preg_match仅在找到时显示替换

Question

The title must be very confusing, but I have to admit, so am I when it comes to regex. My problem is the following:

I have an input string like:

{a href=www.google.com}Google{/a} {b}boldText{/b}

What I want to achieve is following:

<a href="www.google.com">Google</a> <b>boldText</b>

I have the following preg_replace function:

$input = preg_replace('/\{(\/)?'.$tag.'( [a-z]*\=)?([a-zA-Z0-9\.\:\/]*)?\}/', '<$1'.$tag.' $2"$3">', $input);

Where the tag is f.e. a or b or script. The code is working so far, the only problem is that a tag without any attributes comes out like this:

<a href="www.google.com">Google</a> <b "">boldText</b>

Can I somehow add an if condition around the $2"$3" in the replacement argument of the preg_match function, and if so, how can I do it?

Answer 1

I'd merge the 2 optional groups into 1 and use a 2nd preg_replace to add double quotation marks around the attribute values inside opening tags. We cannot do it in 1 go as the replacement pattern is different for tags with attributes.

$input = preg_replace('/\{(\/)?'.$tag.'( [a-z]*\=[a-zA-Z0-9\.\:\/]*)?\}/', '<$1' . $tag . '$2>', $input);

See demo on IDEONE

<?php
$tag = "a";
$str = "{a href=www.google.com at=val}Google{/a} {b}boldText{/b}";
$input = preg_replace('#\{(\/)?' . $tag . '\b((?:\s+[a-z]*\=[a-zA-Z0-9\.\:\/]*)*)\}#', '<$1' . $tag . '$2>', $str);
$result = preg_replace("#(?:(<".$tag.")\b|(?<!^)\G)(\s*[^\s=>]+?)=([^=>\s]*)(?=.*?>)#", "$1$2=\"$3\"", $input);
echo $result;

Output: <a href="www.google.com" at="val">Google</a> {b}boldText{/b}