php - Typecho如何单纯输出上一篇和下一篇文章的链接和标题。

Question

上一篇文章$this->thePrev()
下一篇文章$this->theNext()
输出中包含有a标签，应该如何获取单纯的链接地址 和 文章标题？
是需要修改系统核心还是可以有更简单的方法？

系统核心代码：

public function theNext($format = '%s', $default = NULL, $custom = array())
    {
        $content = $this->db->fetchRow($this->select()->where('table.contents.created > ? AND table.contents.created < ?',
            $this->created, $this->options->gmtTime)
            ->where('table.contents.status = ?', 'publish')
            ->where('table.contents.type = ?', $this->type)
            ->where('table.contents.password IS NULL')
            ->order('table.contents.created', Typecho_Db::SORT_ASC)
            ->limit(1));

        if ($content) {
            $content = $this->filter($content);
            $default = array(
                'title' => NULL,
                'tagClass' => NULL
            );
            $custom = array_merge($default, $custom);
            extract($custom);

            $linkText = empty($title) ? $content['title'] : $title;
            $linkClass = empty($tagClass) ? '' : 'class="' . $tagClass . '" ';
            $link = '<a ' . $linkClass . 'href="' . $content['permalink'] . '" title="' . $content['title'] . '">' . $linkText . '</a>';

            printf($format, $link);
        } else {
            echo $default;
        }
    }

    public function thePrev($format = '%s', $default = NULL, $custom = array())
    {
        $content = $this->db->fetchRow($this->select()->where('table.contents.created < ?', $this->created)
            ->where('table.contents.status = ?', 'publish')
            ->where('table.contents.type = ?', $this->type)
            ->where('table.contents.password IS NULL')
            ->order('table.contents.created', Typecho_Db::SORT_DESC)
            ->limit(1));

        if ($content) {
            $content = $this->filter($content);
            $default = array(
                'title' => NULL,
                'tagClass' => NULL
            );
            $custom = array_merge($default, $custom);
            extract($custom);

            $linkText = empty($title) ? $content['title'] : $title;
            $linkClass = empty($tagClass) ? '' : 'class="' . $tagClass . '" ';
            $link = '<a ' . $linkClass . 'href="' . $content['permalink'] . '" title="' . $content['title'] . '">' . $linkText . '</a>';

            printf($format, $link);
        } else {
            echo $default;
        }
    }

Answer 1

估计是要改的了。

Answer 2

实在不适合你的需求的话，可以在 functions.php 文件中重写两个代码。