duanlie3187 2014-08-28 17:32
浏览 66

如何将数据库查询结果存储在视图中的变量中

I want to store my database query result in my view section in a variable.I am trying but not work.

My View code

$cd=''.base_url().'/video/series/';
$count = count($cd);
for ($i = 0; $i < $count; $i++) 
{
  print'"'.$cd[$i][1].'",';  

}

My Controller code

public function series() {

    $result= $this->video_model->series_list();

    return $result;


}

My Model Code

function series_list()
{

    $string = trim($this->input->get_post('term'));
    $query = $this->db->query("SELECT name FROM `series` WHERE name LIKE '%".$string."%'");


    return $query->row_array();

}

$cd=''.base_url().'/video/series/'; not get any array data ,only get blank data
  • 写回答

2条回答 默认 最新

  • dtvdz911959 2014-08-28 17:36
    关注

    You can get post variable value in controller only. But you trying to get it in model. That's wrong.

    Controller:

    public function series() {
    $string = trim($this->input->get_post('term'));    
    $data['result']= $this->video_model->series_list($string);
    $this->load->view('folder/filename', $data);  
    // in your case i think folder= video and filename = series
    // in this way you can pass value from controller to view
    
    }
    

    Model:

    function series_list($string = null)
    {
        if($string != ''){
           $query = $this->db->query("SELECT name FROM `series` WHERE name LIKE '%".$string."%'");
           return $query->row_array();
        }
        else
           return false;
    }
    

    View:

    <?php
     var_dump($data);
    ?>
    

    you can get your resultset in view and can play with it as you want.

    评论

报告相同问题?

悬赏问题

  • ¥15 虚幻5 UE美术毛发渲染
  • ¥15 CVRP 图论 物流运输优化
  • ¥15 Tableau online 嵌入ppt失败
  • ¥100 支付宝网页转账系统不识别账号
  • ¥15 基于单片机的靶位控制系统
  • ¥15 真我手机蓝牙传输进度消息被关闭了,怎么打开?(关键词-消息通知)
  • ¥15 下图接收小电路,谁知道原理
  • ¥15 装 pytorch 的时候出了好多问题,遇到这种情况怎么处理?
  • ¥20 IOS游览器某宝手机网页版自动立即购买JavaScript脚本
  • ¥15 手机接入宽带网线,如何释放宽带全部速度