Php preg_replace, regex pattern doesn't match

Question

I'm trying to develop a basic template engine, so I have to use preg_replace so much. I've a problem about below subject:

$subject = "{%content%} %content%";
$pattern = '/matched_regex/';
$replace = 'OK';
echo preg_replace($pattern,$replace,$subject);

and the output must be like this:

{%content%} OK

in other words it will be just matched with %content%

What should I do regex pattern?

I've tried this $pattern = '/[^{]%\s*(.*?)\s*%[^}]/';

Kadir Emin İslamoğlu
– Kadir Emin İslamoğlu

2015-03-02 21:31:32 +00:00
Commented Mar 2, 2015 at 21:31 — Kadir Emin İslamoğlu
– Kadir Emin İslamoğlu, Commented Mar 2, 2015 at 21:31
and this $pattern = '/{{0}?%\s*(.*?)\s*%}{0}/';

Kadir Emin İslamoğlu
– Kadir Emin İslamoğlu

2015-03-02 21:35:09 +00:00
Commented Mar 2, 2015 at 21:35 — Kadir Emin İslamoğlu
– Kadir Emin İslamoğlu, Commented Mar 2, 2015 at 21:35

mhall · Accepted Answer · 2015-03-02 21:43:12Z

1

This will match only %content% that is not following an { or is at the beginning of the subject string. Any character that was before the %content% is put back with the \1 in the replacement string:

$subjects = [
    '{%content%} %content%',
    'Foo {%content%} bar %content% baz',
    'Foo{%content%}bar%content%baz',
    'Foo{%content%}bar%content%',
    '{%content%}%content%',
    '%content%{%content%}',
];

$replace = 'OK';
foreach ($subjects as $subject) {
    $pattern = '/(^|[^{])%content%/';
    echo preg_replace($pattern, '\1'.$replace, $subject), PHP_EOL;
}

Output:

{%content%} OK
Foo {%content%} bar OK baz
Foo{%content%}barOKbaz
Foo{%content%}barOK
{%content%}OK
OK{%content%}

edited Mar 2, 2015 at 21:43

answered Mar 2, 2015 at 21:24

mhall

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