Avoid possible statement with regex

Question

How to replace a part of string with avoid a year numbers (f.e. 2019 or 2019-2020) before the first slash occurance with Regex

//something is wrong here
preg_replace('/^[a-z0-9\-]+(-20[0-9]{2}(-20[0-9]{2})?)?/', '$1', $input_lines);

Needed:

~~abc-def/~~something/else/ [incl. slash if there is not character before it]

~~abc-def-~~2019/something/else/

~~abc-def-~~2019-2020/something/else/

~~abc-def-125-~~2019/something/else/

That is usually done like this [(?m)^(?:(?!20(?:19|20))\[a-z0-9\-\])+/?](https://regex101.com/r/j9qA27/1) Expanded https://regex101.com/r/5JT2Sk/1 — , Oct 27 '19 at 22:33

score 1 · Accepted Answer · answered Oct 27 '19 at 22:29

My initial closure was insufficient to handle all requirements. Yes, you have a greedy quantifier problem, but there is more to handle.

Code: (Demo) (Regex101 Demo)

$tests = [
    'abc-def/something/else/',
    'abc-def-2019/something/else/',
    'abc-def-2019-2020/something/else/',
    'abc-def-125-2019/something/else/'
];

var_export(
    preg_replace('~^(?:[a-z\d]+-?)*?(?:/|(?=20\d{2}-?){1,2})~', '', $tests)
);

Output:

array (
  0 => 'something/else/',
  1 => '2019/something/else/',
  2 => '2019-2020/something/else/',
  3 => '2019/something/else/',
)

My pattern matches alpha-numeric sequences, optionally followed by a hyphen -- a subpattern than may be repeated zero or more times ("giving back", aka non-greedy, when possible).

Then the first non-capturing group must be followed by a slash (which is matched) or a your year substrings which also may have a trailing hyphen (this is not matched, but found via a lookahead).

If this doesn't suit your real projects data, you will need to provide more and more accurate samples to test against which reveal the fringe cases.

score 0 · Answer 2 · answered Oct 29 '19 at 16:08

If the forward slash has to be present and it should stop after the first occurrence of 2019 or 2020, you might use:

^(?=[a-z\d-]*/)[a-zA-Z013-9-]+(?>2(?!0(?:19|20)(?!\d))|[a-zA-Z013-9-]+)*/?

In separate parts that would look like

^ Start of string
(?=[a-z\d-]*/) Assert that a / is present
[a-zA-Z013-9-]+ Match 1+ times any of the listed (Note that the 2 is not listed)
(?> Atomic group
- 2(?!0(?:19|20)(?!\d)) Match 2 and assert what is on the right is not 019 or 020
- | Or
- [a-zA-Z013-9-]+ Match 1+ times any of the listed
)* Close group and repeat 0+ times
/? Match optional /

Regex demo | Php demo

Your code might look like

preg_replace('~^(?=[a-z\d-]*/)[a-zA-Z013-9-]+(?>2(?!0(?:19|20)(?!\d))|[a-zA-Z013-9-]+)*/?~', '', $input_lines);

Avoid possible statement with regex

2 Answers2