我想在输入错误的网址时显示404页面
TRUE URL
https://localhost.com/posts/hello_world
错误的网址
https://localhost.com/posts/dsadasd
虚假网址到404
控制器
public function read($slug_posts)
{
$options = $this->options_model->listing();
$posts = $this->posts_model->read($slug_posts);
#$listing = $this->posts_model->home();
$data = array( 'title' => $posts->post_title.' – '.$options->option_title,
'description' => $posts->post_description,
'posts' => $posts,
#'listing' => $listing,
'content' => 'posts/read'
);
$this->load->view('layout/wrapper', $data, FALSE);
}
答案 0 :(得分:0)
试试这个
public function read($slug_posts)
{
$options = $this->options_model->listing();
$posts = $this->posts_model->read($slug_posts);
if ( ! $posts)
{
show_404();
}
#$listing = $this->posts_model->home();
$data = array( 'title' => $posts->post_title.' – '.$options->option_title,
'description' => $posts->post_description,
'posts' => $posts,
#'listing' => $listing,
'content' => 'posts/read'
);
$this->load->view('layout/wrapper', $data, FALSE);
}
如果真正了解您的问题是什么,当根据您通过网址获取的slug没有帖子时,此功能会显示404页面。