通过这种方式，您可以不依赖于编译时泛型解析(因此您可以随时创建枚举类实例，即使是使用class . formame创建的类型)

public Object getInstance(Class enumClazz, int ordinal) throws Exception {
    Object[] allEnums = enumClazz.getDeclaredMethod("values", Object[].class).invoke(null, null);
    return allEnums[ordinal];
}

public enum Suit implements java.io.Serializable, Comparable<Suit>{
  spades, hearts, diamonds, clubs;
  private static final Suit [] lookup  = Suit.values();
  public Suit fromOrdinal(int ordinal) {
    if(ordinal< 1 || ordinal> 3) return null;
    return lookup[value-1];
  }
}

测试类

public class MainTest {
    public static void main(String[] args) {
        Suit d3 = Suit.diamonds;
        Suit d3Test = Suit.fromOrdinal(2);
        if(d3.equals(d3Test)){
            System.out.println("Susses");
        }else System.out.println("Fails");
    }
}

我很感激你与我们分享，如果你有一个更有效的代码，我的enum是巨大的，不断调用数千次。

我同意大多数人的观点，使用序数可能是一个坏主意。我通常通过给枚举一个私有构造函数来解决这个问题，该构造函数可以以一个DB值为例，然后创建一个静态fromDbValue函数，类似于Jan的回答。

public enum ReportTypeEnum {
    R1(1),
    R2(2),
    R3(3),
    R4(4),
    R5(5),
    R6(6),
    R7(7),
    R8(8);

    private static Logger log = LoggerFactory.getLogger(ReportEnumType.class);  
    private static Map<Integer, ReportTypeEnum> lookup;
    private Integer dbValue;

    private ReportTypeEnum(Integer dbValue) {
        this.dbValue = dbValue;
    }


    static {
        try {
            ReportTypeEnum[] vals = ReportTypeEnum.values();
            lookup = new HashMap<Integer, ReportTypeEnum>(vals.length);

            for (ReportTypeEnum  rpt: vals)
                lookup.put(rpt.getDbValue(), rpt);
         }
         catch (Exception e) {
             // Careful, if any exception is thrown out of a static block, the class
             // won't be initialized
             log.error("Unexpected exception initializing " + ReportTypeEnum.class, e);
         }
    }

    public static ReportTypeEnum fromDbValue(Integer dbValue) {
        return lookup.get(dbValue);
    }

    public Integer getDbValue() {
        return this.dbValue;
    }

}

现在您可以在不更改查找的情况下更改顺序，反之亦然。

有一种简单和坏的方法，也有一种相当简单和正确的方法。

首先，简单的和坏的(这些通常很受欢迎)。枚举类方法通过values()方法返回所有可用实例的数组，您可以通过数组索引访问枚举对象。

RenderingMode mode = RenderingMode.values()[index];

//Enum Class somewhere else
public enum RenderingMode
{
    PLAYING,
    PREVIEW,
    VIEW_SOLUTION;    
    
}
    

//RenderingMode.values()[0] will return RenderingMode.PLAYING
//RenderingMode.values()[1] will return RenderingMode.PREVIEW
//Why this is bad? Because it is linked to order of declaration.

//If you later changed the order here, it will impact all your existing logic around this.
public enum RenderingMode
    {
        
        PREVIEW,
        VIEW_SOLUTION,
        PLAYING;    
        
    }
 //Now
//RenderingMode.values()[0] will return RenderingMode.PREVIEW
//RenderingMode.values()[1] will return RenderingMode.VIEW_SOLUTION

这是正确的做法。 在枚举类中创建一个静态方法fromInt。

public enum RenderingMode
{
    PLAYING,
    PREVIEW,
    VIEW_SOLUTION;

    public static RenderingModefromInt(int index)
    {
       //this is independent of order of declaration
        switch (index)
        {
            case 0: return PLAYING;
            case 1: return PREVIEW;
            case 2: return VIEW_SOLUTION;
        }
        //Consider throwing Exception here

        return null;
    }
}

这是我在Android上使用Proguard所做的:

public enum SomeStatus {
    UNINITIALIZED, STATUS_1, RESERVED_1, STATUS_2, RESERVED_2, STATUS_3;//do not change order

    private static SomeStatus[] values = null;
    public static SomeStatus fromInteger(int i) {
        if(SomeStatus.values == null) {
            SomeStatus.values = SomeStatus.values();
        }
        if (i < 0) return SomeStatus.values[0];
        if (i >= SomeStatus.values.length) return SomeStatus.values[0];
        return SomeStatus.values[i];
    }
}

它很短，我不需要担心在Proguard有一个异常

如果我要大量使用values():

enum Suit {
   Hearts, Diamonds, Spades, Clubs;
   public static final Suit values[] = values();
}

同时wherever.java:

Suit suit = Suit.values[ordinal];

如果你想要数组是私有的，请自便:

private static final Suit values[] = values();
public static Suit get(int ordinal) { return values[ordinal]; }

...

Suit suit = Suit.get(ordinal);

注意数组的边界。