php 下面的函数怎么改？

protected function doCodeBlocks($text)
    {
        #
        #   Process Markdown `<code>` blocks.
        #
        preg_match('/\{\{\{/i', $text, $_m_c_open);
        preg_match('/\}\}\}/i', $text, $_m_c_close);

        if (count($_m_c_open) == count($_m_c_close))
        {
            $text = preg_replace_callback('/\{\{\{[ \n]*(.*?)\}\}\}/is', array(
                &$this,
                '_doCodeBlocks_callback'
            ), $text);
        }

        return $text;
    }

    protected function _doCodeBlocks_callback($matches) {
        $codeblock = $matches[1];

        $codeblock = $this->outdent($codeblock);
        $codeblock = htmlspecialchars($codeblock, ENT_NOQUOTES);

        # trim leading newlines and trailing newlines
        $codeblock = preg_replace('/\A\n+|\n+\z/', '', $codeblock);

        $codeblock = str_replace(array(
            '&amp;',
            "\t",
            ' '
        ), array(
            '&',
            '&nbsp;&nbsp;&nbsp;&nbsp;',
            '&nbsp;'
        ), $codeblock);

        $codeblock = "<pre class=\"prettyprint\">$codeblock\n</pre>";

        return $this->hashBlock($codeblock);
    }

这第一个个函数中的参数$text是markdown格式文章中的代码内容，markdown 格式的内容是这种形式的

{{{ 代码 }}}

如果markdown格式的代码块是这种形式的 {{{javascript: 代码}}}
我想把javascript:用变量$lang_code的形式提交给$codeblock里面，之后返回下面的html内容：

$codeblock = "<pre><code class=\"$lang_code\">$codeblock\n</code></pre>";

---

在这个后面

# trim leading newlines and trailing newlines
$codeblock = preg_replace('/\A\n+|\n+\z/', '', $codeblock);

加上下面这两行试试:

$lang_pos = strpos($codeblock,":");
$lang_code_pre = substr($codeblock, 3,$lang_pos - 3);

 switch ($lang_code_pre) {
    case 'php':
        $lang_code = "php";
        break;

    case 'javascript':
        $lang_code = "javascript";
        break;

    default:
        $lang_code = "default value";
        break;
 }