java regex for date format extracts wrong information

Question

I am using regex to figure out what format the input date is. This is one of the patterns i am using

    ^((18[5-9]|19[0-9]|20[0-9])\\d)(0?[1-9]|1[012])(0?[1-9]|[12][0-9]|3[01])$

so the constraint is to have the year between 1850 and 2099. If I pass for instance this string as date 20011212 when I am extracting the year, month and day from it, this is what I get: year: 2001, month: 200, day :12. Any Idea why?

    pattern = Pattern.compile(PATTERN);
    matcher = pattern.matcher(dateString);
    if (matcher.matches()){
       matcher.reset();
       if (matcher.find()){
          Integer.parseInt(matcher.group(1));
          Integer.parseInt(matcher.group(2));
          Integer.parseInt(matcher.group(3));
       }
    }

The code is simplified, but even on this simplified version, it returns erroneous results. Thank you for any suggestions/solutions.

what results are you getting?

The Cat
– The Cat

2012-10-25 09:21:40 +00:00
Commented Oct 25, 2012 at 9:21 — The Cat
– The Cat, Commented Oct 25, 2012 at 9:21
year: 2001, month: 200, day :12

Marius
– Marius

2012-10-25 09:31:28 +00:00
Commented Oct 25, 2012 at 9:31 — Marius
– Marius, Commented Oct 25, 2012 at 9:31

enrico.bacis · Accepted Answer · 2012-10-25 09:33:51Z

4

In a regex everything you put inside (...) is a capturing group. You have two groups in the year, those are both capturing groups:

group(1) = ((18[5-9]|19[0-9]|20[0-9])\\d)
group(2) = (18[5-9]|19[0-9]|20[0-9])
group(3) = (0?[1-9]|1[012])
group(4) = (0?[1-9]|[12][0-9]|3[01])

You can also use non capturing blocks like this: (?:...)

So your pattern should be:

^((?:18[5-9]|19[0-9]|20[0-9])\\d)(0?[1-9]|1[012])(0?[1-9]|[12][0-9]|3[01])$

edited Oct 25, 2012 at 9:33

answered Oct 25, 2012 at 9:27

enrico.bacis

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3 Comments

Marius Over a year ago

The parenthesis are good there, to allow another digit for the year, if not it considers the year on 3 digits.

enrico.bacis Over a year ago

Yes, but you need to specify that that is a non-capturing group.

Marius Over a year ago

+1 You are also right. Thank you for clarifying. I didn't knew the trick for non capturing

rretzbach · Accepted Answer · 2012-10-25 09:28:30Z

4

The second group is the first three digits of the year, use a non capturing group for it:

^((?:18[5-9]|19[0-9]|20[0-9])\\d)(0?[1-9]|1[012])(0?[1-9]|[12][0-9]|3[01])$

answered Oct 25, 2012 at 9:28

rretzbach

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Comments

halex · Accepted Answer · 2012-10-25 09:29:17Z

2

Change your regex to ^(18[5-9]\\d|19[0-9]\\d|20[0-9]\\d)(0?[1-9]|1[012])(0?[1-9]|[12][0-9]|3[01])$. You had a pair of parentheses too much around the first 3 digits of the year that created the second matching group.

answered Oct 25, 2012 at 9:29

halex

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1 Comment

Marius Over a year ago

Yup, you were right. I thought I can put the \\d after all the constraints. Thanks.

John Dvorak · Accepted Answer · 2012-10-25 09:29:52Z

1

This is because you have a capturing group (a pair of parentheses) inside the year regex. You can either:

count the left brackets and select the correct ones. Hard to maintain if you are ever going to change the regex in the future.
use named groups. Not all regex flavors support this. I think Java is one of those who don't.
use non-capturing groups.

A non-capturing group is denoted by ?: at the start of the group:

^((?:18[5-9]|19[0-9]|20[0-9])\\d)(0?[1-9]|1[012])(0?[1-9]|[12][0-9]|3[01])$
   ^^--- here

Note that look-arounds ((?= ... ), ...), are non-capturing as well.

answered Oct 25, 2012 at 9:29

John Dvorak

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