Question

我是CI的新手，我通过视图文件传递参数，例如

<a href="<?php echo base_url('projects/project_detail/'.$project_info['project_id']);?>" class="button small">More Details</a>

我试图通过控制器查看特定id的内容。

    public function project_detail($id){
        $this->load->helper('url');
        $data['content_for_layout'] = 'projects/project_detail/'.$this->id;
        $data['project'] = $this->project_model->get_project_detail($this->id);  //folder/filename in which data is present.
        $this->load->view('layouts/default', $data);
}

projects/project_detail/1 =内容存在的地方 layouts / default想要在此文件中显示内容。

我收到错误 -

Unable to load the requested file: projects/project_detail/.php

在路线档案中，我已经进行了类似的更改

$route['projects/project_detail/(:num)'] = "projects/project_detail/$1";

帮我打开projects / project_detail

中id = 1的内容

Answer 1

改变这一点;

$data['content_for_layout'] = 'projects/project_detail/'.$this->id;
$data['project'] = $this->project_model->get_project_detail($this->id);

对此;

$data['content_for_layout'] = 'projects/project_detail/'.$id;
$data['project'] = $this->project_model->get_project_detail($id);

删除＆＃34; $ this＆＃34;来自身份证。您正在使用您传递给该函数的ID，而不是该对象的ID。

另外，我不认为您需要使用的路线。

无法加载请求的文件：projects / project_detail / .php

1 个答案: