Java中有没有办法将整数转换为序数？

Question

我想取一个整数并得到它的序数,即:

1 -> "First"
2 -> "Second"
3 -> "Third"
...

Answer 1

如果您对"1st","2nd","3rd"等没问题,这里有一些能够正确处理任何整数的简单代码:

public static String ordinal(int i) {
    String[] sufixes = new String[] { "th", "st", "nd", "rd", "th", "th", "th", "th", "th", "th" };
    switch (i % 100) {
    case 11:
    case 12:
    case 13:
        return i + "th";
    default:
        return i + sufixes[i % 10];

    }
}

以下是边缘情况的一些测试:

public static void main(String[] args) {
    int[] edgeCases = { 0, 1, 2, 3, 4, 5, 10, 11, 12, 13, 14, 20, 21, 22, 23, 24, 100, 101, 102, 103, 104, 111, 112, 113, 114 };
    for (int edgeCase : edgeCases) {
        System.out.println(ordinal(edgeCase));
    }
}

输出:

0th
1st
2nd
3rd
4th
5th
10th
11th
12th
13th
14th
20th
21st
22nd
23rd
24th
100th
101st
102nd
103rd
104th
111th
112th
113th
114th

Answer 2

使用优秀的ICU4J(还有一个优秀的C版本)你也可以做到这一点,并将普通话作为简单的单词;

RuleBasedNumberFormat nf = new RuleBasedNumberFormat(Locale.UK, RuleBasedNumberFormat.SPELLOUT);
for(int i = 0; i <= 30; i++)
{
    System.out.println(i + " -> "+nf.format(i, "%spellout-ordinal"));
}

例如生产

0 -> zeroth
1 -> first
2 -> second
3 -> third
4 -> fourth
5 -> fifth
6 -> sixth
7 -> seventh
8 -> eighth
9 -> ninth
10 -> tenth
11 -> eleventh
12 -> twelfth
13 -> thirteenth
14 -> fourteenth
15 -> fifteenth
16 -> sixteenth
17 -> seventeenth
18 -> eighteenth
19 -> nineteenth
20 -> twentieth
21 -> twenty-first
22 -> twenty-second
23 -> twenty-third
24 -> twenty-fourth
25 -> twenty-fifth
26 -> twenty-sixth
27 -> twenty-seventh
28 -> twenty-eighth
29 -> twenty-ninth
30 -> thirtieth

Answer 3

另一种方法

public static String ordinal(int i) {
    int mod100 = i % 100;
    int mod10 = i % 10;
    if(mod10 == 1 && mod100 != 11) {
        return i + "st";
    } else if(mod10 == 2 && mod100 != 12) {
        return i + "nd";
    } else if(mod10 == 3 && mod100 != 13) {
        return i + "rd";
    } else {
        return i + "th";
    }
}

Pro:不需要初始化数组(减少垃圾)
Con:不是单行...

Answer 4

我已经想出了如何在Android中以一种非常简单的方式做到这一点.您需要做的就是将依赖项添加到您app的build.gradle文件中:

implementation "com.ibm.icu:icu4j:53.1"

接下来,创建此方法:

科特林:

fun Number?.getOrdinal(): String? {
    if (this == null) {
        return null
    }

    val format = "{0,ordinal}"

    return if (Build.VERSION.SDK_INT >= Build.VERSION_CODES.N) {
        android.icu.text.MessageFormat.format(format, this)
    } else {
        com.ibm.icu.text.MessageFormat.format(format, this)
    }
}

Java的:

public static String getNumberOrdinal(Number number) {
        if (number == null) {
            return null;
        }

        String format = "{0,ordinal}";

        if (Build.VERSION.SDK_INT >= Build.VERSION_CODES.N) {
            return android.icu.text.MessageFormat.format(format, number);
        } else {
            return com.ibm.icu.text.MessageFormat.format(format, number);
        }
    }

然后,您可以像这样简单地使用它:

科特林:

val ordinal = 2.getOrdinal()

Java的:

String ordinal = getNumberOrdinal(2)

这个怎么运作

从Android N(API 24)开始Android使用icu.text而不是常规java.text(此处更多信息),其中已包含序数的国际化实现.所以解决方案显然很简单 - 将icu4j库添加到项目中并在下面的版本上使用它Nougat

Answer 5

一行:

public static String ordinal(int i) {
    return i % 100 == 11 || i % 100 == 12 || i % 100 == 13 ? i + "th" : i + new String[]{"th", "st", "nd", "rd", "th", "th", "th", "th", "th", "th"}[i % 10];
}

Answer 6

这是转换的代码,这是另一个.看看这个答案如何在java中将数字转换为单词