This will work:
decimal source = 2.4200m;
string output = ((double)source).ToString();
Or if your initial value is string
:
string source = "2.4200";
string output = double.Parse(source).ToString();
Pay attention to this comment.
Use the hash (#
) symbol to only display trailing 0's when necessary. See the tests below.
decimal num1 = 13.1534545765;
decimal num2 = 49.100145;
decimal num3 = 30.000235;
num1.ToString("0.##"); //13.15%
num2.ToString("0.##"); //49.1%
num3.ToString("0.##"); //30%
I ran into the same problem but in a case where I do not have control of the output to string, which was taken care of by a library. After looking into details in the implementation of the Decimal type (see http://msdn.microsoft.com/en-us/library/system.decimal.getbits.aspx), I came up with a neat trick (here as an extension method):
public static decimal Normalize(this decimal value)
{
return value/1.000000000000000000000000000000000m;
}
The exponent part of the decimal is reduced to just what is needed. Calling ToString() on the output decimal will write the number without any trailing 0. E.g.
1.200m.Normalize().ToString();
how about this:
public static string TrimEnd(this decimal d)
{
string str = d.ToString();
if (str.IndexOf(".") > 0)
{
str = System.Text.RegularExpressions.Regex.Replace(str.Trim(), "0+?$", " ");
str = System.Text.RegularExpressions.Regex.Replace(str.Trim(), "[.]$", " ");
}
return str;
}
Very simple answer is to use TrimEnd(). Here is the result,
double value = 1.00;
string output = value.ToString().TrimEnd('0');
Output is 1 If my value is 1.01 then my output will be 1.01
You can just set as:
decimal decNumber = 23.45600000m;
Console.WriteLine(decNumber.ToString("0.##"));
I found an elegant solution from http://dobrzanski.net/2009/05/14/c-decimaltostring-and-how-to-get-rid-of-trailing-zeros/
Basically
decimal v=2.4200M;
v.ToString("#.######"); // Will return 2.42. The number of # is how many decimal digits you support.
In case you want to keep decimal number, try following example:
number = Math.Floor(number * 100000000) / 100000000;
Trying to do more friendly solution of DecimalToString (https://stackoverflow.com/a/34486763/3852139):
private static decimal Trim(this decimal value)
{
var s = value.ToString(CultureInfo.InvariantCulture);
return s.Contains(CultureInfo.InvariantCulture.NumberFormat.NumberDecimalSeparator)
? Decimal.Parse(s.TrimEnd('0'), CultureInfo.InvariantCulture)
: value;
}
private static decimal? Trim(this decimal? value)
{
return value.HasValue ? (decimal?) value.Value.Trim() : null;
}
private static void Main(string[] args)
{
Console.WriteLine("=>{0}", 1.0000m.Trim());
Console.WriteLine("=>{0}", 1.000000023000m.Trim());
Console.WriteLine("=>{0}", ((decimal?) 1.000000023000m).Trim());
Console.WriteLine("=>{0}", ((decimal?) null).Trim());
}
Output:
=>1
=>1.000000023
=>1.000000023
=>
This is simple.
decimal decNumber = Convert.ToDecimal(value);
return decNumber.ToString("0.####");
Tested.
Cheers :)
The following code could be used to not use the string type:
int decimalResult = 789.500
while (decimalResult>0 && decimalResult % 10 == 0)
{
decimalResult = decimalResult / 10;
}
return decimalResult;
Returns 789.5
try this code:
string value = "100";
value = value.Contains(".") ? value.TrimStart('0').TrimEnd('0').TrimEnd('.') : value.TrimStart('0');
I use this code to avoid "G29" scientific notation:
public static string DecimalToString(this decimal dec)
{
string strdec = dec.ToString(CultureInfo.InvariantCulture);
return strdec.Contains(".") ? strdec.TrimEnd('0').TrimEnd('.') : strdec;
}
EDIT: using system CultureInfo.NumberFormat.NumberDecimalSeparator :
public static string DecimalToString(this decimal dec)
{
string sep = CultureInfo.CurrentCulture.NumberFormat.NumberDecimalSeparator;
string strdec = dec.ToString(CultureInfo.CurrentCulture);
return strdec.Contains(sep) ? strdec.TrimEnd('0').TrimEnd(sep.ToCharArray()) : strdec;
}
try like this
string s = "2.4200";
s = s.TrimStart('0').TrimEnd('0', '.');
and then convert that to float
In my opinion its safer to use Custom Numeric Format Strings.
decimal d = 0.00000000000010000000000m;
string custom = d.ToString("0.#########################");
// gives: 0,0000000000001
string general = d.ToString("G29");
// gives: 1E-13
Depends on what your number represents and how you want to manage the values: is it a currency, do you need rounding or truncation, do you need this rounding only for display?
If for display consider formatting the numbers are x.ToString("")
http://msdn.microsoft.com/en-us/library/dwhawy9k.aspx and
http://msdn.microsoft.com/en-us/library/0c899ak8.aspx
If it is just rounding, use Math.Round overload that requires a MidPointRounding overload
http://msdn.microsoft.com/en-us/library/ms131274.aspx)
If you get your value from a database consider casting instead of conversion: double value = (decimal)myRecord["columnName"];
A very low level approach, but I belive this would be the most performant way by only using fast integer calculations (and no slow string parsing and culture sensitive methods):
public static decimal Normalize(this decimal d)
{
int[] bits = decimal.GetBits(d);
int sign = bits[3] & (1 << 31);
int exp = (bits[3] >> 16) & 0x1f;
uint a = (uint)bits[2]; // Top bits
uint b = (uint)bits[1]; // Middle bits
uint c = (uint)bits[0]; // Bottom bits
while (exp > 0 && ((a % 5) * 6 + (b % 5) * 6 + c) % 10 == 0)
{
uint r;
a = DivideBy10((uint)0, a, out r);
b = DivideBy10(r, b, out r);
c = DivideBy10(r, c, out r);
exp--;
}
bits[0] = (int)c;
bits[1] = (int)b;
bits[2] = (int)a;
bits[3] = (exp << 16) | sign;
return new decimal(bits);
}
private static uint DivideBy10(uint highBits, uint lowBits, out uint remainder)
{
ulong total = highBits;
total <<= 32;
total = total | (ulong)lowBits;
remainder = (uint)(total % 10L);
return (uint)(total / 10L);
}
Source: Stackoverflow.com