Remove JavaScript comments using Regex - PHP [duplicate]

Question

I want to minify HTML code using PHP && Regex code

The minify function :

public static function sanitize_output($buffer) {

    $search = array(
        '/ {2,}/',
        '/<!--.*?-->|\t|(?:\r?\n[ \t]*)+/s'
    );
    $replace = array(
        ' ',
        ''
    );
    $buffer = preg_replace($search, $replace, $buffer);

    return $buffer;
}

I use this function when I rendering the template (HTML file), but I get always this error on console

SyntaxError: missing } in compound statement note: { opened at line

The minify code delete some " { " on Js code, is there any solution the fix this problem?

After some search, I found that the error is in JS comments, when I minify the HTML code the comments concatenate with the code.

I try to use this regex

/\/\*[\s\S]*?\*\/|([^:]|^)\/\/.*$/

but it's doesn't work

There are pre-made minification solutions for PHP available, why go through the strain of coding one yourself? RegEx for HTML is a bad idea, even for minification it would be a better idea to use a full-fledged parser instead. — Sarah Groß
– Sarah Groß, Commented Apr 6, 2019 at 16:07
With some search I found that error caused from the JS comment, is there any regex to delete the JS comment ? — ThaTest
– ThaTest, Commented Apr 6, 2019 at 18:29
You can't use RegEx to parse HTML lest you summon the One who should Not Be Named stackoverflow.com/questions/1732348/… — ChrisM
– ChrisM, Commented Apr 6, 2019 at 21:28

sayou · Accepted Answer · 2019-04-07 13:21:01Z

0

Like you say, when you Minify your code using the function, all the HTML code be in single line, for that any code after a JS comments well be a comment to. So You can use this PHP regex to remove inline JavaScript comment :

Your new function :

 public static function sanitize_output($buffer) {

        $search = array(
            '#^\s*//.+$#m', //remove JS comment,
            '/ {2,}/',
            '/<!--.*?-->|\t|(?:\r?\n[ \t]*)+/s'
        );
        $replace = array(
            '',
            ' ',
            ''
        );
        $buffer = preg_replace($search, $replace, $buffer);

        return $buffer;
    }

edited Apr 7, 2019 at 13:21

answered Apr 6, 2019 at 21:34

sayou

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