Regular expression accepts partial match

Question

I have some validation code that checks a string against a regular expression.

Regex regex = new Regex(RegexPattern);
if (!regex.IsMatch(value))
{
    errorMessage = "The data is not in the correct format.";
    return false;
}

If I set the regular expression pattern to ^[0-9]*.[0-9]*.[0-9]*.[0-9]*$, it correctly accepts 1.0.0.0; however, it also accepts 1.0.0..

How can I modify the pattern so that 1.0.0.0 is accepted but 1.0.0. is rejected?

Ashkan Mobayen Khiabani · Accepted Answer · 2017-08-31T17:16:16.603

5

[0-9]* means 0 or more occurrence of [0-9]

[0-9]+ means 1 or more occurrence of [0-9]

^[0-9]*.[0-9]*.[0-9]*.[0-9]*$

Change * to +:

^[0-9]+.[0-9]+.[0-9]+.[0-9]+$

edited Aug 31 '17 at 17:16

answered Aug 31 '17 at 17:12

Ashkan Mobayen Khiabani

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Yup, that's what I was missing. Thanks. – Jonathan Wood Aug 31 '17 at 17:14
This happens, the thing that you have done 100+ times. you do a simple and stupid mistake and the funny part is that you don't find it. I do it a lot ;). Glad could help. – Ashkan Mobayen Khiabani Aug 31 '17 at 17:15
Is this supposed to match `1234567` – Aug 31 '17 at 19:49
@sln nope. because there must be for numbers separated by 3 dots – Ashkan Mobayen Khiabani Aug 31 '17 at 19:51
`^[0-9]+.[0-9]+.[0-9]+.[0-9]+$` matches `1234567` – Aug 31 '17 at 19:53

score 2 · Answer 2 · answered Aug 31 '17 at 17:23

Just a slight misunderstanding about * and +. The former accepts either no occurrences or more (>=0), the latter only matches if such vocabulary occurs at least once (>=1).

^[0-9]+\.[0-9]+\.[0-9]+\.[0-9]+$

I usually also escape the dot for safety, not entirely sure if it's necessary but I still do it. :)

Aleks Andreev · Answer 3 · 2017-08-31T17:21:23.720

0

You can use this pattern:

^\d+\.\d+\.\d+\.\d+$

Explanation:

 ^ - begin of string
   \d - any digit
   + - at least one char
   \. - exactly a dot char
 $ - end of string

By the way, your input looks like an IP address. If so you can modify your reges like this:

^\d{1,3}\.\d{1,3}\.\d{1,3}\.\d{1,3}$

{1,3} means: 'from 1 to 3 chars'

Demo: regex101

edited Aug 31 '17 at 17:21

answered Aug 31 '17 at 17:15

Aleks Andreev

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score 0 · Answer 4 · answered Aug 31 '17 at 19:52

0

To match 4 sets of digits separated by a period

^\d+(?:\.\d+){3}$

answered Aug 31 '17 at 19:52

Regular expression accepts partial match

4 Answers4