getName());
Object o = c.newInstance();
Class[] paramTypes = new Class[1];
paramTypes[0]=String.class;
String methodName = "countWord";
Method m = c.getDeclaredMethod(methodName, paramTypes);
m.invoke(o, "testparam");
}
public void countWord(String input){
System.out.println("My input "+input);
}
}
输出:
我的输入testparam
我可以通过将其名称传递给另一个方法(如main)来调用该方法。