Question

我有两张桌子 - 图片和类别看起来像这样：

    CREATE TABLE `categories` (
  `id` int(11) NOT NULL,
  `title` varchar(255) NOT NULL,
  `user_id` int(11) NOT NULL
) ENGINE=InnoDB DEFAULT CHARSET=utf8;


CREATE TABLE `images` (
  `id` int(11) NOT NULL,
  `categories_id` int(11) NOT NULL,
  `file` text NOT NULL,
  `caption` text NOT NULL,
  `description` text NOT NULL,
  `user_id` int(11) NOT NULL
) ENGINE=InnoDB DEFAULT CHARSET=utf8;

我有4个类别，我需要过滤并分别显示每个类别的图像

我尝试这样但显示所有图像

**Model:**
public function org_filter()//filter categories
    {
        $this->db->select('i.file,i.categories_id,c.id,c.title',false);
        $this->db->from('images as i');
        $this->db->join('categories as c','i.categories_id = c.id','inner');//join tables based on the foreign key
        $this->db->where('c.title','organizers');//set filter 
        $query = $this->db->get();//get data*/
        return $query;
    }

 **Controller**
 $data=array(
    'r_images'   => $this->Gallery_model->org_filter(),
 );

         $this->load->view('layouts/main', $data);

  **View**
     <h3 class="svgbg">ORGANIZERS</h3><!--name of the first category-->
    <?php foreach($r_images->result() as $img) : ?>
        <div>
            <div class="thumbnail">
                <?=img($img->file)?>
            </div>
        </div>
    <?php endforeach; ?>

所以我的目标是从db中进行dinamical fetch来查看。

P.S：抱歉日本人不好

Answer 1

您可以尝试使用此代码来解决问题..

<强> Category_model.php

在笼子模型中添加此功能

 var bl = require('bl');
var http = require('http')


var dataOne = http.get(process.argv[2], function(response){
    response.pipe(bl(function(err, data) {
        if (err) 
            return console.error(err)
        data = data.toString()
        return data
    }))
})
console.log(dataOne)

<强> Gallery_model.php

在gallary模型中添加此函数。

//Model Category
public function get_category(){
   $q = "select * from categories";
   $rs = $this->db->query($q);
   return $rs->result();
}

<强> Gallery.php

以下代码添加到图库控制器中。

//Gallery_model
public function org_filter($categoryId)//filter categories
{
         $q =  "SELECT * FROM images INNER JOIN categories on images.categories_id = categories.id WHERE categories.id =".$categoryId;
         $q = $this->db->query($q)->result();
         return $q;
}

<强> main.php

以下代码添加到视图文件main.php

中

function index()
    {
        $category = $this->dashboard_model->get_category();

        $data = array();
        //Controller
        //$category = $this->Category_model->get_category();
        foreach ($category as $key => $value) {
            $data['category'][$key]['title'] = $value->title;
            $data['category'][$key]['r_images'] = $this->dashboard_model->org_filter($value->id);
        }
        $this->load->view('layouts/main', $data);
}

首先代码添加你的模型，控制器和视图。你可以获得完美的输出。

Answer 2

您似乎忘记了where子句，这就是为什么返回所有图像。

尝试更改函数org_filter，如下所示：

public function org_filter($categoryId)//filter categories
{
    $this->db->select('*');
    $this->db->from('images');
    $this->db->join('categories','images.categories_id = categories.id','inner');//join tables based on the foreign key
    $this->db->where('categories.id',$categoryId);
    $this->db->on('images.id',10);//set filter
    $query = $this->db->get();//get data
    return $query;
}

如何使用Codeigniter过滤数据库数据？

2 个答案: