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Home/ Questions/Q 7941279
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Editorial Team
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Editorial Team
Asked: June 3, 20262026-06-03T23:37:16+00:00 2026-06-03T23:37:16+00:00

I have one date like 12/05/2012 now i would like to change that format

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I have one date like 12/05/2012 now i would like to change that format in to simple string.

for ex.

string newdate = new string();
newdate = "12/05/2012";
DateTime Bdate = DateTime.ParseExact(Newdate, "dd/MM/yyyy", System.Globalization.CultureInfo.InvariantCulture);

now my BDate is DateTime
ie. BDate= 2012/05/12

now i want to do something like

if my Bdate is 12/05/2012
so i want a string which is similar like “Twelve May two thousand twelve”

How can i do this?

Please help me…

Thanks in advance….

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-06-03T23:37:18+00:00Added an answer on June 3, 2026 at 11:37 pm

    You’ll need to look at each date part and use a function to get the written equivalent. I’ve included a class below that converts integers to written text, and extended it to support DateTime conversion as well:

    public static class WrittenNumerics
    {
        static readonly string[] ones = new string[] { "", "One", "Two", "Three", "Four", "Five", "Six", "Seven", "Eight", "Nine" };
        static readonly string[] teens = new string[] { "Ten", "Eleven", "Twelve", "Thirteen", "Fourteen", "Fifteen", "Sixteen", "Seventeen", "Eighteen", "Nineteen" };
        static readonly string[] tens = new string[] { "Twenty", "Thirty", "Forty", "Fifty", "Sixty", "Seventy", "Eighty", "Ninety" };
        static readonly string[] thousandsGroups = { "", " Thousand", " Million", " Billion" };
    
        private static string FriendlyInteger(int n, string leftDigits, int thousands)
        {
            if (n == 0)
                return leftDigits;
    
            string friendlyInt = leftDigits;
            if (friendlyInt.Length > 0)
                friendlyInt += " ";
    
            if (n < 10)
                friendlyInt += ones[n];
            else if (n < 20)
                friendlyInt += teens[n - 10];
            else if (n < 100)
                friendlyInt += FriendlyInteger(n % 10, tens[n / 10 - 2], 0);
            else if (n < 1000)
                friendlyInt += FriendlyInteger(n % 100, (ones[n / 100] + " Hundred"), 0);
            else
                friendlyInt += FriendlyInteger(n % 1000, FriendlyInteger(n / 1000, "", thousands + 1), 0);
    
            return friendlyInt + thousandsGroups[thousands];
        }
    
        public static string DateToWritten(DateTime date)
        {
            return string.Format("{0} {1} {2}", IntegerToWritten(date.Day), date.ToString("MMMM"), IntegerToWritten(date.Year));
        }
    
        public static string IntegerToWritten(int n)
        {
            if (n == 0)
                return "Zero";
            else if (n < 0)
                return "Negative " + IntegerToWritten(-n);
    
            return FriendlyInteger(n, "", 0);
        }
    }
    

    Disclaimer: Basic functionality courtesy of @Wedge

    Using this class, just call the DateToWritten method:

    var output = WrittenNumerics.DateToWritten(DateTime.Today);
    

    The output of the above is: Twelve May Two Thousand Twelve

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