public static String GetOrdinal(int i)Īlso Shahzad Qureshi's answer using string manipulation works fine, however it does have a performance penalty. Here is a version based off a modified version of samjudson's accepted answer that does. Left this answer here for the comparison timings.Īn awful lot of these don't work for numbers > 999, as nickf pointed out in a comment (EDIT: now missing). Or even as direct as 1.DisplayWithSuffix() ĮDIT: 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. If (num.ToString().EndsWith('3')) return num.ToString() + 'rd' If (num.ToString().EndsWith('2')) return num.ToString() + 'nd' If (num.ToString().EndsWith('1')) return num.ToString() + 'st' If (num.ToString().EndsWith('13')) return num.ToString() + 'th' If (num.ToString().EndsWith('12')) return num.ToString() + 'th' If (num.ToString().EndsWith('11')) return num.ToString() + 'th' Public static string DisplayWithSuffix(this int num) Or better yet, as an extension method public static class IntegerExtensions If (num.ToString().EndsWith('3')) return 'rd' If (num.ToString().EndsWith('2')) return 'nd' If (num.ToString().EndsWith('1')) return 'st' If (num.ToString().EndsWith('13')) return 'th' If (num.ToString().EndsWith('12')) return 'th' If (num.ToString().EndsWith('11')) return 'th' Simple, clean, quick private static string GetOrdinalSuffix(int num) Public void Ordinal_ReturnsExpectedResults() Negative and zero have no ordinal representation Public static string Ordinal(this int number) / Ordinal value of positive integers, or if less than 1. / Only works for english-based cultures. / Get the ordinal value of positive integers.

Included unit test to show that the accepted answer is incorrect when number My version of Jesse's version of Stu's and samjudson's versions :)

Ordinal numbers in different languages