如何模仿堆栈溢出自动链接行为

Question

使用PHP,我如何模仿Stack Overflow的自动链接行为(BTW非常酷)？

例如,以下URL:

http://www.stackoverflow.com/questions/1925455/how-to-mimic-stackoverflow-auto-link-behavior

转换成这个:

<a title="how to mimic stackoverflow auto link behavior" rel="nofollow" href="http://www.stackoverflow.com/questions/1925455/how-to-mimic-stackoverflow-auto-link-behavior">stackoverflow.com/questions/1925455/…</a>

title在这种情况下,我并不真正关心属性.

---

还有这个:

http://pt.php.net/manual/en/function.base-convert.php#52450

转换成这个:

<a rel="nofollow" href="http://pt.php.net/manual/en/function.base-convert.php#52450">pt.php.net/manual/en/…</a>

如何在PHP中创建类似的功能？

PS:查看我对这个问题的评论,了解更多示例和行为.

Answer 1

试试吧.URL匹配的正则表达式模式来自Daring Fireball.

/**
 * Replace links in text with html links
 *
 * @param  string $text
 * @return string
 */
function auto_link_text($text)
{
   // a more readably-formatted version of the pattern is on http://daringfireball.net/2010/07/improved_regex_for_matching_urls
   $pattern  = '(?i)\b((?:[a-z][\w-]+:(?:/{1,3}|[a-z0-9%])|www\d{0,3}[.]|[a-z0-9.\-]+[.][a-z]{2,4}/)(?:[^\s()<>]+|\(([^\s()<>]+|(\([^\s()<>]+\)))*\))+(?:\(([^\s()<>]+|(\([^\s()<>]+\)))*\)|[^\s`!()\[\]{};:\'".,<>?«»“”‘’]))';

   $callback = create_function('$matches', '
       $url       = array_shift($matches);
       $url_parts = parse_url($url);

       $text = parse_url($url, PHP_URL_HOST) . parse_url($url, PHP_URL_PATH);
       $text = preg_replace("/^www./", "", $text);

       $last = -(strlen(strrchr($text, "/"))) + 1;
       if ($last < 0) {
           $text = substr($text, 0, $last) . "&hellip;";
       }

       return sprintf(\'<a rel="nofollow" href="%s">%s</a>\', $url, $text);
   ');

   return preg_replace_callback($pattern, $callback, $text);
}

输入文本:

This is my text.  I wonder if you know about asking questions on StackOverflow:
 Check This out http://www.stackoverflow.com/questions/1925455/how-to-mimic-stackoverflow-auto-link-behavior

 Also, base_convert php function?
http://pt.php.net/manual/en/function.base-convert.php#52450

http://pt.php.net/manual/en/function.base-convert.php?wtf=hehe#52450

输出文字:

This is my text.  I wonder if you know about asking questions on StackOverflow:
 Check This out <a rel="nofollow" href="http://www.stackoverflow.com/questions/1925455/how-to-mimic-stackoverflow-auto-link-behavior">stackoverflow.com/questions/1925455/&hellip;</a>

 Also, base_convert php function?
<a rel="nofollow" href="http://pt.php.net/manual/en/function.base-convert.php#52450">pt.php.net/manual/en/&hellip;</a>

<a rel="nofollow" href="http://pt.php.net/manual/en/function.base-convert.php?wtf=hehe#52450">pt.php.net/manual/en/&hellip;</a>

Answer 2

这是基于相同的daringfireball.net正则表达式,但添加了比Eric Coleman的示例更多的逻辑,以及最大URL深度的配置(SO似乎是50),URL被截断时的最大路径深度(SO似乎是2),和省略号字符(…).

据我所知,这复制了所有SO URL重写功能,至少就目前在评论和回复中所讨论的内容而言.

function auto_link_text($text) {
    $pattern  = '#\b(([\w-]+://?|www[.])[^\s()<>]+(?:\([\w\d]+\)|([^[:punct:]\s]|/)))#';
    return preg_replace_callback($pattern, 'auto_link_text_callback', $text);
}

function auto_link_text_callback($matches) {
    $max_url_length = 50;
    $max_depth_if_over_length = 2;
    $ellipsis = '&hellip;';

    $url_full = $matches[0];
    $url_short = '';

    if (strlen($url_full) > $max_url_length) {
        $parts = parse_url($url_full);
        $url_short = $parts['scheme'] . '://' . preg_replace('/^www\./', '', $parts['host']) . '/';

        $path_components = explode('/', trim($parts['path'], '/'));
        foreach ($path_components as $dir) {
            $url_string_components[] = $dir . '/';
        }

        if (!empty($parts['query'])) {
            $url_string_components[] = '?' . $parts['query'];
        }

        if (!empty($parts['fragment'])) {
            $url_string_components[] = '#' . $parts['fragment'];
        }

        for ($k = 0; $k < count($url_string_components); $k++) {
            $curr_component = $url_string_components[$k];
            if ($k >= $max_depth_if_over_length || strlen($url_short) + strlen($curr_component) > $max_url_length) {
                if ($k == 0 && strlen($url_short) < $max_url_length) {
                    // Always show a portion of first directory
                    $url_short .= substr($curr_component, 0, $max_url_length - strlen($url_short));
                }
                $url_short .= $ellipsis;
                break;
            }
            $url_short .= $curr_component;
        }

    } else {
        $url_short = $url_full;
    }

    return "<a rel=\"nofollow\" href=\"$url_full\">$url_short</a>";
}

样本输入:

This is my text.  I wonder if you know about asking questions on StackOverflow:
Check This out http://www.stackoverflow.com/questions/1925455/how-to-mimic-stackoverflow-auto-link-behavior

Also, base_convert php function?
http://pt.php.net/manual/en/function.base-convert.php#52450

http://pt.php.net/manual/en/function.base-convert.php?wtf=hehe#52450

http://a.b/c/d/e/f/test

and http://a.b/c/d/e/f/g/h/i/j/k/l/m/n/o/p/q/r/s/t/u/v/z/y/w/z/test

样本输出:

This is my text.  I wonder if you know about asking questions on StackOverflow:
Check This out <a rel="nofollow" href="http://www.stackoverflow.com/questions/1925455/how-to-mimic-stackoverflow-auto-link-behavior">http://stackoverflow.com/questions/1925455/&hellip;</a> 

Also, base_convert php function?
<a rel="nofollow" href="http://pt.php.net/manual/en/function.base-convert.php#52450">http://pt.php.net/manual/en/&hellip;</a> 

<a rel="nofollow" href="http://pt.php.net/manual/en/function.base-convert.php?wtf=hehe#52450">http://pt.php.net/manual/en/&hellip;</a> 

<a rel="nofollow" href="http://a.b/c/d/e/f/test">http://a.b/c/d/e/f/test</a> 

and <a rel="nofollow" href="http://a.b/c/d/e/f/g/h/i/j/k/l/m/n/o/p/q/r/s/t/u/v/z/y/w/z/test">http://a.b/c/d/&hellip;</a>

Answer 3

这会将示例字符串转换为您所使用的字符串.我遗漏了,title因为它来自不同的来源,而不仅仅是一个独立的URL,你说这并不重要.

<?php
$urlInput="http://www.stackoverflow.com/questions/1925455/how-to-mimic-stackoverflow-auto-link-behavior";
preg_match('@http://(?:www\.)?(\S+/)\S*(?:\s|$)@i', $urlInput, $matches);
print('<a rel="nofollow" href="' . trim($matches[0]) . '">' . $matches[1] . '...</a>');
?>

根据需要进行扩展以扫描文本.

如果您只想匹配一定数量的URL路径元素,请使用此RE:

'@http://(?:www\.)?((?:\S+?/){1,3})\S*(?:\s|$)@i'

这将最多提取3个路径元素(主机和最多两个目录).您可以更改上限{1,3}以定义所需的最大路径元素数.

更改结尾\S以允许零匹配.