How to make html link with php regex

Question

My string

[[https://example.com|link]]

to convert

<a href="https://example.com>link</a>

My regex is

/\[{2}(.*?)\|(.*?)\]{2}/s

But it's not working.I am new to php regex.

This should help: stackoverflow.com/questions/206059/php-validation-regex-for-url — dale landry
– dale landry, Commented Apr 7, 2020 at 4:18
Does this answer your question? PHP validation/regex for URL — dale landry
– dale landry, Commented Apr 7, 2020 at 4:19
Use the capturing groups in the replacement regex101.com/r/riLKnf/1 You could also use negated character classes instead \[{2}([^][|]*)\|([^][]*)\]{2} regex101.com/r/qvY503/1 — The fourth bird
– The fourth bird, Commented Apr 7, 2020 at 6:51

Wiktor Stribiżew · Accepted Answer · 2020-04-07 09:25:19Z

1

You may use

preg_replace('~\[\[((?:(?!\[\[).)*?)\|(.*?)]]~s', '<a href="$1">$2</a>', $string)

See the regex demo

Details

\[\[ - a [[ substring
((?:(?!\[\[).)*?) - Group 1 ($1 in the replacement pattern refers to the value inside this group): any char (.), 0 or more occurrences but as few as possible (*?), that does not start a [[ char sequence ((?!\[\[))
\| - a | char
(.*?) - Group 2 ($2):
]] - a ]] substring.

See the PHP demo:

$string = "[[some_non-matching_text]] [[https://example.com|link]] [[this is not matching either]] [[http://example2.com|link2]]";
echo preg_replace('~\[\[((?:(?!\[\[).)*?)\|(.*?)]]~s', '<a href="$1">$2</a>', $string);
// => [[some_non-matching_text]] <a href="https://example.com">link</a> [[this is not matching either]] <a href="http://example2.com">link2</a>

answered Apr 7, 2020 at 9:25

Wiktor Stribiżew

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