Some times we face the problem if we want to convert the date format in to words then we have many options to do this task, i will give the answer using C# Class.
Solution :-
For example : we get the date in c# with code (DateTime.Today) and convert it in words.
First Create a Class with Name class 'DateInWords' and Code Like this :-
using System;
using System.Collections.Generic;
using System.Linq;
using System.Web;
public static class DateInWords
{
static int firstTime = 0;
static readonly string[] ones = new string[] { "", "One", "Two", "Three", "Four", "Five", "Six", "Seven", "Eight", "Nine" };
static readonly string[] firsts = new string[] { "", "First", "Second", "Third", "Fourth", "Fifth", "Sixth", "Seventh", "Eighth", "Ninth" };
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;
}
else
{
string friendlyInt = leftDigits;
if (friendlyInt.Length > 0)
friendlyInt += " ";
if (n < 10 & firstTime == 0)
{
friendlyInt += firsts[n];
}
else if (n < 10 & firstTime != 0)
{
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, (firsts[n / 100] + " Hundred"), 0);
else
friendlyInt += FriendlyInteger(n % 1000, FriendlyInteger(n / 1000, "", thousands + 1), 0);
firstTime = firstTime + 1;
return friendlyInt + thousandsGroups[thousands];
}
}
public static string DateToWritten(DateTime date)
{
firstTime = 0;
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);
}
}
and this class will return the date format in word .. So firstly we must pass the parameter in this class with any date format given in standards ...
Example :
var outputDate = DateInWords.DateToWritten(DateTime.Today);// Current date is passing in DateToWritten function of 'DateInWords' static class.
dateLabel.Text = outputDate;
the output will display in 'dateLabel'
That’s it!!…..Happy Programming...
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