Question

这是我的模特外观

public function fetchimage(){
        $this->db->select('u_id,file_name');
        $this->db->from('image');
        if($this->session->userdata('id')){
            $this->db->where('u_id',$this->session->userdata('id'));
            $query = $this->db->get();
            $query->result_array();
            $result = $query->row_array();
            echo '<pre>';
            print_r($query);
            die();
        }
        return !empty($result)?$result:false;
    }

这是我的视图

<?php 
            $this->load->model('user');
            $data['uploadData'][] = $this->user->fetchimage();
            ?>
            <div class="row">
                <ul class="gallery">
                    <?php if($data['uploadData']){ foreach($data['uploadData'] as $data['uploadData']){
                    ?>
                    <li class="item">
                        <img src="<?php echo base_url('/assets/uploads/files/'.$data['uploadData']['file_name']); ?>" >
                    </li>
                    <?php } }else{ ?>
                    <p>Image(s) not found.....</p>
                    <?php } ?>
                </ul>
            </div><br>

图像无法全部打印。它仅打印一张图像。当我采取foreach循环。在Model i中，打印查询，即打印所有图像，但在视图中不打印图像。

Answer 1

尝试一下

row_array()将为您提供具有数组格式数组的单行，而result_array()将为您提供具有数组格式数组的多行。

public function fetchimage(){
        $this->db->select('u_id,file_name');
        $this->db->from('image');
        if($this->session->userdata('id')){
            $this->db->where('u_id',$this->session->userdata('id'));
            $query = $this->db->get();
            $result = $query->result_array();
            echo '<pre>';print_r($query);die();
        }
        return !empty($result)?$result:false;
    }

查看：

<?php
$this->load->model('user');
$data['uploadData'] = $this->user->fetchimage();
?>
<div class="row">
     <ul <?php if ($data['uploadData']) {
           foreach ($data['uploadData'] as $rowData) { ?>
               <li class="item">
                   <img src="<?php echo base_url('/assets/uploads/files/' .$rowData['file_name']); ?>" >
               </li>
           <?php } } else { ?>
           <p>Image(s) not found.....</p>
         <?php } ?>
    </ul>
</div>
<br>

我无法从数据库中获取所有图像只有一个图像从数据库中获取

1 个答案: