ASP.Net RegularExpressionValidator Currency Require 2 decimals for validation

Question

I have a very standard ASP.NET RegularExpressionValidator and I just need the ValidationExpression to make it so 2 decimal places are required!

Good Examples..

Bad Examples

I tried /^\d{0,4}(\.\d{0,2})?$/ and I can't get it to work.

Did you try anything yourself? It is a basic regex, actually. — Wiktor Stribiżew, Oct 21 '15 at 22:15
i tried... i even tried /^\d{0,4}(\.\d{0,2})?$/ and i can't get it to work — eqiz, Oct 21 '15 at 22:18

score 4 · Accepted Answer · answered Oct 21 '15 at 22:19

4

You can use

^[0-9]+[.][0-9]{2}$

See regex demo

Regex breakdown:

^ - start of string
[0-9]+ - 1 or more digits
[.] - a literal .
[0-9]{2} - exactly 2 digits
$ - end of string.

Your regex - ^\d{0,4}(\.\d{0,2})?$ - matches 0 to 4 digits at the beginning, and then optionally there can be a period followed by 0 to 2 digits before the end of string. It looks like a live validation regex to me, but it cannot be used for final validation.

answered Oct 21 '15 at 22:19

Wiktor Stribiżew

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works... i'm glad you can make it look so simple lol. Things i read talked about using "d" im not sure why though – eqiz Oct 21 '15 at 22:22
There is always a problem with escaping regex metacharacters. Using character classes (those `[...]`) relieves that pain in many cases. – Wiktor Stribiżew Oct 21 '15 at 22:23
I see that there are already 2 answers here that look identical to mine. `^\d+\.\d{2}$` is the same as `^[0-9]+[.][0-9]{2}$` in JavaScript ES5 - BUT not in .NET, where `\d` may match Persian and some other numeric characters. Perhaps, you would like to avoid that. Check [this SO answer](http://stackoverflow.com/a/16621778/3832970). – Wiktor Stribiżew Oct 21 '15 at 22:51

score 1 · Answer 2 · answered Oct 21 '15 at 22:25

1

Try this:

/\d+\.\d{2}$/

One or more digits, followed by a single dot and followed just for two other digits at the end of the string.

https://regex101.com/r/lF9uU2/1

answered Oct 21 '15 at 22:25

mayo

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Check [what `\d` may match in .NET](http://stackoverflow.com/a/16621778/3832970). – Wiktor Stribiżew Oct 21 '15 at 22:55

score 0 · Answer 3 · answered Oct 21 '15 at 22:24

0

Please try this

/^[0-9]+\.?[2]*$/

answered Oct 21 '15 at 22:24

Osama Aftab

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This regex matches 1 or more digits from the beginning of the string, then 1 or 0 dots, and then 0 or more `2` symbols up to the end. I doubt it is what OP needs. – Wiktor Stribiżew Oct 21 '15 at 22:56

score 0 · Answer 4 · answered Oct 21 '15 at 22:37

0

You can set the RegEx value of the validator to the following verbatim string in your server-side code:

@"^\d+\.\d{2}$"

answered Oct 21 '15 at 22:37

gpersell

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I tested mine and it gives the expected result for all of your test cases. – gpersell Oct 21 '15 at 22:41
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Check [what `\d` may match in .NET](http://stackoverflow.com/a/16621778/3832970). – Wiktor Stribiżew Oct 21 '15 at 22:54
That's a good point @stribizhev . If you need to allow Unicode digits, then \d is the way to go. If you need to allow only 0 to 9, then use [0-9]. – gpersell Oct 21 '15 at 23:00

ASP.Net RegularExpressionValidator Currency Require 2 decimals for validation

4 Answers4