regular expression in .net for validating a number

Question

I need a regular expression in .net which

will only allow a maximum of two digits after "."
will only allow a maximum of two digits before "."
has a maximum length of 5
represents a number greater than 0
represents a number not greater than 100
has a number not greater than 11 after ".", like 5.11, 5.10, 5.03, 5.01, 5.06, etc (not 5.12, 5.2, 5.3)

Examples:

1, 100, 100.00, 58, 58.10, 1.1, 0.02, .2, .02, 1.02, 11.11, 12.00, 01.09, 1.0 etc

Invalid numbers:

1.12, 101.12, 100.1, 100.03, 100.65, etc. and all negative values

Your specification seems inconsistent. 100 cannot be done with a maximum of 2 digits before the '.'... — Lucero
– Lucero, Commented Sep 11, 2009 at 6:35
How can .2 not be invalid. .2 > .11 so surely this should be a failure — Xetius
– Xetius, Commented Sep 12, 2009 at 11:16

Xetius · Accepted Answer · 2009-09-11 20:04:43Z

2

^100\.00|\d{1,2}\.(?:11|10|0\d)$

OK, was simpler than I thought.

^100(?:\.00?)?$|^(?:\d{1,2})?(?:\.(?:1|11|10|0\d?))?$

This matches the new set of data. It also matches .0 at the moment.

edited Sep 11, 2009 at 20:04

answered Sep 11, 2009 at 9:01

Xetius

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1 Comment

Asaph Over a year ago

Not so fast. This regular expression doesn't match "100", "100.0", "1.1", "1.0"

Thomas · Accepted Answer · 2009-09-12 16:18:57Z

1

this regex is based on the one from Xetius:

^(100(?:\.00?)?|\d{1,2}(?:\.(?:11|10|0\d|0|1))?)$

below is the code how I Tested it:

class Program
{
    const string Pattern = @"^(100(?:\.00?)?|\d{1,2}(?:\.(?:11|10|0\d|0|1))?)$";
    static void Main(string[] args)
    {
        for (int before = 0; before < 100; before++)
        {
            Test(before.ToString());
            for (int after = 0; after < 12; after++)
            {
                Test(string.Format("{0}.{1:d2}", before, after));
                Test(string.Format("{0:d2}.{1:d2}", before, after));
                Test(string.Format(".{0:d2}", after));
            }
            Test(string.Format("{0}.{1:d}", before, 0));
            Test(string.Format("{0:d2}.{1:d}", before, 1));
        }
        // Special cases:
        Test("100");
        Test("100.0");
        Test("100.00");

        // intended to fail
        Test("00.20");
        Test("00.12");
        Test("00.00x");
        Test("000.00");
        Console.WriteLine("done.");
        Console.ReadLine();
    }

    private static void Test(string input)
    {
        Regex r = new Regex(Pattern);
        if (!r.IsMatch(input))
        {
            Console.WriteLine("no match: " + input);
        }
    }
}

Edit:

Tweaked the regex so that 100.0 also works
added a comment for the tests intended to fail

Edit2:

Added Test for .00 - .11 and was surprised that the regex already matched them. Thanks to Xetius

edited Sep 12, 2009 at 16:18

answered Sep 11, 2009 at 6:37

Thomas

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5 Comments

Asaph Over a year ago

so what? 0.5 is supposed to fail because you can't have a number greater than 1 as the first digit after the decimal place

Thomas Over a year ago

The real question here is, what does 0.5 mean: 0.50 or 0.05?

Havenard Over a year ago

When I say "failed" I mean it returned valid when it shouldn't.

Xetius Over a year ago

0.5 means 0.5. 0.05 is a completely different number. That is like confusing 5 and 50

Xetius Over a year ago

From the supplied data (Which Im sure has changed) it has to match with .01 to .11 meaning that the whole number is optional.

Asaph · Accepted Answer · 2009-09-11 06:34:28Z

0

^(100(\.00?)?|\d\d?(\.(0\d|1[01]?))?)$

edited Sep 11, 2009 at 6:34

answered Sep 11, 2009 at 6:22

Asaph

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Comments

Lucero · Accepted Answer · 2009-09-11 06:35:49Z

0

^(?:\d{1,2}(?:\.(?:0?\d|1[01]))?|100)$

edited Sep 11, 2009 at 6:35

answered Sep 11, 2009 at 6:26

Lucero

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2 Comments

Asaph Over a year ago

0.5 is supposed to fail. 0.5 = 0.50 > 0.11 and we can't exceed 11 after the decimal place.

Havenard Over a year ago

When I say "failed" I meant it returned valid when it shouldn't. You placed a ? in the wrong place there.

Havenard · Accepted Answer · 2009-09-11 13:24:33Z

0

"^100(\.00?)?|\d\d?(\.(1[01]?|0\d?))?$"

edited Sep 11, 2009 at 13:24

answered Sep 11, 2009 at 6:34

Havenard

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