PHP正则表达式 - 替换一个反向引用[重复]

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• PHP preg_replace_callback, replace only 1 backreference? 3 answers

Is there a way in preg_replace() or preg_replace_callback() to replace only one backreference? For instance I have this regular expression: /(\.popup)([\s\.{])/ and this string .popup .popup-option. The backreferences that will be generated are as follows: $0 = .popup\s, $1 = .popup, $2 = \s. I want to replace only the $1 by another string. How do I manage to do this?

Thanks.

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Answer 1

You can use preg_replace_callback like this:

$s = '.popup .popup-option'; // original string

echo preg_replace_callback('/(\.popup)([\s\.{])/', function ($m) {
    return 'foo' . $m[2]; }, $s);
//=> foo .popup-option

In the callback we are returning some replacement string foo concatenated with $m[2] thus getting only $m[1] replaced.

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Do note that using lookahead you can do the same in preg_replace:

echo preg_replace('/\.popup(?=[\s\.{])/', 'foo', $s);
//=> foo .popup-option