Regular Expression needed

Question

Regular Expressions have always seemed like black magic to me and I have never been able to get my head around building them.

I am now in need of a Reg Exp (for validation putsposes) that checks that the user enters a number according to the following rules.

- no alpha characters
- can have decimal
- can have commas for the thousands, but the commas must be correctly placed

Some examples of VALID values:

1.23

100

1,234

1234

1,234.56

0.56

1,234,567.89

INVALID values:

1.ab

1,2345.67

0,123.45

1.24,687

Thanks
Gary

Dorababu Meka · Answer

Try this ^([1-9]\d{02}(\d{3})+|[1-9]\d*|0)(\.\d+)?$

Vulpes · Answer

Yeah mine didn't deal with this one either: 0000123.45 However Dorababu's expression seems to be dealing with all cases correctly so I'd use that.

Vulpes · Answer

The following seems to be working but I'd test it thoroughly before use. I've assumed there should not be more than two digits after a decimal point if there is one: class Test { static void Main() { string[] numbers = new string[] { 1.23 100 1234 1234 1234.56 0.56 1234567.89 1.ab 12345.67 0123.45 1.24687 0.5 }; string regExp = @^(?!0{13})\d{13}(?\d{3})*(.\d{12})?$; Regex r = new Regex(regExp); foreach(string number in numbers) { Console.WriteLine(r.IsMatch(number)); } Console.ReadKey(); } }

Dorababu Meka · Answer

Yeah thanks for ur comment vulpes

Gary King · Answer

Excellent - thank you very much! However it is not perfect - for example 1234567 (missing comma after the 1) is deemed valid by the Reg Exp. Having said that I think that the solution provided is good enough Thanks Gary

Gary King · Answer

yes 0.5 should be allowed. Other examples of allowed numbers include: 1234567 1234567.89 12345678 12345678.9 123456789 and so on (basically any number that we as humans see as valid)

Dorababu Meka · Answer

will 0.5 is allowed or not