Replacing comments in php with preg_replace - php

Question

I need to replace all block comments with preg_replace() in php.
For example:
/**asdfasdf
fasdfasdf*/
echo "hello World\n";
For this:
echo "hello World\n";
I tried some solutions from this site, but no one works for me.
My code:
$file = file_get_contents($fileinput);
$file = preg_replace('/\/\*([^\\n]*[\\n]?)*\*\//', '', $file);
echo $file;
My output for example is same as input.Link to my regex test

Answer 1

Use the http://www.php.net/manual/en/function.token-get-all.php:
$file = file_get_contents($fileinput);
$tokens = token_get_all($file); // prepend an open tag if your file doesnt have one
$plain = '';
foreach ($tokens as $token) {
if (is_array($token)) {
list($number, $string) = $token;
if (!in_array($number, [T_OPEN_TAG, T_COMMENT])) { // add all tokens you dont want
$plain .= $string;
}
} else {
$plain .= $token;
}
}
print_r($plain);
Output:
echo "hello World\n";
Here is a list of all PHP tokens:
http://www.php.net/manual/en/tokens.php

Answer 2

Try this
$file = preg_replace('/^\s*?\/\*.*?\*\//m', '', $file);

Answer 3

The best way to parse PHP code is to use the tokenizer.
However it is not so difficult to do it with a regex. You must only skip all strings:
$pattern = <<<'EOD'
~
(?(DEFINE)
(?<sq> ' (?>[^'\\]++|\\{2}|\\.)* ' ) # single quotes
(?<dq> " (?>[^"\\]++|\\{2}|\\.)* " ) # double quotes
(?<hd> <<< \s* (["']?)(\w+)\g{-2} \R .*? (?<=\n) \g{-1} ;? (\R|$) ) # heredoc like
(?<string> \g<sq> | \g<dq> | \g<hd>)
)
\g<string> (*SKIP)(*FAIL) | /\* .*? \*/
~xs
EOD;
$result = preg_replace($pattern, '', $data);