将多个空格，制表符和换行符替换为除注释文本之外的一个空格

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

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.

Answer 4

I'd do the following:

1. split the input into comment and non-comment parts
2. do replacement on the non-comment parts
3. 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);