.Net 中有没有办法获取 int 单词的字符串值？

Question

例如：

(1).SomeFunction().Equals("one")
(2).SomeFunction().Equals("two")

在我正在使用的情况下，我真的只需要数字 1-9，我应该只使用开关/选择情况吗？

更新在这种情况下我也不需要本地化。

更新 2 这是我最终使用的内容：

Private Enum EnglishDigit As Integer
    zero
    one
    two
    three
    four
    five
    six
    seven
    eight
    nine
End Enum

(CType(someIntThatIsLessThanTen, EnglishDigit)).ToString()

Answer 1

枚举怎么样？

enum Number
{
    One = 1, // default value for first enum element is 0, so we set = 1 here
    Two,
    Three,
    Four,
    Five,
    Six,
    Seven,
    Eight,
    Nine,
}

然后您可以输入诸如...

((Number)1).ToString()

如果您需要本地化，那么您可以添加 DescriptionAttribute 到每个枚举值。该属性的 Description 属性将存储资源项的键的名称。

enum Number
{
    [Description("NumberName_1")]
    One = 1, // default value for first enum element is 0, so we set = 1 here 

    [Description("NumberName_2")]
    Two,

    // and so on...
}

以下函数将从属性中获取 Description 属性的值

public static string GetDescription(object value)
{
    DescriptionAttribute[] attributes = null;
    System.Reflection.FieldInfo fi = value.GetType().GetField(value.ToString());
    if (fi != null)
    {
        attributes = (DescriptionAttribute[])fi.GetCustomAttributes(typeof(DescriptionAttribute), false);
    }

    string description = null;
    if ((attributes != null) && (attributes.Length > 0))
    {
        description = attributes[0].Description;
    }

    return description;
}

这可以通过以下方式调用：

GetDescription(((Number)1))

然后您可以从资源文件中提取相关值，或者只需调用 .ToString() 如果返回 null。

编辑

各种评论者指出（我必须同意），仅使用枚举值名称来引用本地化字符串会更简单。

Answer 2

创建字符串字典：

string[] digits = new string[] 
{
   "zero",
   "one",
   "two",
   ...
};

string word = digits[digit];

Answer 3

使用查找表；一个数组就可以了。它并不比枚举慢，而且更容易本地化。

编辑

Andrey 的代码示例就是我所建议的，尽管我认为称其为字典有点令人困惑。

Answer 4

如果您不需要本地化，我建议使用 Richard Ev 的解决方案。不过，对于本地化，我建议将十位数字名称添加到资源文件中，例如 NumberName_0 到 NumberName_9。这样，当查找号码时，您只需加载名称为 String.Format("NumberName_{0}", mydigit) 的资源即可。

顺便说一句，同样的技术也适用于可本地化的枚举名称或描述。

Answer 5

为什么停在 1-9...

C# 版本的 Squeak Smalltalk 对所有数字进行处理的方式：

    public static String AsWords(this int aNumber) {
        var answer = "";
        if (aNumber == 0) return "zero";
        if (aNumber < 0) {
            answer = "negative";
            aNumber = Math.Abs(aNumber);
        }

        var thousands = new[] {"", "thousand", "million", "billion", "trillion", "quadrillion", "quintillion", "sextillion", "septillion","octillion", "nonillion", "decillion", "undecillion", "duodecillion", "tredecillion", "quattuordecillion", "quindecillion", "sexdecillion", "septendecillion", "octodecillion", "novemdecillion", "vigintillion"};
        var thousandCount = 0;
        while (aNumber > 0) {
            var underOneThousandName = ThreeDigitName(aNumber % 1000);
            aNumber = aNumber / 1000;
            if(underOneThousandName != "") {
                if (answer != "") answer = "," + answer;
                answer = underOneThousandName + " " + thousands[thousandCount] + answer;
            }
            thousandCount += 1;
        }
        return answer;
    }

    private static string ThreeDigitName(int aNumberLessThanOneThousand) {
        if (aNumberLessThanOneThousand == 0) return "";
        var units = new[] {"one", "two", "three", "four", "five", "six", "seven", "eight", "nine", "ten", "eleven", "twelve", "thirteen", "fourteen", "fifteen", "sixteen", "seventeen", "eightteen", "nineteen"};
        var answer = "";
        if (aNumberLessThanOneThousand > 99) {
            answer = units[(aNumberLessThanOneThousand / 100) - 1] + " hundred";
            if (aNumberLessThanOneThousand % 100 != 0)
                answer += " " + ThreeDigitName(aNumberLessThanOneThousand % 100);
            return answer;
        }
        if (aNumberLessThanOneThousand < 20) return units[aNumberLessThanOneThousand -1];
        var multiplesOfTen = new[] {"twenty", "thirty", "forty", "fifty", "sixty", "seventy", "eighty", "ninety"};
        answer += multiplesOfTen[(aNumberLessThanOneThousand / 10)-2];
        if (aNumberLessThanOneThousand % 10 != 0) answer += "-" + units[(aNumberLessThanOneThousand % 10)-1];
        return answer;
    }

Answer 6

我认为没有任何内置函数可以做到这一点。我会使用选择案例。

Answer 7

如果您使用的是 2008：

public static String AsWord(this int aNumber) {
    return ((Number) aNumber).ToString();
}

enum Number {
    One = 1,
    Two,
    Three,
    Four,
    Five,
    Six,
    Seven,
    Eight,
    Nine,
}