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我在这里有这个自动链接正则表达式代码：

// turn any url into url bbcode that doesn't have it already - so we can auto link urls- thanks stackoverflow

$URLRegex = '/(?:(?<!(\[\/url\]|\[\/url=))(\s|^))'; // No [url]-tag in front and is start of string, or has whitespace in front
$URLRegex.= '(';                                    // Start capturing URL
$URLRegex.= '(https?|ftps?|ircs?):\/\/';            // Protocol
$URLRegex.= '\S+';                                  // Any non-space character
$URLRegex.= ')';                                    // Stop capturing URL
$URLRegex.= '(?:(?<![.,;!?:\"\'()-])(\/|\s|\.?$))/i';      // Doesn't end with punctuation and is end of string, or has whitespace after

$body = preg_replace($URLRegex,"$2[url=$3]$3[/url]$5", $body);

问题是，如果 url 在引用标签内，并且结束引用标签正好在链接上，那么结束引用标签就会包含在链接中，这当然会搞砸！

如何调整该正则表达式以在链接中不包含 [ 和 ] 内的任何内容？

样本输入：

[quote=liamdawe] Have you had a look at [url=http://pcgamingwiki.com/wiki/Serious_Sam_II#Linux_Installation]this howto[/url]? :)

http://pcgamingwiki.com/wiki/Serious_Sam_II#Linux_Installation[/quote]
Testing

正确的输出是：

<div class="quote"><strong>Quote from liamdawe</strong><br />  Have you had a look at <a href="http://pcgamingwiki.com/wiki/Serious_Sam_II#Linux_Installation" target="_blank">this howto</a>? <img src="/jscripts/sce/emoticons/smile.png" alt="" /><br />
<br />
<a href="http://pcgamingwiki.com/wiki/Serious_Sam_II#Linux_Installation" target="_blank">http://pcgamingwiki.com/wiki/Serious_Sam_II#Linux_Installation</a></div><br />
Testing

但我得到的输出是：

<div class="quote"><strong>Quote from liamdawe</strong><br />  Have you had a look at <a href="http://pcgamingwiki.com/wiki/Serious_Sam_II#Linux_Installation" target="_blank">this howto</a>? <img src="/jscripts/sce/emoticons/smile.png" alt="" /><br />
<br />
<a href="http://pcgamingwiki.com/wiki/Serious_Sam_II#Linux_Installation </div>" target="_blank">http://pcgamingwiki.com/wiki/Serious_Sam_II#Linux_Installation[/quote]</a><br />
Testing<br />

如您所见，它在链接中包含了 [/quote] 标记，因为它没有忽略自动链接器正则表达式中的 bbcode 标记。

如果需要，下面是执行该类型引用的代码： // 引用一个真实的人、书或任何东西 $pattern = '/[quote\=(.+?)](.+?)[/quote]/是';

$replace = "<div class=\"quote\"><strong>Quote from $1</strong><br />$2</div>";

while(preg_match($pattern, $body))
{
    $body = preg_replace($pattern, $replace, $body);
}
4

1 回答 1

2

试试这个

$URLRegex = '/(?:(?<!(\[\/url\]|\[\/url=))(\s|^))'; // No [url]-tag in front and is start of string, or has whitespace in front
$URLRegex.= '(';                                    // Start capturing URL
$URLRegex.= '(https?|ftps?|ircs?):\/\/';            // Protocol
$URLRegex.= '[\w\d\.\/#\_\-\?:=]+';                        // Any non-space character
$URLRegex.= ')';                                    // Stop capturing URL
$URLRegex.= '(?:(?<![.,;!?:\"\'()-])(\/|\[|\s|\.?$))/i';      // Doesn't end with punctuation and is end of string, or has whitespace after

$body = preg_replace($URLRegex,"$2[url=$3]$3[/url]$5", $body);
于 2013-06-12T12:00:49.687 回答