php preg_match - search for specific number/character code

Question

I'm trying to search for a number/character pattern inside of a string.

The String can look like this

"any text CA-2019-6-000000 any text" 
"any text KA 2019-2-929029" // note: no "-" between the "KA and 2019" 
"KA-2019-11-929029"

What I can definetly say, There is always a year, like 2000/2019/2055. After the year is always a minus sign, two possible numbers from 1-12 and another minus sign.

Which is the month.

And after the "-<num>-", is a 6 digits long number which can be at least 000000 or max 999999

Before the year, can be max. two characters long string followed by a minus. Between this two characters long strong could be a minus sign or a space letter.

Examples:

"AA 2019"
"ZZ-2018"

I found out that I could get the 6 numbers with /[0-9]{6}/.

The year by /[0-9]{4}. I would like to add, that it only can be between 2000 and 2100

And I can get the number between the two minus signs with: /(?<=\-)(.*?)(?=\-)/ or

/\-(.*?)\-/

For example, I had the idea to look for the number between the two "-" characters and store it in a variable. Then to say I want to have the numbers between the ones after this variable that are 6 characters long and between 000000 and 999999.

A similar game with that year. I'd want to say I'd get the number which is before the variable with the "--" The maximum length is 4 numbers and is between 2000 and 2100.

If I then have stored the year in a variable I can theoretically say I'm looking for two letters which precede "-$yearvariable" or " [space]$yearvariable

Please give a proper example of what you actually want as result here, based on that shown input data. — 04FS
– 04FS, Commented Jul 1, 2019 at 12:07
Oh as a result I want to have the thos example options at the top of the questions. Like KA-2019-11-929029 or CA 2019-6-000000 . — blyadjs
– blyadjs, Commented Jul 1, 2019 at 12:09
Like the text which could possible be after and before this code, should be removed and I want the code which starts with the two characters and ends with the 6 letters. — blyadjs
– blyadjs, Commented Jul 1, 2019 at 12:12

Wiktor Stribiżew · Accepted Answer · 2019-07-01 12:10:43Z

1

You may use

\b([A-Z]{2})[-\s](20[0-9]{2}|2100)-(0?[1-9]|1[0-2])-(\d{6})(?!\d)

See the regex demo

Details

\b - word boundary
([A-Z]{2}) - two uppercase letters
[-\s] - a hyphen or whitespace
(20[0-9]{2}|2100) - number from 2000 to 2100
- - a hyphen
(0?[1-9]|1[0-2]) - a month from 1 to 12
- - a hyphen
(\d{6})(?!\d) - 6 digits (no 7 or more).

See the regex graph:

PHP demo:

$s = "any text CA-2019-6-000000 any text";
if (preg_match('~\b([A-Z]{2})[-\s](20[0-9]{2}|2100)-(0?[1-9]|1[0-2])-(\d{6})(?!\d)~', $s, $m)) { 
    print_r($m);
}

Output:

Array
(
    [0] => CA-2019-6-000000
    [1] => CA
    [2] => 2019
    [3] => 6
    [4] => 000000
)

edited Jul 1, 2019 at 12:10

answered Jul 1, 2019 at 12:06

Wiktor Stribiżew

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1 Comment

Wiktor Stribiżew Over a year ago

@blyadjs I thought you also needed all those parts of the regex match, so I used capturing groups ((...)). If you do not need them, turn all the capturing groups into non-capturing by replacing (...) with (?:...). Also, use this generator for numeric ranges, it looks helpful (though sometimes, the patterns are a bit redundant).

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