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,我应该只使用一个开关/选择案例吗?
更新在这种情况下我也不需要本地化。
更新2 以下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()
答案 0 :(得分:11)
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,
}
然后你可以输入像......这样的东西。
((Number)1).ToString()
如果您需要本地化,则可以为每个枚举值添加DescriptionAttribute
。属性的Description
属性将存储资源项的密钥名称。
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;
}
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,或者只要在返回.ToString()
时调用null
。
各种评论者指出(我必须同意)只使用枚举值名称来引用本地化字符串会更简单。
答案 1 :(得分:4)
创建字符串字典:
string[] digits = new string[]
{
"zero",
"one",
"two",
...
};
string word = digits[digit];
答案 2 :(得分:1)
使用查找表;一个数组会做。它不比枚举慢,并且更容易本地化。
修改的
安德烈的代码示例是我的建议,虽然我认为把它称为字典有点令人困惑。
答案 3 :(得分:1)
如果您不需要本地化,我建议使用Richard Ev的解决方案。但是,对于本地化,我建议将十位数名称添加到资源文件中,例如NumberName_0
到NumberName_9
。这样,在查找数字时,您只需加载名为String.Format("NumberName_{0}", mydigit)
的资源。
顺便说一句,相同的技术也适用于可本地化的枚举名称或描述。
答案 4 :(得分:1)
为什么要停在1-9 ......
C#的版本Squeak Smalltalk为所有数字做了一个警告:
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;
}
答案 5 :(得分:0)
我认为没有任何内置函数可以做到这一点。我会用select case。
答案 6 :(得分:0)
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,
}