0

我有一个类，我在静态块中初始化了 hashmap。通过密钥，我检索了一个类的值。为了为这个类创建对象。我已经使用构造函数类来获取构造函数并传递参数和创建的对象。

我在哈希图中有两个班级。要为 EchoExpression 创建对象，我需要传递两个参数，而对于 OutExpression 类，我只需要传递一个参数（字符串）。

问题：

根据键返回的类，我需要执行要获取和实现的构造函数，无论是带有一个参数还是两个参数的构造函数。

public class ExampleFactory {

  private static HashMap<String,Class<?>> hmap = new HashMap<String,Class<?>>();

  static
  {                   
      hmap.put("echo", EchoExpression.class);         
      hmap.put("Out", OutExpression.class);                       
  }

  public void getExpo(String key,String expression)
  {
    Class aClass =map.get(key);

    //Constructor implementation for OutExpression where only one argument string is passed

    Constructor constructor = aClass.getConstructor(new Class[]{String.class});

    Object object= constructor.newInstance(expression);

    //constructor for passing two arguments string for EchoExpression

    Constructor constructor = aClass.getConstructor(new Class[]{String.class,Class.class});

    Object object= constructor.newInstance(expression, Boolean.class);


    return null;        
  }                
}

如何在不使用 if else 的情况下从值（类）中选择要实现的类？

4

2 回答 2

1

使用枚举并打开它。这是一个可执行存根，没有深入了解示例的反射或语法：

package com.trip.test;

import java.util.HashMap;
import java.util.Map;

public class ExampleFactory {

    private static Map<String, Class<?>> hmap = new HashMap<String, Class<?>>();

    static {
        hmap.put("echo", EchoExpression.class);
        hmap.put("Out", OutExpression.class);
    }

    public static void getExpo(String key, String expression) {

        Class aClass = hmap.get(key);

        ClassMappingEnum myType = ClassMappingEnum.getClassMappingEnum(aClass);

        switch (myType) {
        case ECHO_EXPRESSION:{
            System.out.println(aClass.getName());
            // do something
            break;
        }
        case OUT_EXPRESSION:{
            System.out.println(aClass.getName());
            // do something
            break;          
        }
        case UNKNOWN:
        default:
            System.out.println("Bummer: " + aClass.getName());          
        }

    }

    public static void main(String[] args) {
        getExpo("echo", "B");
        getExpo("Out", "B");
    }   
}

enum ClassMappingEnum {
    ECHO_EXPRESSION(EchoExpression.class), OUT_EXPRESSION(OutExpression.class), UNKNOWN(null);

    private Class typeDes;

    private ClassMappingEnum(Class typeDes) {
        this.typeDes = typeDes;
    }

    public static ClassMappingEnum getClassMappingEnum(Class compare) {
        for (ClassMappingEnum cme : ClassMappingEnum.values()) {
            if (cme.typeDes.equals(compare)) {
                return cme;
            }
        }
        return UNKNOWN;
    }


}

class EchoExpression<T> {
    private String someString;
    private Class<T> someClass;
    public EchoExpression(String someString, Class<T> someClass) {
        super();
        this.someString = someString;
        this.someClass = someClass;
    }
    public String getSomeString() {
        return someString;
    }
    public void setSomeString(String someString) {
        this.someString = someString;
    }
    public Class<T> getSomeClass() {
        return someClass;
    }
    public void setSomeClass(Class<T> someClass) {
        this.someClass = someClass;
    }


}

class OutExpression {
    private String someString;

    public OutExpression(String someString) {
        super();
        this.someString = someString;
    }

    public String getSomeString() {
        return someString;
    }

    public void setSomeString(String someString) {
        this.someString = someString;
    }

}
于 2012-04-05T00:02:40.420 回答
0

如果您可以修改类，以便两个构造函数具有相同的签名（以相同的顺序接受相同数量/类型的参数），您可以这样做

Constructor constructor = aClass.getConstructor(new Class[]{String.class,Class.class});
Object object= constructor.newInstance(expression, Boolean.class);

两个班级。

这当然意味着现在不需要额外参数的类将不得不在更改之后忽略它之前未使用的传入参数

更新：这是使用工厂类实现该想法的一种可能方法：

public interface ObjectFactory
{
  Object create(String expr, Class cls);
}

public class EchoExpressionFactory implements ObjectFactory
{
  public EchoExpression create(String expr, Class cls)
  {
    return new EchoExpression(expr, cls);
  }
}

public class OutExpressionFactory implements ObjectFactory
{
  public OutExpression create(String expr, Class cls)
  {
    return new OutExpression(expr);
  }
}

public class ExampleFactory { 

  private static HashMap<String,ObjectFactory> hmap = new HashMap<String,ObjectFactory>(); 

  static 
  {                    
      hmap.put("echo", new EchoExpressionFactory());          
      hmap.put("Out", new OutExpressionFactory());                        
  } 

  public void getExpo(String key,String expression) 
  { 
    ObjectFactory factory = map.get(key); 

    //Constructor implementation for Expression
    Object object = factory.create(expression); 

    Object object= constructor.newInstance(expression, Boolean.class); 

    return;
  }                 
}
于 2012-04-04T22:47:56.190 回答