I am using next code to format double value:
String.Format("${0:0,0.#}",...);
It working great, but when numbers are less than 10, I got problem. Numbers are displayed as $03, $06 for example.
Please advise me correct string to have a double number in next format ddd,ddd,ddd,ddd.dd
Try this instead:
string result = string.Format("${0:#,##0.00}", d);
If your double represents a currency you should use:
string result = string.Format(CultureInfo.GetCultureInfo("en-US"), "{0:c}", d);
Note that if you omit the CultureInfo.InvariantCulture it could display using something other than $ on some computers. For example on my computer string.Format("{0:c}", d) gives 2,00 kr which might not be what you wanted.
In your example you don't actually need to use string.Format at all. You could use this instead:
string s = d.ToString("c", CultureInfo.GetCultureInfo("en-US"));
As well as being clearer and more concise, it also has the advantage of avoiding boxing. Of course if your format string is more complex than in your example then it would make sense to use string.Format.
And as a final remark I'd recommend against using doubles to store currency. A decimal type is probably more appropriate.
Use currency formatting:
String.Format("{0:C}", money);
String.Format("{0:C}", myDecimal);
or
myDecimal.ToString("C");
will display to two decimal places, include the comma separator and include the dollar sign (based on culture settings) in one fell swoop. If you want it to go to 1 or more than 2 decimal places, include a number after the C (i.e. C3)
Digits after decimal point
// just two decimal places
String.Format("{0:0.00}", 123.4567); // "123.46"
String.Format("{0:0.00}", 123.4); // "123.40"
String.Format("{0:0.00}", 123.0); // "123.00"
// max. 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"
// at least two digits before decimal point
String.Format("{0:00.0}", 123.4567); // "123.5"
String.Format("{0:00.0}", 23.4567); // "23.5"
String.Format("{0:00.0}", 3.4567); // "03.5"
String.Format("{0:00.0}", -3.4567); // "-03.5"
Thousands separator
String.Format("{0:0,0.0}", 12345.67); // "12,345.7"
String.Format("{0:0,0}", 12345.67); // "12,346"
Zero
Following code shows how can be formatted a zero (of double type).
String.Format("{0:0.0}", 0.0); // "0.0"
String.Format("{0:0.#}", 0.0); // "0"
String.Format("{0:#.0}", 0.0); // ".0"
String.Format("{0:#.#}", 0.0); // ""
Align numbers with spaces
String.Format("{0,10:0.0}", 123.4567); // " 123.5"
String.Format("{0,-10:0.0}", 123.4567); // "123.5 "
String.Format("{0,10:0.0}", -123.4567); // " -123.5"
String.Format("{0,-10:0.0}", -123.4567); // "-123.5 "
Custom formatting for negative numbers and zero
String.Format("{0:0.00;minus 0.00;zero}", 123.4567); // "123.46"
String.Format("{0:0.00;minus 0.00;zero}", -123.4567); // "minus 123.46"
String.Format("{0:0.00;minus 0.00;zero}", 0.0); // "zero"
Some funny examples
String.Format("{0:my number is 0.0}", 12.3); // "my number is 12.3"
String.Format("{0:0aaa.bbb0}", 12.3); // "12aaa.bbb3"
String.Format("${0:#,0.#}",...);
should do it.
See Custom Numeric Format Strings
Related
Which string format can convert this:
1 to 0.01
12 to 0.12
123 to 1.23
1234 to 12.34
12345 to 123.45
Inside my xamarin forms if a user types this number to an entry then it will automatically format that number. What is the string format of that?
For example: Text={Binding Price,StringFormat='${0:F0}'}
Seems like you just want division
var result = value / 100m;
Console.WriteLine(result);
--
Additional Resources
Division operator /
You can divide by 100, as others are saying, followed by applying the desired numeric formats.
Just as Michael Randall said above it's simple as that, in addiction you can check if it can be parsed first, something like this:
bool success = decimal.TryParse(value , out number);
if(success)
var result = value / (decimal)100
else
//Handle incorrect user input
C#: I have a decimal number, I want to separate the thousand numbers by a comma but not round the number. As answers of stack overflow, I used .ToString("#,##0") but it rounds the number. It's not acceptable for me to round the number!
In my model I have:
public decimal? ADecimalNumber { get; set; }
In Controller I have:
ViewBag.ADecimalNumber = result?.ADecimalNumber;
And in View I have:
#(ViewBag.ADecimalNumber != null ? ViewBag.ADecimalNumber.ToString("#,##0") : '-')
The ViewBag value is 11106.8314M in quick-watch mode, and the ViewBag.ADecimalNumber.ToString("#,##0") result is "11,107".
It's default behavior of custom specifier. See The "#" Custom Specifier.
The "##" format string causes the value to be rounded to the nearest digit preceding the decimal, where rounding away from zero is always used. For example, formatting 34.5 with "##" would result in the value 35.
Rounding is part of the behavior of the # custom digit formatter. Use Math.Truncate() to cut off the digits after the period if you do not want rounding:
Math.Truncate(ViewBag.ADecimalNumber).ToString("#,##0")
If you want to print n decimal places though, multiply the number with the nth-power of 10, truncate, then divide by that same power, for example: Math.Truncate(123456.78M * 10M)/10M will give 123456.7 which you could print using the format string #,###.#
Looking for:
95,4545454545455 -> 95 %
I tried using:
String resultAsPercentage = result.ToString("##0 %");
But, it shows
9545 %
Then, I solved my problem using regex:
Question: Why my ToString method hasn't worked? And how to fix it to avoid using regex?
Thanks in advance.
As documented on Custom Numeric Format Strings, the % modifier multiplies the value by 100 before inserting the %. It's intended to be used with fractions. To disable this special meaning of %, escape it by preceding it with #"\".
Alternatively, you could take the % out of the format string, and append it manually: result.ToString("##0") + " %".
If you don't care about rounding, you can use the following:
double result = 95.4545454545;
String resultAsPercentage = (int)result + " %";
System.out.println(resultAsPercentage);
Output is: 95 %
Casting to an int drops the decimal places without rounding
You can use thew P(ercentage) format specifier, you need to divide through 100 because the specifier multiplies it by 100:
decimal value = 95.4545454545455m;
String resultAsPercentage = (value / 100).ToString("P0"); // 95%
If you need the space between the value and the percentage symbol you could use this approach:
NumberFormatInfo nfi = (NumberFormatInfo)NumberFormatInfo.CurrentInfo.Clone();
nfi.PercentSymbol = " %";
String resultAsPercentage = (value / 100).ToString("P0", nfi); // 95 %
One way can be Clone a culture (like InvariantCulture), set it's PercentPositivePattern to 0, divide your value by 100 and get it's string representation using The percent ("P") format specifier with 0 precision and that cloned culture as;
var clone = (CultureInfo)CultureInfo.InvariantCulture.Clone();
clone.NumberFormat.PercentNegativePattern = 0;
Console.WriteLine(((int)95.4545454545455 / 100.0).ToString("P0", clone)); // 95 %
You can see all associated patterns on Remarks section on that page.
You can guaranteed to set PercentNegativePattern property as well for negative values.
double c, d, e;
double a = (c - d) / e;
double b = Math.Floor(a);
Debug.WriteLine(a.ToString() + " " + b.ToString());
Code above outputs "3 2" at one configuration where all numbers are double. How is this possible? Is it because of fractional error resulting from double operations? However I think a.ToString() should give the whole number with its fractional part.
It's just a matter of what double.ToString() does. Here's a short but complete program demonstrating the same thing:
using System;
public class Test
{
static void Main(string[] args)
{
// Find the largest double less than 3
long bits = BitConverter.DoubleToInt64Bits(3);
double a = BitConverter.Int64BitsToDouble(bits - 1);
double b = Math.Floor(a);
// Print them using the default conversion to string...
Console.WriteLine(a.ToString() + " " + b.ToString());
// Now use round-trip formatting...
Console.WriteLine(a.ToString("r") + " " + b.ToString("r"));
}
}
Output:
3 2
2.9999999999999996 2
Now double.ToString() is documented with:
This version of the ToString method implicitly uses the general numeric format specifier ("G") and the NumberFormatInfo for the current culture.
... and the general numeric format specifier docs state:
The precision specifier defines the maximum number of significant digits that can appear in the result string. If the precision specifier is omitted or zero, the type of the number determines the default precision, as indicated in the following table.
... where the table shows that the default precision for double is 15. If you consider 2.9999999999999996 rounded to 15 significant digits, you end up with 3.
In fact, the exact value of a here is:
2.999999999999999555910790149937383830547332763671875
... which again, is 3 when regarded with 15 significant digits.
I need all values to rounded to two decimal places. So 1.401 should round to 1.40, but Math.Round(value, 2) rounds to 1.4.
How can I force the trailing zero?
1.4 is the same as 1.40 - you just want to display it differently. Use a format string when calling ToString - like value.ToString("0.00")
1.4 == 1.40 the only time you'd ever need a trailing 0 is when you display the number..i.e. format it to string.
.ToString("N2");
The trailing zero is more of a formatting than a value issue, so use
foo.ToString("0.00")
I know this is an old question, but might help someone!
I am using a c# xml class to populate and then serialise to xml. One of the values is a double. If I assign a '7' to the value this gets serialised to '7' when I actually need '7.00'. Easiest way round this was just to do:
foo = doubleValue + 0.00M
And that makes the value 7.00 instead of just 7. Thought this was better then doing a ToString and then parsing it back.
The trailing zero is just a presentation. Math-wise, 1.40 and 1.4 are equivalent.
Use formatting instead to present it with the 2 decimal places:
String.Format("{0:0.00}", 1.4);
or
yourNumber.ToString("0.00");
It has to do with whether you use a decimal or a double.
While internally (as it appears from the Source Code) Math.Round() preserves the trailing zeros even on a double, still the fact that it is saved as a double in memory causes automatically to remove all trailing zeros
So if you do want tailing zeros, you can either use the string display functions to format it as others have answered, or make sure to pass in the original value as a decimal (causing to use internally Decimal.Math.Round which will deal only with decimals), and make sure to not cast the result to a double and also not to save it in a double variable.
Similarly if you have a decimal and you don't want trailing zeros, just cast it to a double (you can either cast the input to Math.Round or the result, it doesn't matter as long as somewhere in the way it is becoming a double).
It is a number (double?), so it doesn't have a trailing zero - you have to make it text and force a trailing zero.
You can use this function instead of round and just use it like you use round function.
import decimal
def printf(x, n):
d = decimal.Decimal(str(x))
d0 = -(d.as_tuple().exponent)
if d0 < n:
print("x = ", x)
else:
d1 = decimal.Decimal(str(round(x, n)))
d2 = d1.as_tuple().exponent
MAX = n + d2
if MAX == 0:
print("x = ", round(x, n))
else:
i = 0
print("x = ", round(x, n), end = '')
while i != MAX:
if i == (MAX - 1):
print("0")
else:
print("0", end = '')
i = i + 1
So you must have something like this.
>>> printf(0.500000000000001, 13)
>>> 0.5000000000000