php - Remove all spaces from a string that are not enclosed in singlequotes or doublequotes

Solution:

You can use

preg_replace('~(?<!\\\\)(?:\\\\{2})*(?:"[^"\\\\]*(?:\\\\.[^"\\\\]*)*"|\'[^\'\\\\]*(?:\\\\.[^\'\\\\]*)*\')(*SKIP)(?!)|\s+~s', '', $str)

See the PHP demo and a regex demo.

Details

• (?<!\\)(?:\\{2})* - a check if there is no escaping \ immediately on the left: any amount of double backslashes not preceded with \
• (?:"[^"\\]*(?:\\.[^"\\]*)*"'[^'\\]*(?:\\.[^'\\]*)*') - either a double- or single-quoted string literal allowing escape sequences
• (*SKIP)(?!) - skip the match and start a new search from the location where the regex failed
• - or
• \s+ - 1 or more whitespaces.

Note that a backslash in a single-quoted PHP string literal is used to form string escape sequences, and thus a literal backslash is "coded" with the help of double backslashes, and to match a literal backslash in text, two such backslashes are required, hence "\\\\" is used.

Solution:

You could capture either " or ' in a group and consume any escaped variants or each until encountering the closing matching ' or " using a backreference \1

(?<!\\)(['"])(?:(?!(?:\1\\)).\\.)*+\1(*SKIP)(*FAIL)\h+

Regex demo Php demo

Explanation

• (?<!\\) Negative lookbehind, assert not a \ directly to the left
• (['"]) capture group 1, match either ' or "
• (?: Non capture group
  • (?!(?:\1\\)). If what is not directly to the right is either the value in group 1 or a backslash, match any char except a newline
  • Or
  • \\. Match an escaped character
• )*+ Close non capture group and repeat 1+ times
• \1 Backreference to what is captured in group 1 (match up either ' or ")
• (*SKIP)(*FAIL) Skip the match until now. Read more about (*SKIP)(*FAIL)
• Or
• \h+ Match 1+ horizontal whitespace chars that you want to remove

As @Wiktor Stribiżew points out in his comment

In some rare situations, this might match at a wrong position, namely, if there is a literal backslash (not an escaping one) before a single/double quoted string that should be skipped. You need to add (?:\{2})* after (?<!\)

The pattern would then be:

(?<!\\)(?:\\{2})*(['"])(?:(?!(?:\1\\)).\\.)*+\1(*SKIP)(*FAIL)\h+

Regex demo

Solution:

Here is a 3 step approach:

1. replace spaces in quote sections with placeholder
2. remove all spaces
3. restore spaces in quote sections

    $str = 'abc | xx ??   "1 x \' 3" d e f \' y " 5 \' x yz';
    echo 'input:  ' . $str . "\n";
    $result = preg_replace_callback( // replace spaces in quote sections with placeholder
        '|(["\'])(.*?)(\1)|',
        function ($matches) {
            $s = preg_replace('/ /', "\x01", $matches[2]);
            return $matches[1] . $s . $matches[3];
        },
        $str
    );
    $result = preg_replace('/ /', '', $result);     // remove all spaces
    $result = preg_replace('/\x01/', ' ', $result); // restore spaces in quote sections
    echo 'result: ' . $result;
    echo "\nexpect: " . 'abc|xx??"1 x \' 3"def\' y " 5 \'xyz';

Output:

input:  abc | xx ??   "1 x ' 3" d e f ' y " 5 ' x yz
result: abc|xx??"1 x ' 3"def' y " 5 'xyz
expect: abc|xx??"1 x ' 3"def' y " 5 'xyz

Explanation:

1. replace spaces in quote sections with placeholder

• use a preg_replace_callback()
• '|(["\'])(.*?)(\1)|' matches quote sections starting and ending with either " or '
• the (\1) makes sure to match the closing quote (either " or ')
• within the callback, use preg_replace() to replace all spaces with a non-printable replacement "\x01"

2. remove all spaces

• use preg_replace() to remove all spaces
• the replace does not match the replacement "\x01", thus misses spaces in quote sections

3. restore spaces in quote sections

• use preg_replace() to restore all spaces from replacement "\x01"