PHP preg_replace：下面的代码具体做什么？

嗨，我正在修改MyBB的源代码

以下代码来自

class\u feedgeneration.php

：

/**
 * Sanitize content suitable for RSS feeds.
 *
 * @param  string The string we wish to sanitize.
 * @return string The cleaned string.
 */
function sanitize_content($content)
{
    $content = preg_replace("#&[^\s]([^\#])(?![a-z1-4]{1,10};)#i", "&$1", $content);
    $content = str_replace("]]>", "]]]]><![CDATA[>", $content);

    return $content;
}

它到底做什么？我知道一点正则表达式，但这个有点太复杂了

有人能给我解释一下吗

非常感谢

"#& -- the char & as is
[^\s] -- one not space character (also \S could be used instead)
([^\#]) -- one not-dash character
(?![a-z1-4]{1,10};) -- and negative lookahead assertion that previous chars
                    -- are not followed by chars in a-z1-4 range
                    -- (only 1 to 10 in a row) with ; after
#i" -- case insensitive

我们从所有的匹配中选择

（[^\#]）

，在前面加上

&并替换

它用于将所有
&xxx
序列替换为
&x26；xxx
这是在rss提要项中写入符号和字符的安全方法

"#& -- the char & as is
[^\s] -- one not space character (also \S could be used instead)
([^\#]) -- one not-dash character
(?![a-z1-4]{1,10};) -- and negative lookahead assertion that previous chars
                    -- are not followed by chars in a-z1-4 range
                    -- (only 1 to 10 in a row) with ; after
#i" -- case insensitive