Question

我正在尝试使用codeigniter从数据库显示图像。当用户搜索位置时，将显示包括图像的信息。但我没有显示图像。我已将我的图像保存在应用程序外的文件中。这是我的代码。

// view.php

<style>
#searchbutton{
position: absolute;
left:300px;
top:30px;
    }
fieldset {
background-color:#EFEAEA;
margin: 0px 0px 10px 0px;
 padding: 20px;
border-radius: 1px;
width:900px;
margin-left:220px;
margin-top:-10px;
}
#user{
font-style:italic;
font-size: 12px;
text-align:right;
}
#titlereview {
font-style: italic;
font-size:20px;
}
#review {
font-size:16px;
}
</style>

<?=form_open_multipart('viewreview/view');?>
<?php $search = array('name'=>'search',);?>
<div id = "searchbutton">
<?=form_input($search);?><input type=submit value="Search" /></p>
</div>
<?=form_close();?>

<div class = "tablestyle">
<fieldset>
<?php foreach ($query as $row): ?>
<div id = "user">
User: <?php echo $row->name; ?><br>
Visited time: <?php echo $row->visitedtime; ?><br>
</div>
<div id = "titlereview">"<?php echo $row->titlereview; ?>"<br></div>
<div id = "review"><?php echo $row->yourreview; ?><br></div>
<div id = "image"><?php echo '<img src="data:image/jpeg, $row[images]"/>' ?<br><hr><br></div>

<?php endforeach; ?>
</fieldset>
</div>

//控制器

<?php

class viewreview extends CI_Controller {

public function view($page = 'viewreview') //writereview page folder name
{
    $this->load->model('viewreview_model');
    $data['query'] = $this->viewreview_model->get_data();
    $this->load->vars($data);

    if ( ! file_exists('application/views/viewreview/'.$page.'.php')) //link
    {
        // Whoops, we don't have a page for that!
        show_404();
    }

    $data['title'] = 'View Review'; 
    //$data['title'] = ucfirst($page); // Capitalize the first letter
    $this->load->helper('html');
    $this->load->helper('url');
    $this->load->helper('form');
    $this->load->view('templates/header', $data);
    $this->load->view('viewreview/'.$page, $data);
    $this->load->view('templates/footer', $data);
 }
}
?>

//模型

<?php
class viewreview_model extends CI_Model {

public function __construct()
{
    $this->load->database();
}
public function get_data()
{
    $match = $this->input->post('search');
    $this->db->like('sitename',$match);
    $this->db->or_like('titlereview',$match);
    $this->db->or_like('yourreview',$match);
    $this->db->or_like('suggestion',$match);

    $query = $this->db->get('review');      //pass data to query
    return $query->result();
 }

}
?>

Answer 1

默认情况下，Code-igniter将以std-object的形式返回数据。您需要使用 - ＆gt;访问它（箭头）。

e.g。 jd回复评论

<?php echo '<img src="data:image/jpeg,'.$row->images.'" />' ?>

或者您需要在模型中的函数中传递参数'array'

在你的模型中使用它：return $query->result('array'); return（Array）

实例：return $query->result();返回（标准对象）

如何从数据库codeigniter显示图像

1 个答案: