有没有办法让这个switch语句更小更专业？答案

【问题标题】：Is there a way to make this switch statement smaller and more professional?有没有办法让这个switch语句更小更专业？
【发布时间】：2021-03-02 02:43:10
【问题描述】：

我正在为圣诞节制作降临日历，需要使用 switch 语句。我最大的困境是每个（daysAway）案例都会打开一个专门为那天设计的新课程。我正在研究 Google 和 Stack Overflow 可以提供的功能。我想知道是否有其他方法可以压缩它？

public void onClick(View v) {

            //Calculate the days between (date - 12/7/20)
            LocalDate dateBefore = java.time.LocalDate.now();
            LocalDate dateAfter = LocalDate.of(2020, Month.DECEMBER, 25);
            int daysAway = (int) ChronoUnit.DAYS.between(dateBefore, dateAfter);

            switch(daysAway){
                case 24:
                    openDay1();
                    break;
                case 23:
                    openDay2();
                    break;
                case 22:
                    openDay3;
                    break;
                case 21:
                    openDay4;
                    break;
                case 20:
                    openDay5;
                    break;
                case 19:
                    openDay6;
                    break;
                case 18:
                    openDay7;
                    break;
                case 17:
                    openDay8;
                    break;
                case 16:
                    openDay9;
                    break;
                case 15:
                    openDay10;
                    break;
                case 14:
                    openDay11;
                    break;
                case 13:
                    openDay12;
                    break;
                case 12:
                    openDay13;
                    break;
                case 11:
                    openDay14;
                    break;
                case 10:
                    openDay15;
                    break;
                case 9:
                    openDay16;
                    break;
                case 8:
                    openDay17;
                    break;
                case 7:
                    openDay18;
                    break;
                case 6:
                    openDay19;
                    break;
                case 5:
                    openDay20;
                    break;
                case 4:
                    openDay21;
                    break;
                case 3:
                    openDay22;
                    break;
                case 2:
                    openDay23;
                    break;
                case 1:
                    openDay24;
                    break;
                case 0:
                    openChristmas;
                    break;
                default:
                    notTime.start();
                    break;
            }
        }

我知道这是一个巨大的混乱，这就是我要解决的问题！感谢您提供任何反馈！

【问题讨论】：

欢迎来到 Stack Overflow。在您提出下一个问题之前，请花点时间阅读 How to Ask 和 this question checklist。
至于您的问题，可能是一个包含 25 个元素的数组，并使用 daysAway 作为索引（当然，首先检查它是否是有效索引）。

标签： java switch-statement

【解决方案1】：

使用 Java 14 switch 表达式语法：

switch (daysAway) {
    case 24 -> openDay1();
    case 23 -> openDay2();
    case 22 -> openDay3();
    case 21 -> openDay4();
    // ...
    case 4  -> openDay21();
    case 3  -> openDay22();
    case 2  -> openDay23();
    case 1  -> openDay24();
    case 0  -> openChristmas();
    default -> notTime.start();
}

由于代码很简单，折叠成一行就行了：

switch (daysAway) {
    case 24: openDay1();  break;
    case 23: openDay2();  break;
    case 22: openDay3();  break;
    case 21: openDay4();  break;
    // ...
    case 4:  openDay21(); break;
    case 3:  openDay22(); break;
    case 2:  openDay23(); break;
    case 1:  openDay24(); break;
    case 0:  openChristmas(); break;
    default: notTime.start();
}

使用 Java 8 方法引用数组（注意颠倒顺序）：

Runnable[] OPEN_METHODS = {
    this::openChristmas,
    this::openDay24,
    this::openDay23,
    this::openDay22,
    this::openDay21,
    // ...
    this::openDay4,
    this::openDay3,
    this::openDay2,
    this::openDay1
};

if (daysAway >= 0 && daysAway <= 24) {
    OPEN_METHODS[daysAway].run();
} else {
    notTime.start();
}

既然你说“每个 (daysAway) 案例都会打开一个为那一天设计的新类”，请使用一个接口（例如Runnable）和一个数组类文字：

Class<?>[] OPEN_CLASSES = {
    OpenChristmas.class,
    OpenDay24.class,
    OpenDay23.class,
    OpenDay22.class,
    OpenDay21.class,
    // ...
    OpenDay4.class,
    OpenDay3.class,
    OpenDay2.class,
    OpenDay1.class
};

if (daysAway < 0 || daysAway > 24) {
    notTime.start();
} else {
    Runnable clazz;
    try {
        clazz = (Runnable) OPEN_CLASSES[daysAway].getConstructor().newInstance();
    } catch (ReflectiveOperationException e) {
        throw new AssertionError("Oops: " + e, e);
    }
    clazz.run();
}

您也可以动态构建类名（无数组或switch 语句）：

if (daysAway < 0 || daysAway > 24) {
    notTime.start();
} else {
    String className = (daysAway == 0 ? "OpenChristmas" : "OpenDay" + (25 - daysAway));
    Runnable clazz;
    try {
        clazz = (Runnable) Class.forName(className).getConstructor().newInstance();
    } catch (ReflectiveOperationException e) {
        throw new AssertionError("Oops: " + e, e);
    }
    clazz.run();
}

【讨论】：

【解决方案2】：

if (daysAway > 0)
  openDay(25-daysAway); // you haven't shown what you would do here,
         //  but presumably you don't need to write 24 separate functions
else
  notTime.start();

【讨论】：

应该是if (daysAway >= 0 && daysAway <= 24)