只匹配单词之间空格的正则表达式[重复]

这种问题很容易解决 preg_replace_callback . 其思想是提取引号之间的子字符串,然后在回调函数中对其进行编辑:

$text = preg_replace_callback('~"[^"]*"~', function ($m) {
    return preg_replace('~\s~', '#', $m[0]);
}, $text);

这是最简单的方法。

用一个单一的模式来做这件事要复杂得多 preg_replace 但有可能:

$text = preg_replace('~(?:\G(?!\A)|")[^"\s]*\K(?:\s|"(*SKIP)(*F))~', '#', $text);

demo

图案细节:

(?:
    \G (?!\A)  # match the next position after the last successful match
  |
    "          # or the opening double quote
)
[^"\s]*        # characters that aren't double quotes or a whitespaces
\K             # discard all characters matched before from the match result
(?:
    \s         # a whitespace
  |
    "           # or the closing quote
    (*SKIP)(*F) # force the pattern to fail and to skip the quote position
                # (this way, the closing quote isn't seen as an opening quote
                # in the second branch.)
)

这种方式使用 \G 锚定以确保所有匹配的空格都位于引号之间。

边缘情况:

• ~(?:\G(?!\A)|"(?=[^"]*"))[^"\s]*\K(?:\s|"(*SKIP)(*F))~

• 双引号可以包含必须忽略的转义双引号:必须这样描述转义字符:

  ~(?:\G(?!\A)|")[^"\s\\\\]*+(?:\\\\\S[^"\s\\\\]*)*+(?:\\\\?\K\s|"(*SKIP)(*F))~

\s(?=[^"]*+(?:"[^"]*"[^"]*)*+")

这是一个短模式,但对于长字符串来说可能会有问题,因为对于每个带有空格的位置,您必须检查字符串,直到最后一个带有lookahead的引号。

请参见以下代码段:

<?php
$text = 'good with spaces "here all spaces should be removed" and here also good';
echo "$text \n";
$regex = '/(\".+?\")|\s/';
$regex = '/"(?!.?\s+.?)/';
$text = preg_replace($regex,'', $text);
echo "$text \n";
?>

我发现 a sample that works

$text = 'good with spaces "here all spaces should be removed" and here also good'
should be 
$text = 'good with spaces "hereallspacesshouldberemoved" and here also good';