删除字符串中未包含在单引号或双引号中的所有空格

Question

I have short strings like this

$str = 'abc | xx ??   "1 x \' 3" d e f \' y " 5 \' x yz';

I want to remove all spaces from a string that are not enclosed in single or double quotes. Any characters enclosed in single or double quotes should not be changed. As a result, I expect:

$expected =  'abc|xx??"1 x \' 3"def\' y " 5 \'xyz';

My current solution based on character-wise comparisons is the following:

function removeSpaces($string){
  $ret = $stop = "";
  for($i=0; $i < strlen($string);$i++){
    $char = $string[$i];
    if($stop == "") {
      if($char == " ") continue;
      if($char == "'" OR $char == '"') $stop = $char;
    }
    else {
      if($char == $stop) $stop = "";
    }
    $ret .= $char;
  }
  return $ret;
}

Is there a solution that is smarter?

Answer 1

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.

Answer 2

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

Answer 3

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"