How to skip first regex match?

Question

Is there anyway to skip the first match when using regex and php.

Or is there some way of achieveing this using str_replace.

Thanks

UPDATE

Answer 1

preg_replace('/((?:^.*?\btest\b)?.*?)\btest\b/', '$1', $string);

The idea is to match and capture whatever precedes each match, and plug it back in. (?:^.*?test)? causes the first instance of test to be included in the capture. (All the \bs are to avoid partial-word matches, like the test in smartest or testify.)

Answer 2

assume 'blah' is your regex pattern, blah(blah) will match and capture the second one

Answer 3

Late answer but it might be usefull to people.

$string = "This is a test string to test something with the word test and replacing test";
$replace = "test";
$tmp = explode($replace, $string);
$tmp[0] .= $replace;
$newString = implode('', $tmp);
echo $newString; // Output: This is a test string to something with the word and replacing 

Answer 4

Easy PHP way:

<?php
    $pattern = "/an/i";
    $text = "banANA";
    preg_match($pattern, $text, $matches, PREG_OFFSET_CAPTURE);
    preg_match($pattern, $text, $matches, 0, $matches[0][1]);
    echo $matches[0];
?>

will give you "AN".

UPDATE: Didn't know it was a replace. Try this:

<?php
    $toRemove = 'test';
    $string = 'This is a test string to test to removing the word test';
    preg_match("/$toRemove/", $string, $matches, PREG_OFFSET_CAPTURE);
    $newString = preg_replace("/$toRemove/", "", $string);
    $newString = substr_replace($newString, $matches[0][0], $matches[0][1], 0);
    echo $newString;
?>

Find the first match and remember where it was, then delete everything, then put whatever was in the first spot back in.