How do I replace multiple characters with the same number of characters with a regular expression?

Question

I\'ve got the following source:

011011000010011011&         


        

Answer 1

This regex will only match 1 | 0 if it's preceded with "white">.
the (?<=...) syntax in regex is called positive lookbehind...

(?<="white">)([10]+)

Answer 2

Try this here

<?php
$string = '<font color="black">0</font><font color="white">1101100001001101</font><font color="black">1</font><font color="white">0110</font>';
$pattern = '/(?<=<font color="white">)( *?)[10](?=.*?<\/font>)/';
$replacement = '$1 ';
while (preg_match($pattern, $string)) {
        $string = preg_replace($pattern, $replacement, $string);
}
echo $string;

I use a positive look behind (?<=<font color="white">) to search for the color part. And a positive look ahead (?=.*?<\/font>) for the end.

Then I match the already replaced spaces and put them into group 1 and then the [10].

Then I do a while loop until the pattern do not match anymore and a replace with the already replaces spaces and the new found space.

Answer 3

$test = '<font color="black">0</font><font color="white">1101100001001101</font><font color="black">1</font><font color="white">0110</font>';

echo preg_replace ('~(<font color="white">)([10]*)(</font>)~e', '"\\1" . str_repeat(" ", strlen ("\\2")) . "\\3"', $test);