我使用的另一种替代方法是基于所有其他建议，但不需要特殊的大小写:

public static string DateSuffix(int day)
{
    if (day == 11 | day == 12 | day == 13) return "th";
    Math.DivRem(day, 10, out day);
    switch (day)
    {
        case 1:
            return "st";
        case 2:
            return "nd";
        case 3:
            return "rd";
        default:
            return "th";
    }
}

public static string OrdinalSuffix(int ordinal)
{
    //Because negatives won't work with modular division as expected:
    var abs = Math.Abs(ordinal); 

    var lastdigit = abs % 10; 

    return 
        //Catch 60% of cases (to infinity) in the first conditional:
        lastdigit > 3 || lastdigit == 0 || (abs % 100) - lastdigit == 10 ? "th" 
            : lastdigit == 1 ? "st" 
            : lastdigit == 2 ? "nd" 
            : "rd";
}

我使用这个扩展类:

public static class Int32Extensions
{
    public static string ToOrdinal(this int i)
    {
        return (i + "th")
            .Replace("1th", "1st")
            .Replace("2th", "2nd")
            .Replace("3th", "3rd");
    }
}

要求“减少冗余”版本的samjudson的回答…

public static string AddOrdinal(int number)
{
    if (number <= 0) return number.ToString();

    string GetIndicator(int num)
    {
        switch (num % 100)
        {
            case 11:
            case 12:
            case 13:
                return "th";
        }

        switch (num % 10)
        {
            case 1:
                return "st";
            case 2:
                return "nd";
            case 3:
                return "rd";
            default:
                return "th";
        }
    }

    return number + GetIndicator(number);
}

FWIW，对于MS-SQL，这个表达式将完成工作。将第一个WHEN (WHEN num % 100 IN (11,12,13) THEN 'th')作为列表中的第一个，因为这依赖于在其他尝试之前尝试。

CASE
  WHEN num % 100 IN (11, 12, 13) THEN 'th' -- must be tried first
  WHEN num % 10 = 1 THEN 'st'
  WHEN num % 10 = 2 THEN 'nd'
  WHEN num % 10 = 3 THEN 'rd'
  ELSE 'th'
END AS Ordinal

对于Excel:

=MID("thstndrdth",MIN(9,2*RIGHT(A1)*(MOD(A1-11,100)>2)+1),2)

表达式(MOD(A1- 11100)>2)对于除以11,12,13结尾的任何数字(FALSE = 0)外的所有数字都是TRUE(1)。因此2 * RIGHT(A1) * (MOD(A1- 11100)>2) +1)对于11/12/13最终为1，否则: 1等于3 2点到5点， 3至7点 其他:9 -所需的2个字符从该位置开始的“第thstndrdth”中选择。

如果你真的想把它直接转换成SQL，这对我来说适用于一些测试值:

DECLARE @n as int
SET @n=13
SELECT SubString(  'thstndrdth'
                 , (SELECT MIN(value) FROM
                     (SELECT 9 as value UNION
                      SELECT 1+ (2* (ABS(@n) % 10)  *  CASE WHEN ((ABS(@n)+89) % 100)>2 THEN 1 ELSE 0 END)
                     ) AS Mins
                   )
                 , 2
                )

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。