How to validate a decimal number to have precision exactly one?

Question

I need to validate user input into a textbox to be a decimal number of precision exactly one? For example below are some of the cases..

if user inputs 1 then it's invalid
if user inputs 0 then its invalid
if user inputs 1.0 then its valid
if user inputs 1.1 then its valid
if user inputs 1.11 then its invalid
if user inputs 100.1 then its valid

How can I achieve this using C#(.net 3.5)?

Its a WPF textbox.

Thanks, -Mike

WinForms, WPF, WebForms, MVC, Silverlight? Client-side validation (i.e. JavaScript) or Server-side or both? — Belogix, Apr 10 '13 at 09:33
see http://stackoverflow.com/questions/6264576/number-of-decimal-digits-in-a-double and check splitted.Length > 1 && splitted[1].Length = 1 — Meirion Hughes, Apr 10 '13 at 09:34

score 2 · Answer 1 · answered Apr 10 '13 at 09:36

2

Easiest way, to me, seems to be to use regular expressions. By interpreting the user input as a string, you can compare it with something like '[0-9]+\.[0-9]'

answered Apr 10 '13 at 09:36

christopher

26,815
5
55
89

score 2 · Accepted Answer · answered Apr 10 '13 at 09:42

2

For a non-regex way of achieving this:

string input;
decimal deci;
if (input.Length >=3 && input[input.Length - 2].Equals('.') && Decimal.TryParse(input , out deci))
{
    // Success
}
else
{
    // Fail
}

answered Apr 10 '13 at 09:42

Jamie Kelly

309
3
15

I would recommend the regex method of doing this. I originally answered this in my naive early days :) – Jamie Kelly Aug 22 '15 at 23:41

score 0 · Answer 3 · answered Apr 10 '13 at 09:41

Yes regular expression is the best way to check your condition:

For exactly NNNN.N (1234.5) use:

/^\d{4}\.\d$/

For optional .N (1234 1234. 1234.5) go with:

/^\d{4}(\.\d?)?$/

For numbers up to size of NNNN.N (1 .5 12.5 123. 1234 1234.5) go with:

/^(?=.*\d)\d{0,4}(\.\d?)?$/

And if you want to allow +/- at the beginning, then replace ^ with ^[-+]?.

How to validate a decimal number to have precision exactly one?

3 Answers3