Best way to display decimal without trailing zeroes

C#FormattingDecimal

C# Problem Overview


Is there a display formatter that will output decimals as these string representations in c# without doing any rounding?

// decimal -> string

20 -> 20
20.00 -> 20
20.5 -> 20.5
20.5000 -> 20.5
20.125 -> 20.125
20.12500 -> 20.125
0.000 -> 0

{0.#} will round, and using some Trim type function will not work with a bound numeric column in a grid.

C# Solutions


Solution 1 - C#

Do you have a maximum number of decimal places you'll ever need to display? (Your examples have a max of 5).

If so, I would think that formatting with "0.#####" would do what you want.

    static void Main(string[] args)
    {
        var dList = new decimal[] { 20, 20.00m, 20.5m, 20.5000m, 20.125m, 20.12500m, 0.000m };

        foreach (var d in dList)
            Console.WriteLine(d.ToString("0.#####"));
    }

Solution 2 - C#

I just learned how to properly use the G format specifier. See the MSDN Documentation. There is a note a little way down that states that trailing zeros will be preserved for decimal types when no precision is specified. Why they would do this I do not know, but specifying the maximum number of digits for our precision should fix that problem. So for formatting decimals, G29 is the best bet.

decimal test = 20.5000m;
test.ToString("G"); // outputs 20.5000 like the documentation says it should
test.ToString("G29"); // outputs 20.5 which is exactly what we want

Solution 3 - C#

This string format should make your day: "0.#############################". Keep in mind that decimals can have at most 29 significant digits though.

Examples:

? (1000000.00000000000050000000000m).ToString("0.#############################")
-> 1000000.0000000000005

? (1000000.00000000000050000000001m).ToString("0.#############################")
-> 1000000.0000000000005

? (1000000.0000000000005000000001m).ToString("0.#############################")
-> 1000000.0000000000005000000001

? (9223372036854775807.0000000001m).ToString("0.#############################")
-> 9223372036854775807

? (9223372036854775807.000000001m).ToString("0.#############################")
-> 9223372036854775807.000000001

Solution 4 - C#

This is yet another variation of what I saw above. In my case I need to preserve all significant digits to the right of the decimal point, meaning drop all zeros after the most significant digit. Just thought it would be nice to share. I cannot vouch for the efficiency of this though, but when try to achieve aesthetics, you are already pretty much damned to inefficiencies.

public static string ToTrimmedString(this decimal target)
{
	string strValue = target.ToString(); //Get the stock string

	//If there is a decimal point present
	if (strValue.Contains("."))
	{
		//Remove all trailing zeros
		strValue = strValue.TrimEnd('0');

		//If all we are left with is a decimal point
		if (strValue.EndsWith(".")) //then remove it
			strValue = strValue.TrimEnd('.');
	}

	return strValue;
}

That's all, just wanted to throw in my two cents.

Solution 5 - C#

Another solution, based on dyslexicanaboko's answer, but independent of the current culture:

public static string ToTrimmedString(this decimal num)
{
    string str = num.ToString();
    string decimalSeparator = CultureInfo.CurrentCulture.NumberFormat.NumberDecimalSeparator;
    if (str.Contains(decimalSeparator))
    {
        str = str.TrimEnd('0');
        if(str.EndsWith(decimalSeparator))
        {
            str = str.RemoveFromEnd(1);
        }
    }
    return str;
}

public static string RemoveFromEnd(this string str, int characterCount)
{
    return str.Remove(str.Length - characterCount, characterCount);
}

Solution 6 - C#

Extension method:

public static class Extensions
{
	public static string TrimDouble(this string temp)
	{
		var value = temp.IndexOf('.') == -1 ? temp : temp.TrimEnd('.', '0');
		return value == string.Empty ? "0" : value;
	}
}

Example code:

double[] dvalues = {20, 20.00, 20.5, 20.5000, 20.125, 20.125000, 0.000};
foreach (var value in dvalues)
	Console.WriteLine(string.Format("{0} --> {1}", value, value.ToString().TrimDouble()));

Console.WriteLine("==================");

string[] svalues = {"20", "20.00", "20.5", "20.5000", "20.125", "20.125000", "0.000"};
foreach (var value in svalues)
	Console.WriteLine(string.Format("{0} --> {1}", value, value.TrimDouble()));

Output:

20 --> 20
20 --> 20
20,5 --> 20,5
20,5 --> 20,5
20,125 --> 20,125
20,125 --> 20,125
0 --> 0
==================
20 --> 20
20.00 --> 2
20.5 --> 20.5
20.5000 --> 20.5
20.125 --> 20.125
20.125000 --> 20.125
0.000 --> 0

Solution 7 - C#

I don't think it's possible out-of-the-box but a simple method like this should do it

public static string TrimDecimal(decimal value)
{
    string result = value.ToString(System.Globalization.CultureInfo.InvariantCulture);
    if (result.IndexOf('.') == -1)
        return result;

    return result.TrimEnd('0', '.');
}

Solution 8 - C#

It's quite easy to do out of the box:

Decimal YourValue; //just as example   
String YourString = YourValue.ToString().TrimEnd('0','.');

that will remove all trailing zeros from your Decimal.

The only thing that you need to do is add .ToString().TrimEnd('0','.'); to a decimal variable to convert a Decimal into a String without trailing zeros, like in the example above.

In some regions it should be a .ToString().TrimEnd('0',','); (where they use a comma instead of a point, but you can also add a dot and a comma as parameters to be sure).

(you can also add both as parameters)

Solution 9 - C#

You can use the G0 format string if you are happy to accept scientific notation as per the documentation:

>Fixed-point notation is used if the exponent that would result from expressing the number in scientific notation is greater than -5 and less than the precision specifier; otherwise, scientific notation is used.

You can use this format string as a parameter to the .ToString() method, or by specifying it as a within an interpolated string. Both are shown below.

decimal hasTrailingZeros = 20.12500m;
Console.WriteLine(hasTrailingZeros.ToString("G0")); // outputs 20.125
Console.WriteLine($"{hasTrailingZeros:G0}"); // outputs 20.125

decimal fourDecimalPlaces = 0.0001m;
Console.WriteLine(fourDecimalPlaces.ToString("G0")); // outputs 0.0001
Console.WriteLine($"{fourDecimalPlaces:G0}"); // outputs 0.0001

decimal fiveDecimalPlaces = 0.00001m;
Console.WriteLine(fiveDecimalPlaces.ToString("G0")); // outputs 1E-05
Console.WriteLine($"{fiveDecimalPlaces:G0}"); // outputs 1E-05

Solution 10 - C#

decimal val = 0.000000000100m;
string result = val == 0 ? "0" : val.ToString().TrimEnd('0').TrimEnd('.');

Solution 11 - C#

I ended up with the following code:

    public static string DropTrailingZeros(string test)
    {
        if (test.Contains(CultureInfo.InvariantCulture.NumberFormat.NumberDecimalSeparator))
        {
            test = test.TrimEnd('0');
        }

        if (test.EndsWith(CultureInfo.InvariantCulture.NumberFormat.NumberDecimalSeparator))
        {
            test = test.Substring(0,
                test.Length - CultureInfo.InvariantCulture.NumberFormat.NumberDecimalSeparator.Length);
        }

        return test;
    }

Solution 12 - C#

I have end up with this variant:

public static string Decimal2StringCompact(decimal value, int maxDigits)
    {
        if (maxDigits < 0) maxDigits = 0;
        else if (maxDigits > 28) maxDigits = 28;
        return Math.Round(value, maxDigits, MidpointRounding.ToEven).ToString("0.############################", CultureInfo.InvariantCulture);
    }

Advantages:

you can specify the max number of significant digits after the point to display at runtime;

you can explicitly specify a round method;

you can explicitly control a culture.

Solution 13 - C#

You can create extension method

public static class ExtensionMethod {
  public static decimal simplify_decimal(this decimal value) => decimal.Parse($"{this:0.############}");
}

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QuestionJC.View Question on Stackoverflow
Solution 1 - C#TobyView Answer on Stackoverflow
Solution 2 - C#SchmallsView Answer on Stackoverflow
Solution 3 - C#Erwin MayerView Answer on Stackoverflow
Solution 4 - C#dyslexicanabokoView Answer on Stackoverflow
Solution 5 - C#David DostalView Answer on Stackoverflow
Solution 6 - C#jgauffinView Answer on Stackoverflow
Solution 7 - C#Tim SkaugeView Answer on Stackoverflow
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