我已经完成了CI文档中包含的众所周知的'新闻'插曲。有时候我的链接字符串会出现双重“新闻/”段,如下所示:'/ codeig / news / news / entry'有时在重新加载页面之后一切正常。我应该提到我已经摆脱了'index.php'片段之后的另一个流行教程。我的代码出了什么问题?
这是我的路线:
$route['news/create'] = 'news/create';
$route['news/update/(:any)'] = 'news/update/$1';
$route['news/delete/(:any)'] = 'news/delete/$1';
$route['news/(:any)'] = 'news/view/$1';
$route['news'] = 'news';
$route['welcome'] = 'welcome';
$route['auth/(:any)'] = 'auth/$1';
$route['auth'] = 'auth';
$route['activate/:num/:any'] = "/auth/activate/$1/$2";
$route['reset_password/:num/:any'] = "/auth/reset_password/$1/$2";
$route['(:any)'] = 'pages/view/$1';
$route['default_controller'] = 'news';
views / news / index.php文件:
<?php foreach ($news as $news_item): ?>
<h2><?php echo $news_item['title'] ?></h2>
<div id="main">
<?php echo $news_item['text'] ?>
</div>
<p><a href="news/<?php echo $news_item['slug'] ?>">View article</a></p>
<?php endforeach ?>
查看方法(新闻控制器)
public function view($slug)
{
$data['news_item'] = $this->news_model->get_news($slug);
if (empty($data['news_item']))
{
show_404();
}
$data['title'] = $data['news_item']['title'];
$this->load->view('templates/header', $data);
$this->load->view('news/view', $data);
$this->load->view('templates/footer');
}
答案 0 :(得分:8)
在HTML中,如果当前的URL是这样的:
http://example.com/news/
你有这样的链接:
<a href="news/article-slug">Link</a>
点击该链接后,您的网址将最终成为:
http://example.com/news/news/article-slug
如果您再次点击它,您将拥有:
http://example.com/news/news/article-slug/news/article-slug
注意:这不完全正确,路径的相对性取决于当前网址和/或链接中是否有尾部斜杠/
。< / p>
href="news/something"
是相对于当前页面的相对网址。您想使用绝对或根网址:
<a href="http://example.com/mysite/news/article-slug">Link</a>
<a href="/news/article-slug">Link</a>
使用以下任何Codeigniter功能使绝对URL更容易:
anchor()
(为您制作完整链接)site_url()
(返回您的绝对基本网址)base_url()
(与上述相同)<?php echo anchor('news/'.$news_item['slug'], 'Link Text'); ?>
<a href="<?php echo base_url('news/'.$news_item['slug']); ?>">Link Text</a>
所以只是为了澄清一下,这与您的路由无关 - 这纯粹是HTML中相对链接的问题。
答案 1 :(得分:2)
不太详细的解释:
在'application / views / news / index.php'文件中
之前(导致/ news / news / in URL):
<p><a href="news/<?php echo $news_item['slug'] ?>">View article</a></p>
之后(已修复):
<p><a href="<?php echo $news_item['slug'] ?>">View article</a></p>
news/
包含在$news_item['slug']
中,因此不需要在HTML中。