PHP REGEX fine-tuning capturing

Question

I have to process text that comes from student essays (texts can be VERY large).

I need in PHP a preg_match for dates inside that strings which may come in this way:

...blah blah blah (1994) blah blah blah ... 
...blah blah blah (nov-1994) blah blah blah ... 
...blah blah blah (november-1994) blah blah blah ...
...blah blah blah (1994-nov) blah blah blah ...
...blah blah blah (1994-november) blah blah blah ...

The dates in the strings may come with '( )' or with '[ ]'

I have done it this way:

if (preg_match('/\w{0,8}-?(19|20)\d{2}-?\w{0,8}/', $string, $s)) {
 # code
}

which is right and do its job but its capturing some unrelated strings like

... blah blah blah (SKU_1956) blah blah blah ...
... blah blah blah [INFERNO2000] blah blah blah ...
... blah blah blah [like-2000-me] blah blah blah ...

I dont seem to be able to do it, so I need help to fine-tuning this regexp to only capture if

start with either ( [
may be a single word and if it exists, MUST end in -
MUST BE a year in the lap 19xx-20xx
may be a single word and if it exists, MUST start with -
end with either ) ]

The word is limited to 8 chars because of the longest month (like december)

There is a huge amount of non-related strings captured, thats why I want to fine-tuning it.

Try something like [([]((\w{1,8})-((?:20|19)\d{2})|(?3)-(?2)|(?3))[])] — Wiktor Stribiżew
– Wiktor Stribiżew, Commented Mar 30, 2018 at 10:10
You will need to specify the months to realize this. Because \w or [a-z] does not make a difference between a word "like" and "nov". You also need to escape the "-"-character. — Bernhard
– Bernhard, Commented Mar 30, 2018 at 10:12

Zenoo · Accepted Answer · 2018-03-30 10:17:01Z

1

You can use the RegEx [(\[](([a-zA-Z]{1,8}-)?(19|20)\d{2}|(19|20)\d{2}-[a-zA-Z]{1,8})[)\]]

[(\[] ... [)\]] matches anything inside () or []
([a-zA-Z]{1,8}-)?(19|20)\d{2} matches month-YEAR with the month being optional
- ([a-zA-Z]{1,8}-)? matches an alphabetical char between 1 and 8 times, and a -
- (19|20)\d{2} matches 19.. or 20..
(19|20)\d{2}-[a-zA-Z]{1,8}) matches YEAR-month

Demo.

edited Mar 30, 2018 at 10:17

answered Mar 30, 2018 at 10:12

Zenoo

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2 Comments

Bernhard Over a year ago

what about "like-1994"?

user3393366 Over a year ago

yes, (like-1994) is also captured... BTW there were a small typo error: it should capture the 4 year digits, so I think Zenoo answer should be [([](([a-zA-Z]{1,8}-)?(19\d{2}|20\d{2})|(19\d{2}|20\d{2})\-[a-zA-Z]{1,8})[)]] regex101.com/r/rGPPzh/1 THUMBS UP FOR (regex101.com) Didn't knew that site !!! Think that close the question, I will be dealing with the text (if it is a month) in PHP.

The fourth bird · Accepted Answer · 2018-03-30 10:50:39Z

You could list all the valid date formats in an array:

$formats = ["M-Y", "Y", "F-Y", "Y-F", "Y-M"];

and then loop them to test if you can create a valid DateTime:

As the regex pattern you could capture what is between the parenthesis in group 1:

/$([^)]+)$/

$strings = [
    "...blah blah blah (1994) blah blah blah ... ",
    "...blah blah blah (nov-1994) blah blah blah ... ",
    "...blah blah blah (november-1994) blah blah blah ...",
    "...blah blah blah (1994-nov) blah blah blah ...",
    "...blah blah blah (1994-november), (1994), (nov-1994) blah blah blah ...",
    "...blah blah blah (1994-november) blah blah blah ..."
];
$formats = ["M-Y", "Y", "F-Y", "Y-F", "Y-M"];
$pattern = '/\(([^)]+)\)/';
foreach ($strings as $string) {
    preg_match_all($pattern, $string, $matches);
    foreach ($matches[1] as $match) {
        foreach ($formats as $format) {
            if (DateTime::createFromFormat($format, $match) !== false) {
                echo "$string contains valid date: <b>$match</b>" . PHP_EOL;
                break;
            }
        }

    }
}

Demo

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