How do I display a decimal value to 2 decimal places? - c#
When displaying the value of a decimal currently with .ToString(), it's accurate to like 15 decimal places, and since I'm using it to represent dollars and cents, I only want the output to be 2 decimal places.
Do I use a variation of .ToString() for this?
decimalVar.ToString("#.##"); // returns ".5" when decimalVar == 0.5m
or
decimalVar.ToString("0.##"); // returns "0.5" when decimalVar == 0.5m
or
decimalVar.ToString("0.00"); // returns "0.50" when decimalVar == 0.5m
I know this is an old question, but I was surprised to see that no one seemed to post an answer that;
Didn't use bankers rounding
Keeps the value as a decimal.
This is what I would use:
decimal.Round(yourValue, 2, MidpointRounding.AwayFromZero);
http://msdn.microsoft.com/en-us/library/9s0xa85y.aspx
decimalVar.ToString("F");
This will:
Round off to 2 decimal places eg. 23.456 → 23.46
Ensure that there
are always 2 decimal places eg. 23 → 23.00; 12.5 → 12.50
Ideal for displaying currency.
Check out the documentation on ToString("F") (thanks to Jon Schneider).
If you just need this for display use string.Format
String.Format("{0:0.00}", 123.4567m); // "123.46"
http://www.csharp-examples.net/string-format-double/
The "m" is a decimal suffix. About the decimal suffix:
http://msdn.microsoft.com/en-us/library/364x0z75.aspx
Given decimal d=12.345; the expressions d.ToString("C") or String.Format("{0:C}", d) yield $12.35 - note that the current culture's currency settings including the symbol are used.
Note that "C" uses number of digits from current culture. You can always override default to force necessary precision with C{Precision specifier} like String.Format("{0:C2}", 5.123d).
If you want it formatted with commas as well as a decimal point (but no currency symbol), such as 3,456,789.12...
decimalVar.ToString("n2");
There's a very important characteristic of Decimal that isn't obvious:
A Decimal 'knows' how many decimal places it has based upon where it came from
The following may be unexpected :
Decimal.Parse("25").ToString() => "25"
Decimal.Parse("25.").ToString() => "25"
Decimal.Parse("25.0").ToString() => "25.0"
Decimal.Parse("25.0000").ToString() => "25.0000"
25m.ToString() => "25"
25.000m.ToString() => "25.000"
Doing the same operations with Double will result in zero decimal places ("25") for all of the above examples.
If you want a decimal to 2 decimal places there's a high likelyhood it's because it's currency in which case this is probably fine for 95% of the time:
Decimal.Parse("25.0").ToString("c") => "$25.00"
Or in XAML you would use {Binding Price, StringFormat=c}
One case I ran into where I needed a decimal AS a decimal was when sending XML to Amazon's webservice. The service was complaining because a Decimal value (originally from SQL Server) was being sent as 25.1200 and rejected, (25.12 was the expected format).
All I needed to do was Decimal.Round(...) with 2 decimal places to fix the problem regardless of the source of the value.
// generated code by XSD.exe
StandardPrice = new OverrideCurrencyAmount()
{
TypedValue = Decimal.Round(product.StandardPrice, 2),
currency = "USD"
}
TypedValue is of type Decimal so I couldn't just do ToString("N2") and needed to round it and keep it as a decimal.
Here is a little Linqpad program to show different formats:
void Main()
{
FormatDecimal(2345.94742M);
FormatDecimal(43M);
FormatDecimal(0M);
FormatDecimal(0.007M);
}
public void FormatDecimal(decimal val)
{
Console.WriteLine("ToString: {0}", val);
Console.WriteLine("c: {0:c}", val);
Console.WriteLine("0.00: {0:0.00}", val);
Console.WriteLine("0.##: {0:0.##}", val);
Console.WriteLine("===================");
}
Here are the results:
ToString: 2345.94742
c: $2,345.95
0.00: 2345.95
0.##: 2345.95
===================
ToString: 43
c: $43.00
0.00: 43.00
0.##: 43
===================
ToString: 0
c: $0.00
0.00: 0.00
0.##: 0
===================
ToString: 0.007
c: $0.01
0.00: 0.01
0.##: 0.01
===================
Math.Round Method (Decimal, Int32)
Very rarely would you want an empty string if the value is 0.
decimal test = 5.00;
test.ToString("0.00"); //"5.00"
decimal? test2 = 5.05;
test2.ToString("0.00"); //"5.05"
decimal? test3 = 0;
test3.ToString("0.00"); //"0.00"
The top rated answer is incorrect and has wasted 10 minutes of (most) people's time.
Mike M.'s answer was perfect for me on .NET, but .NET Core doesn't have a decimal.Round method at the time of writing.
In .NET Core, I had to use:
decimal roundedValue = Math.Round(rawNumber, 2, MidpointRounding.AwayFromZero);
A hacky method, including conversion to string, is:
public string FormatTo2Dp(decimal myNumber)
{
// Use schoolboy rounding, not bankers.
myNumber = Math.Round(myNumber, 2, MidpointRounding.AwayFromZero);
return string.Format("{0:0.00}", myNumber);
}
None of these did exactly what I needed, to force 2 d.p. and round up as 0.005 -> 0.01
Forcing 2 d.p. requires increasing the precision by 2 d.p. to ensure we have at least 2 d.p.
then rounding to ensure we do not have more than 2 d.p.
Math.Round(exactResult * 1.00m, 2, MidpointRounding.AwayFromZero)
6.665m.ToString() -> "6.67"
6.6m.ToString() -> "6.60"
The top-rated answer describes a method for formatting the string representation of the decimal value, and it works.
However, if you actually want to change the precision saved to the actual value, you need to write something like the following:
public static class PrecisionHelper
{
public static decimal TwoDecimalPlaces(this decimal value)
{
// These first lines eliminate all digits past two places.
var timesHundred = (int) (value * 100);
var removeZeroes = timesHundred / 100m;
// In this implementation, I don't want to alter the underlying
// value. As such, if it needs greater precision to stay unaltered,
// I return it.
if (removeZeroes != value)
return value;
// Addition and subtraction can reliably change precision.
// For two decimal values A and B, (A + B) will have at least as
// many digits past the decimal point as A or B.
return removeZeroes + 0.01m - 0.01m;
}
}
An example unit test:
[Test]
public void PrecisionExampleUnitTest()
{
decimal a = 500m;
decimal b = 99.99m;
decimal c = 123.4m;
decimal d = 10101.1000000m;
decimal e = 908.7650m
Assert.That(a.TwoDecimalPlaces().ToString(CultureInfo.InvariantCulture),
Is.EqualTo("500.00"));
Assert.That(b.TwoDecimalPlaces().ToString(CultureInfo.InvariantCulture),
Is.EqualTo("99.99"));
Assert.That(c.TwoDecimalPlaces().ToString(CultureInfo.InvariantCulture),
Is.EqualTo("123.40"));
Assert.That(d.TwoDecimalPlaces().ToString(CultureInfo.InvariantCulture),
Is.EqualTo("10101.10"));
// In this particular implementation, values that can't be expressed in
// two decimal places are unaltered, so this remains as-is.
Assert.That(e.TwoDecimalPlaces().ToString(CultureInfo.InvariantCulture),
Is.EqualTo("908.7650"));
}
You can use system.globalization to format a number in any required format.
For example:
system.globalization.cultureinfo ci = new system.globalization.cultureinfo("en-ca");
If you have a decimal d = 1.2300000 and you need to trim it to 2 decimal places then it can be printed like this d.Tostring("F2",ci); where F2 is string formating to 2 decimal places and ci is the locale or cultureinfo.
for more info check this link
http://msdn.microsoft.com/en-us/library/dwhawy9k.aspx
https://msdn.microsoft.com/en-us/library/dwhawy9k%28v=vs.110%29.aspx
This link explains in detail how you can handle your problem and what you can do if you want to learn more. For simplicity purposes, what you want to do is
double whateverYouWantToChange = whateverYouWantToChange.ToString("F2");
if you want this for a currency, you can make it easier by typing "C2" instead of "F2"
The most applicable solution is
decimalVar.ToString("#.##");
Double Amount = 0;
string amount;
amount=string.Format("{0:F2}", Decimal.Parse(Amount.ToString()));
If you need to keep only 2 decimal places (i.e. cut off all the rest of decimal digits):
decimal val = 3.14789m;
decimal result = Math.Floor(val * 100) / 100; // result = 3.14
If you need to keep only 3 decimal places:
decimal val = 3.14789m;
decimal result = Math.Floor(val * 1000) / 1000; // result = 3.147
var arr = new List<int>() { -4, 3, -9, 0, 4, 1 };
decimal result1 = arr.Where(p => p > 0).Count();
var responseResult1 = result1 / arr.Count();
decimal result2 = arr.Where(p => p < 0).Count();
var responseResult2 = result2 / arr.Count();
decimal result3 = arr.Where(p => p == 0).Count();
var responseResult3 = result3 / arr.Count();
Console.WriteLine(String.Format("{0:#,0.000}", responseResult1));
Console.WriteLine(String.Format("{0:#,0.0000}", responseResult2));
Console.WriteLine(String.Format("{0:#,0.00000}", responseResult3));
you can put as many 0 as you want.
Related
C# decimal remembers precision: how to adjust
Decimal remembers its number of decimal places: Console.WriteLine(1.1m); // 1.1 Console.WriteLine(1.10m); // 1.10 I am in a situation where I do not control the serialization (so I cannot use a more explicitly formatted stringify), but there is a requirement to output exactly 2 decimal places. So, I am trying to force any decimal into the same value but with 2 decimal places. var d = Decimal.Round(1.1m, 2, MidpointRounding.AwayFromZero); // Try to increase precision Console.WriteLine(d); // Is 1.1, but I need 1.10 It is achievable with ToString("F") and then Decimal.Parse(). But is there a neater way than via ToString()?
In order to ensure/increase "precision" up to 2 digits after the decimal point you can add 0.00m: 1.1m + 0.00m == 1.10m The same idea generalized: private static decimal FormatOut(decimal value, int afterDecimal) { // rounding up and adding 0.00…0 - afterDecimal zeroes return Decimal.Round(value, afterDecimal, MidpointRounding.AwayFromZero) + new decimal(0, 0, 0, false, (byte)afterDecimal); } ... decimal result = FormatOut(1.1m, 2); // 1.10 Console.Write(result);
Compare decimals to 1 decimal point?
I'm new to C# (Come from Java/C++ at uni so it's not really that new I guess) but for a project I'm needing to compare decimals. e.g. a = 1234.123 b = 1234.142 Decimal.Compare() will of course say they're not the same as a is smaller than b. What I want to do is compare it to that first decimal place (1 and 1) so it would return true. The only way I've been able to think of is to convert it to use Decimal.GetBits() but I was hoping there is a simpler way I just haven't thought of yet.
You can round a decimal to one fractional digit and then compare them. if (Decimal.Round(d1,1) == Decimal.Round(d2,1)) Console.WriteLine("Close enough."); And, if rounding (with default midpoint handling) is not what you want, Decimal types can also be used with all the other options, like those I covered in this earlier answer.
You can use Math.Truncate(Decimal) (MSDN) Calculates the integral part of a specified decimal number. Coding example. Decimal a = 1234.123m; Decimal b = 1234.142m; Decimal A = Math.Truncate(a * 10); Console.WriteLine(A);// <= Will out 12341 Decimal B = Math.Truncate(b * 10); Console.WriteLine(B);// <= Will out 12341 Console.WriteLine(Decimal.Compare(A, B)); // Will out 0 ; A and B are equal. Which means a,b are equal to first decimal place Note : This was tested and posted . Also simple one line comparison : Decimal a = 1234.123m; Decimal b = 1234.142m; if(Decimal.Compare(Math.Truncate(a*10),Math.Truncate(b*10))==0){ Console.WriteLine("Equal upto first decimal place"); // <= Will out this for a,b }
Converting a Decimal to a string equivalent
I am working with a decimal type in my application and have to submit that value as a string to a legacy web service which accepts values as strings. Its a payment value, so: 1000 => "100000" (1000 dollars and 00 cents) 131.11 => "13111" I thought I'd multiply by 100 initially but ran into some cases that don't work as expected. EDIT: OK I will clarify: decimal val = 145.99m; Console.WriteLine((val * 100).ToString()); results in: 14599.00 but I really need 14599 without the decimal points, since the value the other side is expecting is 145 dollars and 99 cents. I was thinking there may be a different way rather than doing something like String.Replace(".00", string.Empty) or is this the only way?
If you want it to be 0 decimal places, you can do myDecimal.ToString("0");. For example: decimal myDecimal = 25.99m; (myDecimal*100m).ToString("0"); // 2599 While if your decimal had more decimal places: decimal myDecimal = 3.14159m; (myDecimal*100m).ToString("0"); // 314 EDIT: If your decimal has extra decimal places, it performs an .5 away round. (in my experience, VS2008).
You could simply convert to float, divide by 100, and convert to string then (supposing you store the number as an integer).
You could multiply by 100 then cast it to an integer based (integer in the mathematical sense) to truncate the decimal portion, and then ToString() the result. decimal x = 45.13m; string output = ((long)(x * 100)).ToString();
two decimal places for decimal/money field
I have a table with money field in my database. I have created entity and created decimal property for that money field. When the value of that field is displayed on My MVC3 view, It has four zeros 0000 after decimal like this : 5489.0000. I want to show only two 00 or decimal places. How can I fix it. Why it is showing four decimal places even I declared property as decimal. Please suggest.
The SQL Server money datatype internally is a 64-bit integer with an implied scale of 4 decimal places. To quote Books Online, it is accurate "to ten-thousandsth of a currency unit." It is, the rough equivalent of a decimal(19,4). The reason for the scale of 4 rather than 2 is to maintain precision in the results of arithmetic. Your ordinary currency value has a scale of 2 (e.g. $3.27) Multiplication or division of two numbers scaled to 2 decimal places gives a results that is precise to 4 decimal places: 9.23 divided by 3.27 yields a result of 2.82262996941896 (approximately). You can carry the result to whatever accuracy (number of decimal places) you desire. However, the result is only precise to 4 decimal places (2.8226) as the original values were only precise to 2 decimal places. That measurement is precise to within 1/2 of the smallest unit specified (+/- 0.005). But I digress. As a result of a SQL Server money value having an implied scale of 4, ADO.Net converts the value to a System.Decimal with a scale of 4. And since System.Decimal tracks scale, when you convert it to string, you get 4 decimal places. To get fewer, you can Round it before conversion, using the appropriate Decimal.Round() overload, or Format it as desired (eg. (3.27M).ToString("0.00") ;. Hope this helps. This program: namespace Sandbox { using System ; class Program { static void Main( string[] args ) { decimal pi = (decimal) Math.PI ; string piText = pi.ToString("0.00"); Console.WriteLine("PI to 2 decimal places is {0} one way, and {1:0.00} another" , piText , pi ) ; return; } } } Produces what you'd expect: PI to 2 decimal places is 3.14 one way, and 3.14 another Cheers, N.
You have to format the string. One thing you can do if it money you want to display is: static void Main () { decimal x = 0.999m; decimal y = 9999999999999999999999999999m; Console.WriteLine("My amount = {0:C}", x); Console.WriteLine("Your amount = {0:C}", y); } } OUTPUT: Output My amount = $1.00 Your amount = $9,999,999,999,999,999,999,999,999,999.00 the {0:C} is the currency Format Hope this helps!
Here is everything you need to know about formatting strings. http://blog.stevex.net/string-formatting-in-csharp/
I used this and it worked: YourDecField.ToString("N")
How do you round a number to two decimal places in C#?
I want to do this using the Math.Round function
Here's some examples: decimal a = 1.994444M; Math.Round(a, 2); //returns 1.99 decimal b = 1.995555M; Math.Round(b, 2); //returns 2.00 You might also want to look at bankers rounding / round-to-even with the following overload: Math.Round(a, 2, MidpointRounding.ToEven); There's more information on it here.
Try this: twoDec = Math.Round(val, 2)
If you'd like a string > (1.7289).ToString("#.##") "1.73" Or a decimal > Math.Round((Decimal)x, 2) 1.73m But remember! Rounding is not distributive, ie. round(x*y) != round(x) * round(y). So don't do any rounding until the very end of a calculation, else you'll lose accuracy.
Personally I never round anything. Keep it as resolute as possible, since rounding is a bit of a red herring in CS anyway. But you do want to format data for your users, and to that end, I find that string.Format("{0:0.00}", number) is a good approach.
Wikipedia has a nice page on rounding in general. All .NET (managed) languages can use any of the common language run time's (the CLR) rounding mechanisms. For example, the Math.Round() (as mentioned above) method allows the developer to specify the type of rounding (Round-to-even or Away-from-zero). The Convert.ToInt32() method and its variations use round-to-even. The Ceiling() and Floor() methods are related. You can round with custom numeric formatting as well. Note that Decimal.Round() uses a different method than Math.Round(); Here is a useful post on the banker's rounding algorithm. See one of Raymond's humorous posts here about rounding...
// convert upto two decimal places String.Format("{0:0.00}", 140.6767554); // "140.67" String.Format("{0:0.00}", 140.1); // "140.10" String.Format("{0:0.00}", 140); // "140.00" Double d = 140.6767554; Double dc = Math.Round((Double)d, 2); // 140.67 decimal d = 140.6767554M; decimal dc = Math.Round(d, 2); // 140.67 ========= // just two decimal places String.Format("{0:0.##}", 123.4567); // "123.46" String.Format("{0:0.##}", 123.4); // "123.4" String.Format("{0:0.##}", 123.0); // "123" can also combine "0" with "#". String.Format("{0:0.0#}", 123.4567) // "123.46" String.Format("{0:0.0#}", 123.4) // "123.4" String.Format("{0:0.0#}", 123.0) // "123.0"
If you want to round a number, you can obtain different results depending on: how you use the Math.Round() function (if for a round-up or round-down), you're working with doubles and/or floats numbers, and you apply the midpoint rounding. Especially, when using with operations inside of it or the variable to round comes from an operation. Let's say, you want to multiply these two numbers: 0.75 * 0.95 = 0.7125. Right? Not in C# Let's see what happens if you want to round to the 3rd decimal: double result = 0.75d * 0.95d; // result = 0.71249999999999991 double result = 0.75f * 0.95f; // result = 0.71249997615814209 result = Math.Round(result, 3, MidpointRounding.ToEven); // result = 0.712. Ok result = Math.Round(result, 3, MidpointRounding.AwayFromZero); // result = 0.712. Should be 0.713 As you see, the first Round() is correct if you want to round down the midpoint. But the second Round() it's wrong if you want to round up. This applies to negative numbers: double result = -0.75 * 0.95; //result = -0.71249999999999991 result = Math.Round(result, 3, MidpointRounding.ToEven); // result = -0.712. Ok result = Math.Round(result, 3, MidpointRounding.AwayFromZero); // result = -0.712. Should be -0.713 So, IMHO, you should create your own wrap function for Math.Round() that fit your requirements. I created a function in which, the parameter 'roundUp=true' means to round to next greater number. That is: 0.7125 rounds to 0.713 and -0.7125 rounds to -0.712 (because -0.712 > -0.713). This is the function I created and works for any number of decimals: double Redondea(double value, int precision, bool roundUp = true) { if ((decimal)value == 0.0m) return 0.0; double corrector = 1 / Math.Pow(10, precision + 2); if ((decimal)value < 0.0m) { if (roundUp) return Math.Round(value, precision, MidpointRounding.ToEven); else return Math.Round(value - corrector, precision, MidpointRounding.AwayFromZero); } else { if (roundUp) return Math.Round(value + corrector, precision, MidpointRounding.AwayFromZero); else return Math.Round(value, precision, MidpointRounding.ToEven); } } The variable 'corrector' is for fixing the inaccuracy of operating with floating or double numbers.
This is for rounding to 2 decimal places in C#: label8.Text = valor_cuota .ToString("N2") ; In VB.NET: Imports System.Math round(label8.text,2)
I know its an old question but please note for the following differences between Math round and String format round: decimal d1 = (decimal)1.125; Math.Round(d1, 2).Dump(); // returns 1.12 d1.ToString("#.##").Dump(); // returns "1.13" decimal d2 = (decimal)1.1251; Math.Round(d2, 2).Dump(); // returns 1.13 d2.ToString("#.##").Dump(); // returns "1.13"
Had a weird situation where I had a decimal variable, when serializing 55.50 it always sets default value mathematically as 55.5. But whereas, our client system is seriously expecting 55.50 for some reason and they definitely expected decimal. Thats when I had write the below helper, which always converts any decimal value padded to 2 digits with zeros instead of sending a string. public static class DecimalExtensions { public static decimal WithTwoDecimalPoints(this decimal val) { return decimal.Parse(val.ToString("0.00")); } } Usage should be var sampleDecimalValueV1 = 2.5m; Console.WriteLine(sampleDecimalValueV1.WithTwoDecimalPoints()); decimal sampleDecimalValueV1 = 2; Console.WriteLine(sampleDecimalValueV1.WithTwoDecimalPoints()); Output: 2.50 2.00
One thing you may want to check is the Rounding Mechanism of Math.Round: http://msdn.microsoft.com/en-us/library/system.midpointrounding.aspx Other than that, I recommend the Math.Round(inputNumer, numberOfPlaces) approach over the *100/100 one because it's cleaner.
You should be able to specify the number of digits you want to round to using Math.Round(YourNumber, 2) You can read more here.
Math.Floor(123456.646 * 100) / 100 Would return 123456.64
string a = "10.65678"; decimal d = Math.Round(Convert.ToDouble(a.ToString()),2)
public double RoundDown(double number, int decimalPlaces) { return Math.Floor(number * Math.Pow(10, decimalPlaces)) / Math.Pow(10, decimalPlaces); }