Is there an easy way to create ordinals in C#?-第3页回答

2楼-- · 2019-01-01 09:01

EDIT: As YM_Industries points out in the comment, samjudson's answer DOES work for numbers over 1000, nickf's comment seems to have gone, and I can't remember what the problem I saw was. Left this answer here for the comparison timings.

An awful lot of these don't work for numbers > 999, as nickf pointed out in a comment (EDIT: now missing).

Here is a version based off a modified version of samjudson's accepted answer that does.

public static String GetOrdinal(int i)
{
    String res = "";

    if (i > 0)
    {
        int j = (i - ((i / 100) * 100));

        if ((j == 11) || (j == 12) || (j == 13))
            res = "th";
        else
        {
            int k = i % 10;

            if (k == 1)
                res = "st";
            else if (k == 2)
                res = "nd";
            else if (k == 3)
                res = "rd";
            else
                res = "th";
        }
    }

    return i.ToString() + res;
}

Also Shahzad Qureshi's answer using string manipulation works fine, however it does have a performance penalty. For generating a lot of these, a LINQPad example program makes the string version 6-7 times slower than this integer one (although you'd have to be generating a lot to notice).

LINQPad example:

void Main()
{
    "Examples:".Dump();

    foreach(int i in new int[] {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 22, 113, 122, 201, 202, 211, 212, 2013, 1000003, 10000013 })
        Stuff.GetOrdinal(i).Dump();

    String s;

    System.Diagnostics.Stopwatch sw = System.Diagnostics.Stopwatch.StartNew();

    for(int iter = 0; iter < 100000; iter++)
        foreach(int i in new int[] {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 22, 113, 122, 201, 202, 211, 212, 2013, 1000003, 1000013 })
            s = Stuff.GetOrdinal(i);

    "Integer manipulation".Dump();
    sw.Elapsed.Dump();

    sw.Restart();

    for(int iter = 0; iter < 100000; iter++)
        foreach(int i in new int[] {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 22, 113, 122, 201, 202, 211, 212, 2013, 1000003, 1000013 })
            s = (i.ToString() + Stuff.GetOrdinalSuffix(i));

    "String manipulation".Dump();
    sw.Elapsed.Dump();
}

public class Stuff
{
        // Use integer manipulation
        public static String GetOrdinal(int i)
        {
                String res = "";

                if (i > 0)
                {
                        int j = (i - ((i / 100) * 100));

                        if ((j == 11) || (j == 12) || (j == 13))
                                res = "th";
                        else
                        {
                                int k = i % 10;

                                if (k == 1)
                                        res = "st";
                                else if (k == 2)
                                        res = "nd";
                                else if (k == 3)
                                        res = "rd";
                                else
                                        res = "th";
                        }
                }

                return i.ToString() + res;
        }

        // Use string manipulation
        public static 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";
        }
}

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明月照影归

3楼-- · 2019-01-01 09:02

Simple, clean, quick

    private static 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";
    }

Or better yet, as an extension method

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

Now you can just call

int a = 1;
a.DisplayWithSuffix();

or even as direct as

1.DisplayWithSuffix();

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看风景的人

4楼-- · 2019-01-01 09:02

FWIW, for MS-SQL, this expression will do the job. Keep the first WHEN (WHEN num % 100 IN (11, 12, 13) THEN 'th') as the first one in the list, as this relies upon being tried before the others.

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

For Excel :

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

The expression (MOD(A1-11,100)>2) is TRUE (1) for all numbers except any ending in 11,12,13 (FALSE = 0). So 2 * RIGHT(A1) * (MOD(A1-11,100)>2) +1) ends up as 1 for 11/12/13, otherwise :
1 evaluates to 3
2 to 5,
3 to 7
others : 9
- and the required 2 characters are selected from "thstndrdth" starting from that position.

If you really want to convert that fairly directly to SQL, this worked for me for a handful of test values :

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
                )

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美炸的是我

5楼-- · 2019-01-01 09:04

My version of Jesse's version of Stu's and samjudson's versions :)

Included unit test to show that the accepted answer is incorrect when number < 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());
    }

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爱死公子算了

6楼-- · 2019-01-01 09:04

You'll have to roll your own. From the top of my head:

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;
}

You can then do

Console.WriteLine(432.Ordinal());

Edited for 11/12/13 exceptions. I DID say from the top of my head :-)

Edited for 1011 -- others have fixed this already, just want to make sure others don't grab this incorrect version.

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Is there an easy way to create ordinals in C#?

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