在c#中有一个简单的方法来创建一个数字的序数吗?例如:

1返回第1位 2返回第2 3返回第3 等

这是否可以通过String.Format()来完成,或者是否有可用的函数来完成?


当前回答

杰西版本的斯图和萨姆贾德森版本的我的版本:)

包含单元测试,以显示接受的答案是不正确的,当数字< 1

/// <summary>
/// Get the ordinal value of positive integers.
/// </summary>
/// <remarks>
/// Only works for english-based cultures.
/// Code from: http://stackoverflow.com/questions/20156/is-there-a-quick-way-to-create-ordinals-in-c/31066#31066
/// With help: http://www.wisegeek.com/what-is-an-ordinal-number.htm
/// </remarks>
/// <param name="number">The number.</param>
/// <returns>Ordinal value of positive integers, or <see cref="int.ToString"/> if less than 1.</returns>
public static string Ordinal(this int number)
{
    const string TH = "th";
    string s = number.ToString();

    // Negative and zero have no ordinal representation
    if (number < 1)
    {
        return s;
    }

    number %= 100;
    if ((number >= 11) && (number <= 13))
    {
        return s + TH;
    }

    switch (number % 10)
    {
        case 1: return s + "st";
        case 2: return s + "nd";
        case 3: return s + "rd";
        default: return s + TH;
    }
}

[Test]
public void Ordinal_ReturnsExpectedResults()
{
    Assert.AreEqual("-1", (1-2).Ordinal());
    Assert.AreEqual("0", 0.Ordinal());
    Assert.AreEqual("1st", 1.Ordinal());
    Assert.AreEqual("2nd", 2.Ordinal());
    Assert.AreEqual("3rd", 3.Ordinal());
    Assert.AreEqual("4th", 4.Ordinal());
    Assert.AreEqual("5th", 5.Ordinal());
    Assert.AreEqual("6th", 6.Ordinal());
    Assert.AreEqual("7th", 7.Ordinal());
    Assert.AreEqual("8th", 8.Ordinal());
    Assert.AreEqual("9th", 9.Ordinal());
    Assert.AreEqual("10th", 10.Ordinal());
    Assert.AreEqual("11th", 11.Ordinal());
    Assert.AreEqual("12th", 12.Ordinal());
    Assert.AreEqual("13th", 13.Ordinal());
    Assert.AreEqual("14th", 14.Ordinal());
    Assert.AreEqual("20th", 20.Ordinal());
    Assert.AreEqual("21st", 21.Ordinal());
    Assert.AreEqual("22nd", 22.Ordinal());
    Assert.AreEqual("23rd", 23.Ordinal());
    Assert.AreEqual("24th", 24.Ordinal());
    Assert.AreEqual("100th", 100.Ordinal());
    Assert.AreEqual("101st", 101.Ordinal());
    Assert.AreEqual("102nd", 102.Ordinal());
    Assert.AreEqual("103rd", 103.Ordinal());
    Assert.AreEqual("104th", 104.Ordinal());
    Assert.AreEqual("110th", 110.Ordinal());
    Assert.AreEqual("111th", 111.Ordinal());
    Assert.AreEqual("112th", 112.Ordinal());
    Assert.AreEqual("113th", 113.Ordinal());
    Assert.AreEqual("114th", 114.Ordinal());
    Assert.AreEqual("120th", 120.Ordinal());
    Assert.AreEqual("121st", 121.Ordinal());
    Assert.AreEqual("122nd", 122.Ordinal());
    Assert.AreEqual("123rd", 123.Ordinal());
    Assert.AreEqual("124th", 124.Ordinal());
}

其他回答

这里是DateTime扩展类。复制,粘贴和享受

public static class DateTimeExtensions
{
    public static string ToStringWithOrdinal(this DateTime d)
    {
        var result = "";
        bool bReturn = false;            
        
        switch (d.Day % 100)
        {
            case 11:
            case 12:
            case 13:
                result = d.ToString("dd'th' MMMM yyyy");
                bReturn = true;
                break;
        }

        if (!bReturn)
        {
            switch (d.Day % 10)
            {
                case 1:
                    result = d.ToString("dd'st' MMMM yyyy");
                    break;
                case 2:
                    result = d.ToString("dd'nd' MMMM yyyy");
                    break;
                case 3:
                    result = d.ToString("dd'rd' MMMM yyyy");
                    break;
                default:
                    result = d.ToString("dd'th' MMMM yyyy");
                    break;
            }

        }

        if (result.StartsWith("0")) result = result.Substring(1);
        return result;
    }
}

结果:

2014年10月9日

你得自己动手了。在我的脑海中:

public static string Ordinal(this int number)
{
  var work = number.ToString();
  if ((number % 100) == 11 || (number % 100) == 12 || (number % 100) == 13)
    return work + "th";
  switch (number % 10)
  {
    case 1: work += "st"; break;
    case 2: work += "nd"; break;
    case 3: work += "rd"; break;
    default: work += "th"; break;
  }
  return work;
}

你可以这样做

Console.WriteLine(432.Ordinal());

针对11/12/13例外进行了编辑。我确实从我的头顶说过:-)

为1011编辑-其他人已经修复了这个问题,只是想确保其他人不会抓取这个错误的版本。

这是dart中的实现,可以根据语言进行修改。

String getOrdinalSuffix(int num){
    if (num.toString().endsWith("11")) return "th";
    if (num.toString().endsWith("12")) return "th";
    if (num.toString().endsWith("13")) return "th";
    if (num.toString().endsWith("1")) return "st";
    if (num.toString().endsWith("2")) return "nd";
    if (num.toString().endsWith("3")) return "rd";
    return "th";
}

简单、干净、快捷

private static string GetOrdinalSuffix(int num)
{
    string number = num.ToString();
    if (number.EndsWith("11")) return "th";
    if (number.EndsWith("12")) return "th";
    if (number.EndsWith("13")) return "th";
    if (number.EndsWith("1")) return "st";
    if (number.EndsWith("2")) return "nd";
    if (number.EndsWith("3")) return "rd";
    return "th";
}

或者更好的是,作为一个扩展方法

public static class IntegerExtensions
{
    public static string DisplayWithSuffix(this int num)
    {
        string number = num.ToString();
        if (number.EndsWith("11")) return number + "th";
        if (number.EndsWith("12")) return number + "th";
        if (number.EndsWith("13")) return number + "th";
        if (number.EndsWith("1")) return number + "st";
        if (number.EndsWith("2")) return number + "nd";
        if (number.EndsWith("3")) return number + "rd";
        return number + "th";
    }
}

现在你可以打电话了

int a = 1;
a.DisplayWithSuffix(); 

甚至直接到

1.DisplayWithSuffix();

c# 8和9中接受的带有开关表达式和模式匹配的答案。

没有不必要的字符串转换或分配。

string.Concat(number, number < 0 ? "" : (number % 100) switch 
{   
    11 or 12 or 13 => "th", 
    int n => (n % 10) switch 
    { 
        1 => "st", 
        2 => "nd", 
        3 => "rd", 
        _ => "th", 
    }
})

或者是不友好的一句话:

$"{number}{(number < 0 ? "" : (number % 100) switch { 11 or 12 or 13 => "th", int n => (n % 10) switch { 1 => "st", 2 => "nd", 3 => "rd", _ => "th" }})}"