How can i format 07/03/2012 to March 7th,2012 in c#

Question

Any one please help i need to show the date 03/03/2012 as March 3rd,2012 etc

Does it have to do the "rd" in "3rd"? Looking at the standard and custom format strings available none of them will do this for you... Its relatively easy without though. :) — Chris, Mar 07 '12 at 12:38

score 58 · Accepted Answer · edited Jul 08 '15 at 08:32

You can create your own custom format provider to do this:

public class MyCustomDateProvider: IFormatProvider, ICustomFormatter
{
    public object GetFormat(Type formatType)
    {
        if (formatType == typeof(ICustomFormatter))
            return this;

        return null;
    }

    public string Format(string format, object arg, IFormatProvider formatProvider)
    {
        if (!(arg is DateTime)) throw new NotSupportedException();

        var dt = (DateTime) arg;

        string suffix;

        if (new[] {11, 12, 13}.Contains(dt.Day))
        {
            suffix = "th";
        }
        else if (dt.Day % 10 == 1)
        {
            suffix = "st";
        }
        else if (dt.Day % 10 == 2)
        {
            suffix = "nd";
        }
        else if (dt.Day % 10 == 3)
        {
            suffix = "rd";
        }
        else
        {
            suffix = "th";
        }

        return string.Format("{0:MMMM} {1}{2}, {0:yyyy}", arg, dt.Day, suffix);
    }
}

This can then be called like this:

var formattedDate = string.Format(new MyCustomDateProvider(), "{0}", date);

Resulting in (for example):

March 3rd, 2012

but need to check if (dt.Day == 11 || dt.Day == 12 || dt.Day == 13) { suffix = "th"; return string.Format("{0:MMMM} {1}{2}, {0:yyyy}", arg, dt.Day, suffix); } — Sreenath Plakkat, Mar 15 '12 at 04:52
oh yes - you'll have to add them as special cases, before the other checks. — Rob Levine, Mar 15 '12 at 08:33
Short version of this answer here http://stackoverflow.com/a/2050854/581414 — Ruskin, Nov 03 '16 at 13:38

regisbsb · Answer 2 · 2015-05-29T10:41:11.070

31

Humanizer meets all your .NET needs for manipulating and displaying strings, enums, dates, times, timespans, numbers and quantities

To install Humanizer, run the following command in the Package Manager Console

PM> Install-Package Humanizer

Ordinalize turns a number into an ordinal string used to denote the position in an ordered sequence such as 1st, 2nd, 3rd, 4th:

1.Ordinalize() => "1st"
5.Ordinalize() => "5th"

Then you can use:

String.Format("{0} {1:MMMM yyyy}", date.Day.Ordinalize(), date)

edited May 29 '15 at 10:41

answered May 27 '14 at 16:18

regisbsb

3,664
2
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41

+1 - I used String.Format("{0} {1}", d.Day.Ordinalize(), d.ToString("MMMM yyyy")) – Tom Oct 21 '14 at 12:56
+1 Humanizer is awesome. btw you can write: string.Format("{0} {1:MMMM yyyy}", date.Day.Ordinalize(), date) – BritishDeveloper Feb 18 '15 at 13:02
Good one @BritishDeveloper, I'll update the example – regisbsb May 29 '15 at 10:41
1

Is there a way to parse a British date with Humanizer? Like parse "8th Nov 2012" into a proper DateTime object. – Costin_T Feb 08 '17 at 11:03

score 13 · Answer 3 · edited May 23 '17 at 12:17

13

Custom Date and Time Format Strings

date.ToString("MMMM d, yyyy")

Or if you need the "rd" too:

string.Format("{0} {1}, {2}", date.ToString("MMMM"), date.Day.Ordinal(), date.ToString("yyyy"))

the Ordinal() method can be found here

edited May 23 '17 at 12:17

Community

1
1

answered Mar 07 '12 at 12:35

CD..

72,281
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This doesn't quite give the right format due to the "rd" in the original spec. – Chris Mar 07 '12 at 12:43
should be string.Format("{0} {1}, {2}", DateTime.Now.ToString("MMMM"), Ordinal(DateTime.Now.Day), DateTime.Now.ToString("yyyy")) – Bobby Bridgeman Mar 13 '23 at 18:35

score 3 · Answer 4 · answered Mar 04 '14 at 02:50

public static class IntegerExtensions
{
    /// <summary>
    /// converts an integer to its ordinal representation
    /// </summary>
    public static String AsOrdinal(this Int32 number)
    {
        if (number < 0)
            throw new ArgumentOutOfRangeException("number");

        var work = number.ToString("n0");

        var modOf100 = number % 100;

        if (modOf100 == 11 || modOf100 == 12 || modOf100 == 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;
    }
}

Proof:

[TestFixture]
class IntegerExtensionTests
{
    [Test]
    public void TestCases_1s_10s_100s_1000s()
    {
        Assert.AreEqual("1st", 1.AsOrdinal());
        Assert.AreEqual("2nd", 2.AsOrdinal());
        Assert.AreEqual("3rd", 3.AsOrdinal());

        foreach (var integer in Enumerable.Range(4, 6))
            Assert.AreEqual(String.Format("{0:n0}th", integer), integer.AsOrdinal());

        Assert.AreEqual("11th", 11.AsOrdinal());
        Assert.AreEqual("12th", 12.AsOrdinal());
        Assert.AreEqual("13th", 13.AsOrdinal());

        foreach (var integer in Enumerable.Range(14, 6))
            Assert.AreEqual(String.Format("{0:n0}th", integer), integer.AsOrdinal());

        Assert.AreEqual("21st", 21.AsOrdinal());
        Assert.AreEqual("22nd", 22.AsOrdinal());
        Assert.AreEqual("23rd", 23.AsOrdinal());

        foreach (var integer in Enumerable.Range(24, 6))
            Assert.AreEqual(String.Format("{0:n0}th", integer), integer.AsOrdinal());

        Assert.AreEqual("31st", 31.AsOrdinal());
        Assert.AreEqual("32nd", 32.AsOrdinal());
        Assert.AreEqual("33rd", 33.AsOrdinal());

        //then just jump to 100

        Assert.AreEqual("101st", 101.AsOrdinal());
        Assert.AreEqual("102nd", 102.AsOrdinal());
        Assert.AreEqual("103rd", 103.AsOrdinal());

        foreach (var integer in Enumerable.Range(104, 6))
            Assert.AreEqual(String.Format("{0:n0}th", integer), integer.AsOrdinal());

        Assert.AreEqual("111th", 111.AsOrdinal());
        Assert.AreEqual("112th", 112.AsOrdinal());
        Assert.AreEqual("113th", 113.AsOrdinal());

        foreach (var integer in Enumerable.Range(114, 6))
            Assert.AreEqual(String.Format("{0:n0}th", integer), integer.AsOrdinal());

        Assert.AreEqual("121st", 121.AsOrdinal());
        Assert.AreEqual("122nd", 122.AsOrdinal());
        Assert.AreEqual("123rd", 123.AsOrdinal());

        foreach (var integer in Enumerable.Range(124, 6))
            Assert.AreEqual(String.Format("{0:n0}th", integer), integer.AsOrdinal());

        //then just jump to 1000

        Assert.AreEqual("1,001st", 1001.AsOrdinal());
        Assert.AreEqual("1,002nd", 1002.AsOrdinal());
        Assert.AreEqual("1,003rd", 1003.AsOrdinal());

        foreach (var integer in Enumerable.Range(1004, 6))
            Assert.AreEqual(String.Format("{0:n0}th", integer), integer.AsOrdinal());

        Assert.AreEqual("1,011th", 1011.AsOrdinal());
        Assert.AreEqual("1,012th", 1012.AsOrdinal());
        Assert.AreEqual("1,013th", 1013.AsOrdinal());

        foreach (var integer in Enumerable.Range(1014, 6))
            Assert.AreEqual(String.Format("{0:n0}th", integer), integer.AsOrdinal());

        Assert.AreEqual("1,021st", 1021.AsOrdinal());
        Assert.AreEqual("1,022nd", 1022.AsOrdinal());
        Assert.AreEqual("1,023rd", 1023.AsOrdinal());

        foreach (var integer in Enumerable.Range(1024, 6))
            Assert.AreEqual(String.Format("{0:n0}th", integer), integer.AsOrdinal());
    }
}

score 3 · Answer 5 · edited May 23 '17 at 11:46

No, there is nothing in string.Format() that will give you ordinals (1st, 2nd, 3rd, 4th and so on).

You can combine the date format like suggested in other answers, with your own ordinal as suggested for example in this answer

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

string Format(DateTime date)
{
    int dayNo = date.Day;
    return string.Format("{0} {1}{2}, {3}", 
                date.ToString("MMMM"), dayNo, AddOrdinal(dayNo), date.Year); 
}

score 2 · Answer 6 · answered Feb 25 '15 at 22:24

Based on Rob Levine's answer and the comments on that answer... I've adapted as an extension method on DateTime so you can just call:

var formattedDate = date.Friendly();

Here's the extension method:

public static class DateFormatter
{
  public static string Friendly(this DateTime dt)
  {
    string suffix;

    switch (dt.Day)
    {
      case 1:
      case 21:
      case 31:
        suffix = "st";
        break;
      case 2:
      case 22:
        suffix = "nd";
        break;
      case 3:
      case 23:
        suffix = "rd";
        break;
      default:
        suffix = "th";
        break;
    }

    return string.Format("{0:MMMM} {1}{2}, {0:yyyy}", dt, dt.Day, suffix);
  }
}

Amine Boulaajaj · Answer 7 · 2019-07-16T21:59:52.703

0

Here is another version of the Ordinalize() extension, short and sweet:

public static string Ordinalize(this int x)
{
    var xString = x.ToString();
    var xLength = xString.Length;
    var xLastTwoCharacters = xString.Substring(Math.Max(0, xLength - 2));
    return xString + 
        ((x % 10 == 1 && xLastTwoCharacters != "11") 
            ? "st"
        : (x % 10 == 2 && xLastTwoCharacters != "12") 
            ? "nd"
        : (x % 10 == 3 && xLastTwoCharacters != "13") 
            ? "rd"
        : "th");
    }

and then call that extension like this

myDate.Day.Ordinalize()

or

myAnyNumber.Ordinalize()

edited Jul 16 '19 at 21:59

answered Jan 23 '18 at 06:20

Amine Boulaajaj

421
4
6

2

What about 211? – Kristopher Jul 16 '19 at 19:30
@Kristopher nice - I think I fixed it this time. here is a tester on .NetFiddle: https://dotnetfiddle.net/hy3OoI – Amine Boulaajaj Jul 16 '19 at 22:00

score -3 · Answer 8 · answered Mar 07 '12 at 12:38

-3

DateTime dt = new DateTime(args);
String.Format("{0:ddd, MMM d, yyyy}", dt);

// "Sun, Mar 9, 2008"

answered Mar 07 '12 at 12:38

score -4 · Answer 9 · answered Mar 07 '12 at 12:37

Use following Code:

DateTime thisDate1 = new DateTime(2011, 6, 10);
Console.WriteLine("Today is " + thisDate1.ToString("MMMM dd, yyyy") + ".");

DateTimeOffset thisDate2 = new DateTimeOffset(2011, 6, 10, 15, 24, 16, 
                                              TimeSpan.Zero);
Console.WriteLine("The current date and time: {0:MM/dd/yy H:mm:ss zzz}", 
                   thisDate2); 
// The example displays the following output:
//    Today is June 10, 2011.
//    The current date and time: 06/10/11 15:24:16 +00:00

This isn't what was asked.... – Stuart.Sklinar Feb 04 '14 at 17:00 — Stuart.Sklinar, Feb 04 '14 at 17:00

How can i format 07/03/2012 to March 7th,2012 in c#

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