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Answer 1

+2 A:

^(\d+(\.\d{2})?|\.\d{2})$

\d+ one or more digits
(\.\d{2})? optional decimal part
| or
(\.\d{2}) or just the decimal part

If you make the integer part \d*, the regex will match an empty string.

If you can use look ahead, you can use it to make sure that the string is not empty.

^(?=.)\d*(\.\d{2})?$

Amarghosh 2010-07-20 04:10:06

I'd personally use `(?!$)` and `\d\d`. Explanation gets you my vote. (Any non-trivial regexp deserves comments.)

strager 2010-07-20 04:19:52

`(?!$)` sounds better than `(?=.)`. I considered `\d\d` but went for `\d{2}` as it had `2` in it - thought it reads better.

Amarghosh 2010-07-20 04:23:10

in my original question I had to use escape char "\" because it was in csharp code, do you fellows know of a better way to put this in csharp code without the escape character?

VoodooChild 2010-07-20 04:23:50

@VoodooChild, Use verbatim string literals: `@"^(\d+(\.\d{2})?|\.\d{2})$"` (including the `@`).

strager 2010-07-20 04:27:25

Thanks guys :) (verbatim string I forgot that word before, thanks)Cheers

VoodooChild 2010-07-20 04:29:16

Answer 2

A:

You could try the following:

"^\d+(\.\d{1,2})?$"

This will match a number followed by an optional decimal point and 1 or two decimal places.

Josiah 2010-07-20 04:10:48

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