Replace last occurrence of a string in a string

This is an ancient question, but why is everyone overlooking the simplest regexp-based solution? Normal regexp quantifiers are greedy, people! If you want to find the last instance of a pattern, just stick .* in front of it. Here's how:

$text = "The quick brown fox, fox, fox, fox, jumps over etc.";
$fixed = preg_replace("((.*)fox)", "$1DUCK", $text);
print($fixed);

This will replace the last instance of "fox" to "DUCK", like it's supposed to, and print:

The quick brown fox, fox, fox, DUCK, jumps over etc.

A short version:

$NewString = substr_replace($String,$Replacement,strrpos($String,$Replace),strlen($Replace));

While using regex is typically less performant than non-regex techniques, I do appreciate the control and flexibility that it affords.

In my snippet, I will set the pattern to be case-insensitive (\i, although my sample input will not challenge this rule) and include word boundaries (\b, although they were not explicitly called for).

I am also going to use the \K metacharacter to reset the fullstring match so that no capture groups / backreferences are needed.

Code: (Demo)

$search = 'The';
$replace = 'A';
$subject = "The Quick Brown Fox Jumps Over The Lazy Dog's Thermos!";

echo preg_replace(
         '/.*\K\b' . preg_quote($search, '/') . '\b/i',
         $replace,
         $subject
     );

Output:

  The Quick Brown Fox Jumps Over A Lazy Dog's Thermos!
# ^^^                            ^            ^^^
# not replaced                replaced        not replaced

Without word boundaries: (Demo)

echo preg_replace(
         '/.*\K' . preg_quote($search, '/') . '/i',
         $replace,
         $subject
     );

Output:

  The Quick Brown Fox Jumps Over The Lazy Dog's Armos!
# ^^^                            ^^^            ^
# not replaced              not replaced        replaced

Below corrected zack's code (https://stackoverflow.com/a/11144999/9996503). Be careful with regular expressions! Consider this:

$string = 'Round brackets (parentheses) "()", square brackets "()"';
$find = '()';
$replace = '[]';
// Zack's code:
$result = preg_replace(strrev("/$find/"),strrev($replace),strrev($string),1);
var_dump($result); // Output: NULL
// Correct code:
$result = strrev(preg_replace('/'.preg_quote(strrev($find)).'/', strrev($replace), strrev($string), 1));
echo $result; //Output: Round brackets (parentheses) "()", square brackets "[]"

For the interested: I've written a function that utilises preg_match so that you're able to replace from right hand side using regex.

function preg_rreplace($search, $replace, $subject) {
    preg_match_all($search, $subject, $matches, PREG_SET_ORDER);
    $lastMatch = end($matches);

    if ($lastMatch && false !== $pos = strrpos($subject, $lastMatchedStr = $lastMatch[0])) {
        $subject = substr_replace($subject, $replace, $pos, strlen($lastMatchedStr));
    }

    return $subject;
}

Or as a shorthand combination/implementation of both options:

function str_rreplace($search, $replace, $subject) {
    return (false !== $pos = strrpos($subject, $search)) ?
        substr_replace($subject, $replace, $pos, strlen($search)) : $subject;
}
function preg_rreplace($search, $replace, $subject) {
    preg_match_all($search, $subject, $matches, PREG_SET_ORDER);
    return ($lastMatch = end($matches)) ? str_rreplace($lastMatch[0], $replace, $subject) : $subject;
}

based on https://stackoverflow.com/a/3835653/3017716 and https://stackoverflow.com/a/23343396/3017716