如何根据控制器中的条件加载视图[codeigniter] [英] How to load a view based on condition in controller [codeigniter]

问题描述

我是codeigniter的新手。我试图做一个简单的网站，将获得其数据库上的设置，如果网站启用（1）或不（0）。

I'm new to codeigniter. I'm trying to make a simple site, that will get its "settings" on the database, if the site is enabled (1) or not (0).

我试图确定网站设置，如果启用（1）或不（0）
并显示索引视图（如果站点值== 1）

I'm trying to determine the site settings if enabled (1) or not (0) and display the index view (if site value == 1)

和maintenace视图（如果网站值是== 0或null）

and maintenace view (if site value is == 0 or null)

我已经写了代码，我可以回复 值。

I have written the code already and I can echo the "site" value in a view.

我的模型

function getsetting() {
$this->db->select("site");
$this->db->from('configuration');
$query = $this->db->get();
$ret = $query->row();
return $ret->site;
}

控制器

    //if site value is == 1 load the normal view
    $this->load->view('index', $data);
    //else load the maintenance view
    $this->load->view('maintenance', $data)

我有一个数据库mysite有一个表配置和一列网站值[null或1]是从网站列，用于确定应显示哪个视图

I have a database "mysite" with one table "configuration" and one column "site" the values [null or 1] are from site column, for determining what view should be displayed.

接受任何帮助。提前感谢。

Any help is accepted. Thanks in advance.

推荐答案

MODEL
这里需要使用 $ query-> row（）; 并传递给控制器​​

MODEL Here you need to fetch single row by using $query->row(); and pass it to controller

function getsetting() {
$this->db->select("site");
$this->db->from('configuration');
$query = $this->db->get();
$ret = $query->row();
return $ret->site;
}

>

As you describe in your question

the values [null or 1]

控制器

function your_controler() {

        $this->load->model('model_file');

        $site = $this->model_file->getsetting();
        if(isset($site) && $site==1){// your condition here
        $this->load->view('index', $data);
        }else{
            $this->load->view('maintenance', $data)

        }

    }

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