replace multiple spaces, tabs and newlines into one space except commented text - php

Question

I need replace multiple spaces, tabs and newlines into one space except commented text in my html.
For example the following code:
<br/> <br>
<!--
this is a comment
-->
<br/> <br/>
should turn into
<br/><br><!--
this is a comment
--><br/><br/>
Any ideas?

Answer 1

The new solution
After thinking a bit, I came up with the following solution with pure regex. Note that this solution will delete the newlines/tabs/multi-spaces instead of replacing them:
$new_string = preg_replace('#(?(?!<!--.*?-->)(?: {2,}|[\r\n\t]+)|(<!--.*?-->))#s', '$1', $string);
echo $new_string;
Explanation
(? # If
(?!<!--.*?-->) # There is no comment
(?: {2,}|[\r\n\t]+) # Then match 2 spaces or more, or newlines or tabs
| # Else
(<!--.*?-->) # Match and group it (group #1)
) # End if
So basically when there is no comment it will try to match spaces/tabs/newlines. If it does find it then group 1 wouldn't exist and there will be no replacements (which will result into the deletion of spaces...). If there is a comment then the comment is replaced by the comment (lol).
Online demo
The old solution
I came up with a new strategy, this code require PHP 5.3+:
$new_string = preg_replace_callback('#(?(?!<!--).*?(?=<!--|$)|(<!--.*?-->))#s', function($m){
if(!isset($m[1])){ // If group 1 does not exist (the comment)
return preg_replace('#\s+#s', ' ', $m[0]); // Then replace with 1 space
}
return $m[0]; // Else return the matched string
}, $string);
echo $new_string; // Output
Explaining the regex:
(? # If
(?!<!--) # Lookahead if there is no <!--
.*? # Then match anything (ungreedy) until ...
(?=<!--|$) # Lookahead, check for <!-- or end of line
| # Or
(<!--.*?-->) # Match and group a comment, this will make for us a group #1
)
# The s modifier is to match newlines with . (dot)
Online demo
Note: What you are asking and what you have provided as expected output are a bit contradicting. Anyways if you want to remove instead of replacing by 1 space, then just edit the code from '#\s+#s', ' ', $m[0] to '#\s+#s', '', $m[0].

Answer 2

It's much simpler to do this in several runs (as is done for instance in php markdown).
Step1: preg_replace_callback() all comments with something unique while keeping their original values in a keyed array -- ex: array('comment_placeholder:' . md5('comment') => 'comment', ...)
Step2: preg_replace() white spaces as needed.
Step3: str_replace() comments back where they originally were using the keyed array.
The approach you're leaning towards (splitting the string and only processing the non-comment parts) works fine too.
There almost certainly is a means to do this with pure regex, using ugly look-behinds, but not really recommended: the regex might yield backtracking related errors, and the comment replacement step allows you to process things further if needed without worrying about the comments themselves.

Answer 3

I’d do the following:
split the input into comment and non-comment parts
do replacement on the non-comment parts
put everything back together
Example:
$parts = preg_split('/(<!--(?:(?!-->).)*-->)/s', $input, -1, PREG_SPLIT_DELIM_CAPTURE);
foreach ($parts as $i => &$part) {
if ($i % 2 === 0) {
// non-comment part
$part = preg_replace('/\s+/', ' ', $part);
} else {
// comment part
}
}
$output = implode('', $parts);

Answer 4

You can use this:
$pattern = '~\s*+(<br[^>]*>|<!--(?>[^-]++|-(?!->))*-->)\s*+~';
$replacement = '$1';
$result = preg_replace($pattern, $replacement, $subject);
This pattern captures br tags and comments, and matches spaces around. Then it replaces the match by the capture group.