How can you make a regex there will only fail if comma is used

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Hello.

Can anyone tell me how to make a regex for this problem. I have an issue with currency where sometimes people by mistake types a '.' instead of a ',' However it should be fine to use the '.' to seperate numbers

1,0
1.0 <- fail
1,92
1.92 <- fail
100,2
100.2 <- fail
100,02
100.02 <-- fail
1.000,00
1.000.00 <- fail
1.000.000,01
1.000.000.02 <- fail

The plan is to use it as a testing on an aspx box. I know how to add a regex but i fail utterly in trying to make a working regex for this. If anyone can help out with an explanation i would truly appreciate it

Posted 14-Oct-13 21:42pm

Member 10121607

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Kenneth Haugland 15-Oct-13 3:47am

How about ToString.RePlace(",",".") or something like that?

Member 10121607 15-Oct-13 3:50am

Hello. I tried to set up some replace. But it will not work for me. Example this: 1.000.00 would then turn into 1,000,00 and that will not do :( This is the reason im looking for some regex expression instead.

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Kenneth Haugland · Answer 1 · 2013-10-14T22:02:00

Using the build in functions in .NET might be the way to go:

C#

using System;
using System.Globalization;
class Sample
{
public static void Main()
{
CultureInfo ciClone = null;
double value = 0;
try {
ciClone = (CultureInfo)CultureInfo.InvariantCulture.Clone();
ciClone.NumberFormat.NumberDecimalSeparator = ",";
ciClone.NumberFormat.NumberGroupSeparator = ".";
value = Double.Parse("1.000,05", ciClone);
Console.WriteLine("The value is {0}", value);
}

catch (Exception e) {
Console.WriteLine(e);
}
}
}

If you want to use regex you could try this:
Expresso - A Tool for Building and Testing Regular Expressions[^]

and build one for yourself.

OriginalGriff · Answer 2 · 2013-10-14T22:13:00

Solution 2

Try this:

\.\d+$

If it matches, it's wrong according to your definitions.

As Kenneth said : get a copy of Expresso! Lovely piece of software...

Posted 14-Oct-13 22:13pm

OriginalGriff

Gabriel Szabo · Answer 3 · 2013-10-14T22:59:00

This should work for you:

"^\d{1,3}(\.\d{3})*,\d{1,2}$"

It matches a number ending with comma followed by one or two digits: ",\d{1,2}", having optional digit triplets separated by a 'dot' thousands separator: "(\.\d{3})*" and starting with one to three digits: "^\d{1,3}".
If you want an optional fraction part, use "(,\d{1,2}){0,1}$" instead of ",\d{1,2}".

How can you make a regex there will only fail if comma is used

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