Question

鉴于我有一个“空”界面：

public interface Color {
}

还有一个实现上述接口的枚举类：

public enum Fruits implements Color  {

    APPLE,
    APRICOT,
    AVOCADO,
    BANANA,
    BLACKBERRY,
    BLUEBERRY,
    CHERRY,
    COCONUT,
    CRANBERRY,
    DATE,
    DURIAN,
    ELDERBERRY,
    FIG,
    JACKFRUIT,
    KIWI,
    LEMON,
    LIME,
    LYCHEE,
    MANGO,
    MELLON,
    ORANGE,
    PAPAYA,
    PASSION_FRUIT,
    PEAR,
    PLUM,
    PINEAPPLE,
    POMELO    
}

经过一段时间的编码，我回到那个界面并添加一个方法，例如：

public interface Color {
    Color getColor();
}

IntelliJ Idea肯定会告诉我，每个枚举值都应实现新添加的方法。我现在能做的就是跳到每个枚举值（使用F2作为跳到每个值的快捷方式），然后转到ALT + ENTER创建它的默认实现。

APPLE {
    @Override
    public Color getColor() {
        return null;
    }

},
APRICOT {
    @Override
    public Color getColor() {
        return null;
    }
},
AVOCADO {
    @Override
    public Color getColor() {
        return null;
    }
},
...

一旦我意识到实现此接口的不仅仅是这个枚举，这将很快失去控制。

说明

我正在 IntelliJ Idea 中搜索 键盘快捷键 ，该IDE提供了生成代码的功能一次创建每个枚举值的空方法树桩。

IntelliJ Idea 确实为每个枚举值分别提供了这种功能，这引出了我的问题：可以一次为所有值完成此操作吗？

Answer 1

可以。只需在界面上创建一个default method：

default String getSomething(Any any) {
    return null;
}

Answer 2

您可以在枚举类本身中提供默认的实现，而不是在每个枚举值中提供：

public enum DirectCallRate implements KeyPerformanceIndicator  {

    COUNT,
    COUNT_ANSWERED,
    COUNT_UNANSWERED,
    COUNT_EXTERNAL,
    COUNT_ANSWERED_EXTERNAL,
    COUNT_UNANSWERED_EXTERNAL,
    COUNT_INTERNAL,
    COUNT_ANSWERED_INTERNAL,
    COUNT_UNANSWERED_INTERNAL,
    COUNT_INBOUND,
    COUNT_ANSWERED_INBOUND,
    COUNT_UNANSWERED_INBOUND,
    COUNT_OUTBOUND,
    COUNT_ANSWERED_OUTBOUND,
    COUNT_UNANSWERED_OUTBOUND,
    COUNT_EXTERNAL_OUTBOUND,
    COUNT_ANSWERED_EXTERNAL_OUTBOUND,
    COUNT_UNANSWERED_EXTERNAL_OUTBOUND,
    COUNT_EXTERNAL_INBOUND,
    COUNT_ANSWERED_EXTERNAL_INBOUND,
    COUNT_UNANSWERED_EXTERNAL_INBOUND,
    COUNT_INTERNAL_OUTBOUND,
    COUNT_ANSWERED_INTERNAL_OUTBOUND,
    COUNT_UNANSWERED_INTERNAL_OUTBOUND,
    COUNT_INTERNAL_INBOUND,
    COUNT_ANSWERED_INTERNAL_INBOUND,
    COUNT_UNANSWERED_INTERNAL_INBOUND;

    @Override
    public String getSomething(Any any) {
        return null;
    }
}

我认为在这种情况下，这比在接口中创建默认方法更好，因为默认实现似乎主要与枚举有关，而不是与接口的所有其他实现者有关。

Answer 3

没有捷径可做。

所有枚举值的IntelliJ实现方法

3 个答案: