Using String Format to show decimal up to 2 places or simple integer

When dealing with decimals coming from a (T-)SQL database, you want to be able to convert nullable and non-nullable decimals with x decimal places and be able to review the code easily against your table definitions - and of course, display the right number of decimals to the user.

Unfortunately, the Entity Framework doesn't automatically convert something like a SQL decimal(18,2) into a .NET equivalent with the same number of decimal places (since there is only decimal with full precision available). You have to truncate the decimal places manually.

So, I did it this way:

public static class Extensions
{
    public static string ToStringDecimal(this decimal d, byte decimals)
    {
        var fmt = (decimals>0) ? "0." + new string('0', decimals) : "0";
        return d.ToString(fmt);
    }

    public static string ToStringDecimal(this decimal? d, byte decimals)
    {
        if (!d.HasValue) return "";
        return ToStringDecimal(d.Value, decimals);
    }
}

Example Usage:

void Main()
{
    decimal d = (decimal)1.2345;
    decimal? d2 = null; 

    Console.WriteLine(d.ToStringDecinal(2)); // prints: "1.23" (2 decimal places)
    Console.WriteLine(d.ToStringDecinal(0)); // prints: "1" (show integer number)
    Console.WriteLine(d2.ToStringDecimal(2)); // prints: "" (show null as empty string)
}

A recent project has a similar requirement. I wrote this decimal extension method, which uses the currency ("C") Format Specifier. In addition to removing zeros, it also has options for decimal digits precision, currency symbol, separator and culture.

public static DecimalExtension{

     public static string ToCurrency(this decimal val, 
                                     int precision = 2, 
                                     bool currencySymbol = false, 
                                     bool separator = false, 
                                     CultureInfo culture = null)
      {     
         if(culture == null) culture = new CultureInfo("en-US");
                                                   
         NumberFormatInfo nfi = culture.NumberFormat;
         nfi.CurrencyDecimalDigits = precision;
            
         string zeros = new String('0', precision);       
                    
         //Remove zeros
         var result = val.ToString("C",fi).Replace(nfi.CurrencyDecimalSeparator + zeros,"");
                     
         if(!separator) result = result.Replace(nfi.CurrencyGroupSeparator,"");
                    
         return currencySymbol? result: result.Replace(nfi.CurrencySymbol,"");      
        }   
}

Examples:

decimal Total = 123.00M;
Console.WriteLine(Total.ToCurrency());  
//output: 123

decimal Total = 1123.12M;
Console.WriteLine(Total.ToCurrency()); 
//Output:  1123.12

Console.WriteLine(Total.ToCurrency(4));
//Output:  1123.1200

Console.WriteLine(Total.ToCurrency(2,true,true));
//output:  $1,123.12
 
CultureInfo culture = new CultureInfo("pt-BR")  //Brazil
Console.WriteLine(Total.ToCurrency(2,true,true, culture));
//output:  R$ 1.123,12

Try this:

var number = 123.4567;
var str = number.ToString("N2");

Try This

string Output = String.Format("{0:0.00}", Decimal.Parse(InputStringValue));

To get one,two,three or four decimal places you can try below formats in place of 123 you can place your own required output

Input                                         Output

String.Format("{0:0.0}", "123.0")               123.0    // For 1 decimal place
String.Format("{0:0.00}", "123.00")           123.00   // For 2 decimal place
String.Format("{0:0.000}", "123.000")         123.000  // For 3 decimal place
String.Format("{0:0.0000}", "123.0000")       123.0000 // For 4 decimal place