删除嵌套引号

Question

我有这段文字，我试图删除所有内部引号，而只保留一个引号级别。引号内的文本包含任何字符，甚至换行符等。是否可以使用正则表达式或者我必须编写一个小解析器？

[quote=foo]I really like the movie. [quote=bar]World 

War Z[/quote] It's amazing![/quote]
This is my comment.
[quote]Hello, World[/quote]
This is another comment.
[quote]Bye Bye Baby[/quote]

这里是我想要的文字：

[quote=foo]I really like the movie.  It's amazing![/quote]
This is my comment.
[quote]Hello, World[/quote]
This is another comment.
[quote]Bye Bye Baby[/quote]

这是我在PHP中使用的正则表达式：

%\[quote\s*(=[a-zA-Z0-9\-_]*)?\](.*)\[/quote\]%si

我也尝试了此变体，但它与.或,不匹配，而且我无法弄清楚在引号中还能找到什么：

%\[quote\s*(=[a-zA-Z0-9\-_]*)?\]([\w\s]+)\[/quote\]%i

问题位于这里：

(.*)

Answer 1

您可以使用此：

$result = preg_replace('~\G(?!\A)(?>(\[quote\b[^]]*](?>[^[]+|\[(?!/?quote)|(?1))*\[/quote])|(?<!\[)(?>[^[]+|\[(?!/?quote))+\K)|\[quote\b[^]]*]\K~', '', $text);

详细信息：

\G(?!\A)              # contiguous to a precedent match
(?>                   ## content inside "quote" tags at level 0
  (                    ## nested "quote" tags (group 1)
    \[quote\b[^]]*]
    (?>                ## content inside "quote" tags at any level
      [^[]+
     |                  # OR
      \[(?!/?quote)
     |                  # OR
      (?1)              # repeat the capture group 1 (recursive)
    )*
    \[/quote]
  )
 |
  (?<!\[)           # not preceded by an opening square bracket
  (?>              ## content that is not a quote tag
    [^[]+           # all that is not a [
   |                # OR
    \[(?!/?quote)   # a [ not followed by "quote" or "/quote"
  )+\K              # repeat 1 or more and reset the match
)
|                   # OR
\[quote\b[^]]*]\K   # "quote" tag at level 0 

Answer 2

使用此模式

\[quote=?[^\]]*\][^\[]*\[/quote\](?=((.(?!\[q))*)\[/)

并且一无所有就像这个example

Answer 3

我认为编写解析器会更容易。

使用正则表达式查找[quote]和[\quote]，然后分析结果。

preg_match_all('#(\[quote[^]]*\]|\[\/quote\])#', $bbcode, $matches, PREG_OFFSET_CAPTURE);
$nestlevel = 0;
$cutfrom = 0;
$cut = false;
$removed = 0
foreach($matches(0) as $quote){
    if (substr($quote[0], 0, 1) == '[') $nestlevel++; else $nestlevel--;
    if (!$cut && $nestlevel == 2){ // we reached the first nested quote, start remove here
        $cut = true;
        $cutfrom = $quote[1];
    }
    if ($cut && $nestlevel == 1){ // we closed the nested quote, stop remove here
        $cut = false;
        $bbcode = substr_replace($bbcode, '', $cutfrom - $removed, $quote[1] + 8 - $removed); // strlen('[\quote]') = 8
        $removed += $quote[1] + 8 - $cutfrom;
    }
);

Answer 4

以上都不对我有用。制定了两个对我有用的解决方案：

~\[quote.*\].*\[\/quote\]*~si

或

~\[\/?quote.*\].*\[\/?quote\]~si