使用 PHP 用正则表达式替换正则表达式

Question

I want to replace hash tags in a string with the same hash tag, but after adding a link to it

Example:

$text = "any word here related to #English must #be replaced."

I want to replace each hashtag with

#English ---> <a href="bla bla">#English</a>
#be ---> <a href="bla bla">#be</a>

So the output should be like that:

$text = "any word here related to <a href="bla bla">#English</a> must <a href="bla bla">#be</a> replaced."

Answer 1

$input_lines="any word here related to #English must #be replaced.";
preg_replace("/(#\w+)/", "<a href='bla bla'>$1</a>", $input_lines);

DEMO

OUTPUT :

any word here related to <a href='bla bla'>#English</a> must <a href='bla bla'>#be</a> replaced.

Answer 2

This should nudge you in the right direction:

echo preg_replace_callback('/#(\w+)/', function($match) {
    return sprintf('<a href="https://www.google.com?q=%s">%s</a>', 
        urlencode($match[1]), 
        htmlspecialchars($match[0])
    );
}, htmlspecialchars($text));

See also: preg_replace_callback()

Answer 3

If you need to refer to the whole match from the string replacement pattern all you need is a $0 placeholder, also called replacemenf backreference.

So, you want to wrap a match with some text and your regex is #\\w+ , then use

$text = "any word here related to #English must #be replaced.";
$text = preg_replace("/#\w+/", "<a href='bla bla'>$0</a>", $text);

Note you may combine $0 with $1 , etc. In case you need to enclose a part of the match with some fixed strings you will have to use capturing groups. Say, you want to get access to both #English and English within one preg_replace call. Then use

preg_replace("/#(\w+)/", "<a href='path/$0'>$1</a>", $text)

Output will be any word here related to <a href='path/#English'>English</a> must <a href='path/#be'>be</a> replace .