类似于Ryan的解决方案，但更基本，我只是使用一个普通数组，并使用日期来查找正确的序数:

private string[] ordinals = new string[] {"","st","nd","rd","th","th","th","th","th","th","th","th","th","th","th","th","th","th","th","th","th","st","nd","rd","th","th","th","th","th","th","th","st" };
DateTime D = DateTime.Now;
String date = "Today's day is: "+ D.Day.ToString() + ordinals[D.Day];

我没有这个需要，但是我假设如果您想要多语言支持，您可以使用多维数组。

根据我在大学的记忆，这种方法只需要服务器做最少的工作。

private static string GetOrd(int num) => $"{num}{(!(Range(11, 3).Any(n => n == num % 100) ^ Range(1, 3).All(n => n != num % 10)) ? new[] { "ˢᵗ", "ⁿᵈ", "ʳᵈ" }[num % 10 - 1] : "ᵗʰ")}";

如果有人在找一句俏皮话。

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());
}

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" }})}"

虽然我还没有对此进行基准测试，但通过避免所有条件case语句，您应该能够获得更好的性能。

这是java，但是移植到c#很简单:

public class NumberUtil {
  final static String[] ORDINAL_SUFFIXES = {
    "th", "st", "nd", "rd", "th", "th", "th", "th", "th", "th"
  };

  public static String ordinalSuffix(int value) {
    int n = Math.abs(value);
    int lastTwoDigits = n % 100;
    int lastDigit = n % 10;
    int index = (lastTwoDigits >= 11 && lastTwoDigits <= 13) ? 0 : lastDigit;
    return ORDINAL_SUFFIXES[index];
  }

  public static String toOrdinal(int n) {
    return new StringBuffer().append(n).append(ordinalSuffix(n)).toString();
  }
}

注意，如果在一个紧密循环中生成大量序数，减少条件和使用数组查找应该会提高性能。然而，我也承认这并不像case语句解决方案那样可读。

我很喜欢Stu和samjudson解决方案中的元素，并将它们结合在一起，形成了我认为可用的组合:

public static string Ordinal(this int number)
{
    const string TH = "th";
    var s = number.ToString();
    
    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;
    }
}