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C# formatting a number with thousand separators but keeping the decimal places count as is

I want to add a thousands separator to a double number but want to keep the decimal places as is i.e. dont want any rounding.
#,# is solving my problem of adding the thousand separator but how do I preserve the decimal places ? #,# strips off the part after ..
I cannot use any culture or something like that & the developer whose function I am calling has only given me a way of changing the format by passing as parameter strFormat.
I did check other posts & even the docs but somehow not able to figure this out.
string strFormat = "#,#";
string str = double.parse("912123456.1123465789").ToString(strFormat);
//Expected here 912,123,456.1123465789
//Actual Output 912,123,456
//912123456.123 should give 912,123,456.123
//912123456.1 should give 912,123,456.1
//912123456.1123465789 should give 912,123,456.1123465789
//912123456 should give 912,123,456

Well, after looking at the documentation on Microsoft, it would appear that there is no particular way to allow a floating point position in a number - all characters in a format string are character placeholders.
I would recommend that you either use a very nasty predetermined number of #s to set the width of the decimal position, or the slightly less (or possibly more, depending on your outlook) nasty option of reading all numbers into an array, determining the longest decimal position, then building a format of #s using the result.
At the end of the day, this is a single format string that you can put into place, test and ensure it works, then come back later and fix if you find a better alternative.
Also, this is one of those things where you could put the string into a configuration setting and change as and when you need to - far more flexible.
To be honest, this is a very slight thing to be worried about in the grand scheme of performance and writing a program.
Technically, Udi Y gets my vote!

If you know the max number of decimal places, e.g. 10, then use:
string strFormat = "#,#0.##########";
Update:
This max number is known.
According to Microsoft documentation a Double value has up to 15 decimal digits of precision (including both before and after the decimal point). More than 15 digits will be rounded.
So if you must invoke that method of 'double.parse' and can only send the format, this is the best you can do:
string strFormat = "#,#0.###############";

You can calculate the formatting dynamically for each number:
public static void Main()
{
var number = 1234.12312323123;
var format = GetNumberFormat(number);
Console.WriteLine(number.ToString(format));
}
public static string GetNumberFormat(double number)
{
var numberAsString = number.ToString();
var decimalPartSize = numberAsString.Substring(numberAsString.LastIndexOf('.') + 1).Length;
return $"N{decimalPartSize}";
}
So
number = 1234.12312323123
will give you 1,234.12312323123. Works for negative numbers as well. Also, as we work with strings, there won't be any rounding errors or precision artifacts.

Format decimal to number of decimal places without scientific notation

I am currently formatting a double using the code:
myDouble.ToString("g4");
To get the first 4 decimal places. However I find this often switches over to scientific notation if the number is very large or very small. Is there an easy format string in C# to just have the first four decimal places, or zero if it is too small to be represented in that number of places?
For example, I would like:
1000 => 1000
0.1234567 => 0.1235
123456 => 123456 (Note: Not scientific notation)
0.000001234 => 0 (Note: Not scientific notation)

You can try like this:
0.1234567.ToString("0.####")
Also check Custom Numeric Format Strings
#
Replaces the "#" symbol with the corresponding digit if one is
present; otherwise, no digit appears in the result string.
Also as Jon as correctly pointed that it will round your number. See the note section
Rounding and Fixed-Point Format Strings
For fixed-point format strings
(that is, format strings that do not contain scientific notation
format characters), numbers are rounded to as many decimal places as
there are digit placeholders to the right of the decimal point.

Use the String.Format() method.
String.Format("{0:0.####}", 123.4567123); //output: 123.4567
Note: Num of #'s indicate the maximum number of digits after decimal that are required.

I agree with kjbartel comment.
I wanted exactly what the original question asked. But his question is slightly ambiguous.
The problem with ### format is it fills the slot if a digit can be represented or not.
So it does what the original question asks for some numbers but not others.
My basic need is, and it's a pretty common one, if the number is big I don't need to show decimal places. If the number is small I do want to show decimal places. Basically X number of significant digits.
The "Gn" Format will do significant digits, but it switches to scientific notation if you go over the number of digits. I don't want E notation, ever (same requirement as the question).
So I used fixed format ("Fn") but I calculate the width on the fly based on how "big" the number is.
var myFloatNumber = 123.4567;
var digits = (int) Math.Log10(myFloatNumber);
var maxDecimalplaces = 3;
var format = "F" + Math.Max(0,(maxDecimalplaces - digits));
I swear there was a way to do this in C++ (Visual Studio Flavor) in teh format statement or in C# and perhaps there is, but I can't find it.
So I came up with this. I could have converted to a string and measured length before decimal point as well. But converting it to a string twice felt wrong.

Rounding decimal number upto two decimal places

I am trying to round decimal number upto two decimal places which is working perfectly.
I am doing as below :
Math.Round(Amount, 2)
So, if I have Amount as 40000.4567, I am getting 40000.46which is exactly what I want.
Now problem is I have decimal number like 40000.0000, when I round it, the result is 40000, and what I really want is 40000.00. So round will always neglect trailing zeros.
To solve this problem, I have the option of converting it to string and use format , but I don't want to do that as that will be inefficient and I believe there must be some way to do it better.
I also tried something like
Decimal.Round(Amount, 2)
Now one way can be to check whether number contains anything in fractional part and use round function accordingly , but that is really bad way to do it.
I can't use truncate as well due to obvious reasons of this being related to amount.
What is the way around?

It is rounding correctly but you fail to understand that the value is not the format. There is no difference between the two values, 40000 and 40000.00, and you'll have a similar issue with something like 3.1.
Simply use formatting to output whatever number you have to two decimal places, such as with:
Console.WriteLine(String.Format("{0:0.00}", value));
or:
Console.WriteLine(value.ToString("0.00"));

You are mixing two things - rounding and output formatting. In order to output a number in a format you want you can use function string.Format with required format, for example:
decimal number = 1234.567m;
string.Format("{0:#.00}", number);
You can read more about custom numeric format strings in MSDN

I think what you're looking for is displaying two decimals, even if they are zero. You can use string.Format for this (I've also combined it with Round):
Console.WriteLine(string.Format("{0:0.00}", Math.Round(Amount, 2));

for rounding decimal number you can use
decimal number=200.5555m;
number= Math.Round(number, 2);
string numString= string.Format("{0:0.00}", number);

Convert.ToInt32 does not take the a zero ( 0 ) as a starting integer

I have a question or actually two regarding the
Convert.ToInt32 function or Int32.Parse.
It seems when I use Int32.Parse("06487965") or the other function it just seems to lose the 0.
So the output will be "6487965".
My questions:
Why is that exactly ?
How can I solve this without getting into crazy hacks ?

Leading-zeroes are purely a formatting idiom. An actual number has no leading zeroes as they make no sense and are superfluous.
You don't really have anything to solve, BUT, if you want to display the number at a later point with a leading zero, then you can use string formatting to achieve this.
int myNum = 6487965;
string formatted = string.Format("{0:00000000}", myNum);

Leading zeroes do not make sense mathematically
You can't have leading zeroes in an integer, if you must have leading zeroes best way is to keep your variable as string.

Because in that form, the leading 0, or 0s is/are nonsensical and redundant.
It can become useful in terms of formatting, for visual representation when you know how it should be formatted, but otherwise it is a non-issue.

Leading zeroes are meaningless when converting to a numeric type like Int32 - it only cares about the actual numeric value represented by the string you're converting from, which is the same regardless of the number of zeroes stuck on the front.
If later on in your program you want to convert your Int32 back to a string or output it to your user then at that point you need to do some formatting to get the number of leading zeroes you want. The MSDN article "How to: Pad a Number with Leading Zeros" would be a good place to start with that.

Leading 0's aren't represented when you ToString an int by default. You want to pass a format like "00000000".
int value = Int32.Parse("06487965");
string stringAgain = value.ToString("000000000");

integer 06487965 equals to integer 6487965

because, nobody uses pading, here is it:
int intValue = Int32.Parse("06487965");
string stringAgain = intValue.ToString().PadLeft(9, '0');

Converting string to int without losing the zero in the beginning

I tried int.parse, and convert class to convert a string to int.
While I'm converting. I'm losing the 0 in the beginning which i don't want.
Ex : 09999 becomes 9999 - I don't want this.
I want to keep it as it is.
How can i do that?

myNumber.ToString("D5");
//D represents 'Decimal', and 5 is the specified amount of digits you want the number to be always. This will pad your value with zeroes until it reaches 5 digits.

You cannot. An int is meant to represent a mathematical integer. The numbers 09999 and 9999 are exactly the same number, mathematically.
Perhaps there is a different problem here. Why do you need to do this? Maybe there's a better way to do what you want to do.

If you want to do something like always print your number with 5 places, it goes like
myNumber.ToString().PadLeft(5, '0');

No, int.Parse("09999") actually returns 0x0000270F. Exactly 32 bits (because that's how big int is), 18 of which are leading zeros (to be precise, one is a sign bit, you could argue there are only 17 leading zeros).
It's only when you convert it back to a string that you get "9999", presence or absence of the leading zero in said string is controlled by the conversion back to string.

you cannot. you will have to maintain the value as a string if you want it to remain that way.

Int values cannot have leading zeros

If you are just converting to int to test the value, keep the original data around and use the string value of it when you want the leading zeor. If you require the integer to have zero padding after mathematically working with it you will have to format it with sprintf or the like whenever you output it.

you cant, but if you need to cast it to int and keep the zeros you can create a copy of it and then cast it to int, then you will have two versions of it one as int and one as string.

// For our requirement that new ID's should be of same length as old ID's.
string strID = "002017";
int number = int.Parse(strID);
string newID = (++number).ToString("D" + strID.Length);

Although this is a old thread, but this can also help:
// convert to big integer
var bigIntBits = BigInteger.Parse(intNumber);
int indexOfOne = intNumber.IndexOf('1');
string backToString = new string('0', indexOfOne) + bigIntBits.ToString();

I had same problem;
Solved it with an if else
Couldn't find another solution for it