带代码指示器的FURLS [英] FURLS with codeigniter

问题描述

我正在开发一个网站，其中包含许多网页，其内容将从数据库中提取。整个网站是用codeigniter完成的。

我不想拥有像www.mywebsite.com/pages/1这样的网址。
而是我想在输入页面内容的时候有管理员要求的URL名称（FURL）。假设管理员输入FURL为：study-programs / animation
，则路径应显示为www.mywebsity.com/study-programs/animation

任何建议？

解决方案

1）Foreach页面存储页面名称（可以使用 uri_title（） / p>

2）在路由中（ application / config / routes ）添加类似

  $ route ['page /（：any）'] =pages / page_lookup; 
  

3）在页面控制器中;

 <？php if（！defined（'BASEPATH'））exit（'不允许直接脚本访问）; 
 
类扩展CI_Controller {
 
 public function __construct（）{
 parent :: __ construct（）; 
} 
 
 public function page_lookup（$ page）{
 //执行数据库查询获取页面id其中页面名称= $ page 
 $ result = $ this- > page_model-> get_page_by_name（$ page）; 
 
 // $ result可以保存页面html 
 //如果是这样，那么只需
 $ this-> load-> view（$ result）; 
} 
} 
  

有关详细信息，请参阅示例。

I am developing a website with many pages whose content will be fetched from database. The whole website is done in codeigniter.

I do not want to have url's like www.mywebsite.com/pages/1 or something. Rather I would like to have url name (FURL) asked by the administrator at the time of entering the contents of the page. Suppose the admin enters FURL as : study-programs/animation then the path should be shown as www.mywebsity.com/study-programs/animation

Any suggestions?

解决方案

1) Foreach page store the page name (could use uri_title())

2) In the routes (application/config/routes) add something like;

$route['page/(:any)'] = "pages/page_lookup";

3) In the pages controller;

<?php if ( ! defined('BASEPATH')) exit('No direct script access allowed');

class Pages extends CI_Controller {

    public function __construct() {
        parent::__construct();
    }

    public function page_lookup($page) {
        // do your database query getting page id where page name = $page
        $result = $this->page_model->get_page_by_name($page);

        // $result could hold the page html
        // if so, then just do
        $this->load->view($result);
    }
}

For more details look at the examples in the Codeigniter manual.

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