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。

So one way is to doExampleEnum valueOfOrdinal = ExampleEnum.values()[ordinal]; which works and its easy, however, as mentioned before, ExampleEnum.values() returns a new cloned array for every call. That can be unnecessarily expensive. We can solve that by caching the array like so ExampleEnum[] values = values(). It is also "dangerous" to allow our cached array to be modified. Someone could write ExampleEnum.values[0] = ExampleEnum.type2; So I would make it private with an accessor method that does not do extra copying.

private enum ExampleEnum{
    type0, type1, type2, type3;
    private static final ExampleEnum[] values = values();
    public static ExampleEnum value(int ord) {
        return values[ord];
    }
}

您可以使用exampleenumeration .value(ordinal)来获取与ordinal关联的枚举值

这几乎肯定是个坏主意。当然，如果序号实际上是持久化的(例如，因为有人已将URL标记为书签)——这意味着将来必须始终保留枚举顺序，这对代码维护者来说可能不太明显。

为什么不使用myEnumValue.name()(和解码通过reporttypeenume . valueof (s))来编码枚举?

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。

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];
}

你可以定义一个简单的方法:

public enum Alphabet{
    A,B,C,D;

    public static Alphabet get(int index){
        return Alphabet.values()[index];
    }
}

像这样使用它:

System.out.println(Alphabet.get(2));