解决java switch……case不能匹配字符串的问题

java1.7已经支持了匹配字符串

方案1.

 enum Animal {  

     dog,cat,bear;

      public static Animal getAnimal(String animal){  

        return valueOf(animal );  

     }   

 }

 public class Client {  

    public void caseAnimal(String animal){  

        switch(Animal.getAnimal(animal)){  

        case cat:  

            System.out.println("this is a cat");  

            break;  

        case dog:  

            System.out.println("this is a dog");  

            break;  

        case bear:  

            System.out.println("this is a bear");  

            break;  

        }  

    }

   public static void main(String[] args) {

       Client client = new Client();  

        client.caseAnimal("cat"); 

 } 

 } 

方案2

 enum Seasons{

          SPRING,SUMMER,AUTUMN,WINTER

 }

 public class TestMultiInterface{

          public static void main(String[] args){

                    Seasons str=Seasons.SPRING;

                    switch(str.ordinal()){

                    case 0:

                    System.out.println(str.toString());

                    break;

                    case 1:

                    System.out.println(str.toString());

                    break;

                    default:

                    System.out.println("没找到");

                    }

          }

 }

方案3

 public void switchCase() { 

   Map<String,Integer> map=new HashMap<String, Integer>(); 

   map.put("中国", 1); 

   map.put("美国", 2); 

   map.put("英国", 3); 

   map.put("法国", 4); 

   String str="中国"; 

   switch(map.get(str)) 

   {       

    case 1: 

         System.out.println("中国"); 

         break; 

     case 2: 

         System.out.println("美国"); 

        break; 

     case 3:

         System.out.println("英国");

        break;

     case 4: 

        System.out.println("法国"); 

        break;   

    default: 

     System.out.println("default"); 

   } 

  } 

方案4

 String str1 = "aa";

         String[] str2 = {"bb", "aa", "cc", "aaa"};

         for (int i = 0; i < str2.length; ++i){

             switch (str1.compareTo(str2[i])) {

               case 0: System.out.println("OK");

                       break;

               default: System.out.println("Error");

                         break;

             }

         }

方案5

 public class Client {

          public Client(String animal) {

                    //使用String的hasCode取得字符串的哈希码,此方法要保证哈希码不重复

                    switch (animal.hashCode()) {

                    case 98262:

                             System.out.println("this is a cat");

                             break;

                    case 99644:

                             System.out.println("this is a dog");

                             break;

                    case 3019700:

                             System.out.println("this is a bear");

                             break;

                    }

          }

          public static void main(String[] args) {

                    Client client = new Client("bear");

          }

方案6  如果是在使用简单工厂设计模式,可以考虑利用java反射技术代替switch

 public static UserDao createUser(){
Animal animal=null;
try {
iu=(Animal) Class.forName(Animal接口下子类的完整类名).newInstance();
}catch (Exception e) {
e.printStackTrace();
}
return animal; }

以上方案,有的是从网上收集来的,有的是自己原创的,因为时间过长,已经不记得从哪个地方摘抄下来的,没有注明转载的地方,十分抱歉!

上一篇:classPath与PATH


下一篇:python全栈开发day34-线程Thread