在PHP preg_replace的替换参数中转义引用外观相似

Question

I would like to replace a pattern string '/r/' within a subject string 'r r' with predetermined replacements like '"\\23"' and '[4]'. So (showing you only a subset of my replacements array) I try:

$subject = 'r r';
$replacements = array('"\\23"', '[4]');
$replacements = array_map('preg_quote', $replacements);
print_r($replacements);

foreach($replacements as $replacement){
    $subject = preg_replace('/r/', $replacement, $subject, 1);
}

echo $subject;

The string I want is '"\\23" [4]', but I get back '"\23" \[4\]', which is both missing characters and has additional characters.

Does PHP have an escape function that would work here? How do I do this?

Note: Without the preg_quote line the result is '"" [4]'.

Answer 1

preg_quote function is needed only for first argument of preg_replace if it should be escaped, so remove it

to replace r with "\\23" you need to use:

$subject = 'r r';
$replacements = array('"\\\\23"', '[4]'); // here is double escaping, but only in code, for input you do no need to escape anyhitng
foreach ($replacements as $replacement)
    $subject = preg_replace('/r/', addcslashes($replacement, '\\'), $subject, 1); // here is the addition of \ only to \, as it is required by preg_replace
echo $subject;

http://php.net/manual/en/function.preg-replace.php

To use backslash in replacement, it must be doubled ("\\\\" PHP string).