problem in validation fix number and decimal numbers

[sh-a]

hi
I use following regular expression to validate my numbers:

^[-]?\d{1,15}(\.\d{1,3})?

my validate numbers is :

0 to 9 max length <= 15
decimal numbers are optional and decimal place is <= 3 example : 1234.123
0.123

my invalidate numbers is :

0 to 9 max length > 15
decimal numbers are optional and decimal place is > 3 example : 1234.12345
1234.
.1234
0.0
1.0
1.00
1.000

4 last invalid number don't validate by my regex
how I can fix my regular expression that (0.0 , 1.0, ..) can be invalid ?

sorry for may bad English

[dunfiddlin]

I don't think you can. The problem is that you're treating these numbers as text so you can't apply any kind of mathematical logic.

[sh-a]

hi
if found this regular expression :
this is work fine

^-?(([1-9]\d*)|0)(\.0*[1-9](0*[0-9])*)?$

but two limitation I want
before decimal point (.) I want only 15 numbers (length), and after it , 3 number as following sample

123456789012345.123

please help me

[sh-a]

hi
if found this regular expression :
this is work fine

^-?(([1-9]\d*)|0)(\.0*[1-9](0*[0-9])*)?$

but two limitation I want
before decimal point (.) I want only 15 numbers (length), and after it , 3 number as following sample

123456789012345.123

please help me

[sh-a]

I found solution myself

^-?(([1-9])([0-9]{1,14})?|0)(\.[0-9]?[0-9]?[1-9])?$

[Shaggy Hiker]

If these are just real numbers, then you would probably be better off not using RegEx at all. RegEx is slow, as are all string manipulations. It's a very convenient tool for validating text, but when the text is just, and always, a number, there may be better alternatives. For example:

1) Use Decimal.TryParse to convert it to a Decimal. If this fails, then the string wasn't a number anyways, and will fail your RegEx, too.

2) Once you have the decimal you can see whether or not it is smaller than your maximum number, which is 1x10^16. If it is greater than or equal to that, it will fail your RegEx because it will be too large.

3) If both of those tests pass then you are pretty close, so a final test might be better performed on the string:

If originalString.IndexOf("."c) < originalString.Length-4

The whole thing might look like this:

Dim dc As Decimal

If Decimal.TryParse(originalString,dc) Then
 If dc < 1000000000000000D AndAlso originalString.IndexOf("."c) < originalString.Length-4 Then
  'It passed.
 Else
  'Fail
Else
 'Fail
End If

I may have the decimal point off a little in there, but that would be close. It wouldn't be quite as compact as the RegEx solution, but it would most likely be faster. It would probably be faster still if you didn't use the IndexOf solution, and there are a couple ways to avoid that.

On the other hand, if you have the RegEx solution, it probably won't make any difference which you use. Also, if the number is part of a larger string, then what I showed wouldn't work anyways.