How to convert datetime field to string like 1st Feb 2011?

Question

How do I convert a datetime field to a string with the format 1st Feb 2011 in C# ? doj is datetime field in sql server.

string DateOfJoin = dt.Rows[0]["DOJ"].ToString();//2011-02-01 00:00:00.000

Answer 1

First of all, 2/1/2011 can be 1st Feb or 2nd Jan not 1st Jan.

Second, let's parse your string to DateTime.

DateTime dt = DateTime.ParseExact(dt.Rows[0]["DOJ"].ToString(),
                                  "M/d/yyyy hh:mm:ss tt",
                                   CultureInfo.InvariantCulture);

or you can explicitly cast it to DateTime

DateTime dt = (DateTime)dt.Rows[0]["DOJ"];

Third, .NET does not have a built-in way in BCL to generate day suffix. But Lazlow write a method for that which works seems okey to me as;

static string GetDaySuffix(int day)
{
    switch (day)
    {
        case 1:
        case 21:
        case 31:
            return "st";
        case 2:
        case 22:
            return "nd";
        case 3:
        case 23:
            return "rd";
        default:
            return "th";
    }
}

and you can this method like;

string DateOfJoin = String.Format("{0}{1} {2}",
                                  dt.Day,
                                  GetDaySuffix(dt.Day),
                                  dt.ToString("MMM yyyy", CultureInfo.InvariantCulture));

which generates

Answer 2

use format below to get the similar without suffix see explanation

 string DateOfJoin = dt.Rows[0]["DOJ"].ToString("dd MMMM yyyy");

However to get the suffix you'll need to break it up so that you get the day separate i.e.

string DateOfJoin = dt.Rows[0]["DOJ"].ToString("dd MMMM yyyy");

I would use below if you really need the suffix

string day = dt.Rows[0]["DOJ"].ToString("dd");
day = GetDaySuffix(Int32.Parse(day));

This using a function to add the suffix that I originally found here

  string GetDaySuffix(int day)
{
    switch (day)
    {
        case 1:
        case 21:
        case 31:
            return "st";
        case 2:
        case 22:
            return "nd";
        case 3:
        case 23:
            return "rd";
        default:
            return "th";
    }
}



 string DateOfJoin = String.Format("{0} {1}", day, dt.Rows[0]["DOJ"].ToString("MMMM yyyy"));

Not tested but should be useful start

Answer 3

Try this

 string DateOfJoin =Convert.ToDateTime( dt.Rows[0]["DOJ"]).ToString("dd MMMM yyyy HH:mm:ss");
 string DateOfJoin =Convert.ToDateTime( dt.Rows[0]["DOJ"]).ToString("dd MMMM yyyy");

Answer 4

Depending on the specified culture, long date strings can be achieved by casting the object to DateTime and then calling ToString() with the format parameter.

string DateOfJoin = ((DateTime)dt.Rows[0]["DOJ"]).ToString("D");

Further reference here

Edit: Formatting with things like '1st', '2nd', requires you to write a custom method which evaluates the day number and appends the correct string to it, something like:

private string GetSuffix(DateTime dt)
{
   if(dt.Days % 10 == 1)
   {
      return dt.Days.ToString() + "st";
   }
   else if(dt.Days % 10 == 2)
   {
      return dt.Days.ToString() + "nd";
   }
   else if(dt.Days % 10 == 3)
   {
      return dt.Days.ToString() + "rd";
   }
   else
   {
      return dt.Days.ToString() + "th";
   } 
}

And then add this part of the string, to a ToString("Y") of the DateTime, like:

string DateOfJoin = GetSuffix((DateTime)dt.Rows[0]["DOJ"]) + " " + ((DateTime)dt.Rows[0]["DOJ"]).ToString("Y");

Answer 5

Var dt = DateTime.Parse(DateOfJoin);
dt.ToString('F');

See bottom of the following page for more formats

Msdn DateTime

Answer 6

This tack can be accomplished with using the following approach:

DateTime date = new DateTime(2015, 01, 01);
// 1st Jan 2015
string s = Ordinal.Add(date.Day) + date.ToString(" MMM yyyy", System.Globalization.CultureInfo.InvariantCulture); 

Here the Ordinal class is implemented as follows:

static class Ordinal {
    public static string Add(int num) {
        if(num <= 0) return num.ToString();
        switch(num % 100) {
            case 11:
            case 12:
            case 13:
                return num + "th";
        }
        switch(num % 10) {
            case 1: return num + "st";
            case 2: return num + "nd";
            case 3: return num + "rd";
            default: return num + "th";
        }
    }
}