Question

我正在使用WAMP服务器，我想使用CI在数据库中上传图像。数据库中的图像变量是blob数据类型。我的问题如下：

1）如何存储图像而不是文件名以及我应该使用哪些数据类型？

2）如何从数据库中检索图像？

我的控制器代码：

<?php class Image_control extends CI_Controller{
function index()
{
    //$this->load->view('image_view');

    //$this->Image_model->do_upload();

    $data['images']=$this->Image_model->get_images();
    $this->load->view('image_view',$data);  
}
function do_upload()
{
    $config = array(
        'allowed_types' => 'jpg|png|bmp', 
        'upload_path'=>'./images1/',
        'max_size'=>2000
    );
    $this->load->library('upload',$config);
    if (!$this->upload->do_upload()) {
        $errors[]=array('error'=>$this->upload->display_errors());
        $this->load->view('image_view',$errors);
    }
    $image_path=$this->upload->data();
    $file_name=$image_path['file_name'];
    $config = array(
        'a_name' => $this->input->post('a_name'),
        'a_details'=>$this->input->post('a_info'),
        'a_photo'=>$file_name
    );
    $insert=$this->db->insert('animalstore',$config);
    return $insert;
}   
}
?>

我的模特代码：

<?php class Image_model extends CI_Model {
function get_images()
{
    $query = $this->db->get('animalstore');
    if($query->num_rows > 0 )
    {
        foreach($query->result() as $rows)
        {
            $data[] = $rows;
        }
        return $data;
    }
}
}
?>

最后这是我的观点的代码：

<?php
    echo form_open_multipart('image_control/do_upload');
    echo form_input('a_name','Animal Name');
    echo form_input('a_info','Animal Information');
    echo form_upload('userfile');
    echo form_submit('upload','Upload');
    echo form_close();
?>

<?php foreach ($images as $image):?>
<h1><?php echo $image->a_name;?></h1>
<h1><?php echo $image->a_details;?></h1>
<img src = "/<?php// echo ltrim($image->a_photo, '/'); ?>" >
<img src="http://localhost/ci_test/images1/<?php echo $image->a_photo;?>"/> 
<img src="<?php //echo sprintf("images/%s", $image['screenshot']);?>" />
<h1><?php// echo $image->a_photo;?></h1>
<?php endforeach; ?>

我尝试用不同的方式解决它并搜索我的问题，但我没有找到任何合适的答案。

Answer 1

不要存储数据库内的文件！

这总是一个糟糕的设计理念。将文件存储在文件系统中，只需存储文件名并指向文件，这将为您节省很多麻烦。

Answer 2

// uploading
public function do_upload(){
...

$image_path=$this->upload->data();
$uploaded_image = $image_path['full_path'];

// Read the file
$fp = fopen($uploaded_image, 'r');
$data = fread($fp, filesize($uploaded_image));
$data = addslashes($data);
fclose($fp);

// here you can easy insert $data to 'a_photo' column.    

}


// Viewing, $image_id is row id
public function getImage($image_id){

// select $row from database as usual and then

$content = $row['a_photo'];
echo '<img src="data:image/jpeg;base64,'.base64_encode($content).'">';
}

在你的模板中：

<?php getImage(12); ?>

其中12是行ID。

Answer 3

试试这段代码

模型代码

function do_upload() {

    $config = array(
            'allowed_types' => 'jpg|png|bmp', 
            'upload_path'=>'./images1/', //make sure you have this folder
            'max_size'=>2000
        );

        $this->load->library('upload',$config);

        if ($this->upload->do_upload()) {
            echo "Upload success!";
        } else {
            echo "Upload failed!";
        }
    $image_data = $this->upload->data();

    }  

function get_images()
    {
        $query = $this->db->get('animalstore');
        return $query;
    }

function Save_gallery($in)
{
$save=$this->db->insert('animalstore',$in);
return $save;
}

控制器代码

function index()
{
    $this->load->model('Image_control'); //call a models

    if ($this->input->post('upload')) {

    $in=array();

    $in['a_name']    = $this->input->post('a_name'),
    $in['a_details'] = $this->input->post('a_info'),
    $in['a_photo']=$_FILES['userfile']['name'];

    if($this->Image_model->do_upload()) {

    echo $this->upload->display_errors();

    }else {

    $this->Image_model->Save_gallery($in);

    header('location:index');
    }

    $data['images']=$this->Image_model->get_images();
    $this->load->view('image_view',$data);  
}

视图

<?php
    echo form_open_multipart('image_control/index');
    echo form_input('a_name','Animal Name');
    echo form_input('a_info','Animal Information');
    echo form_upload('userfile');
    echo form_submit('upload','Upload');
    echo form_close();
?>

<?php foreach ($images as $image):?>
<h1><?php echo $image['a_name'];?></h1>
<h1><?php echo $image['a_details'];?></h1>
<?php echo '<img src ="'. base_url().'images1/'.$image['a_photo'].'" >";
endforeach; ?>

Answer 4

这是我用于小型PNG缩略图的快速内容。它们存储在名为＆＃34; coreg＆＃34;的表中。在名为＆＃34; IMAGE＆＃34;的字段中。字段类型是LONGBLOB。每次上传都会覆盖上一张图片。在实际应用中，视图文件显示为iframe：

查看上传文件（当然，最好使用CI特定的表单标签，但是你明白了）

add_image.php：

    Current image:
   <img src='/media/png/coreg/<?=$coregID?>' />
    <? if(isset($error)){ echo $error; }?>
    <form method='post' enctype="multipart/form-data" action='add_image/<?=$coregID?>'>
      <input name="userfile" type="file"  class='vLink' />
      <input name="submitbtn" type="submit" value=" upload &amp; overwrite " class='eLink' />
    </form>

控制器显示图像

更优雅的是使用视图而不是回声并将DB逻辑移动到模型中，但这显示了更好的功能恕我直言：

忽略原始

    require_once dirname(__FILE__) . "/base.php";

        class Media extends BaseController {

        function __construct() {  
            parent::__construct();
        }


        function png($table,$id)  {
             $this->db->where('ID',$id);
            $r = $this->db->get($table);
            if($r->num_rows){
                $r = $r->result_array();
                header("Content-Type: image/png"); 
                echo $r[0]['IMAGE'];
            }   
        }


    }

控制器上传图片：

function add_image($coregID){
        $data['coregID'] = $coregID;
        $data['error'] = '';
        if(isset($_POST['submitbtn'])){
            $config['upload_path'] = './assets/img/coreg/';
            $config['allowed_types'] = 'png';
            $config['max_size'] = '100';
            $config['max_width']    = '350';
            $config['max_height']  = '350';
            $config['file_name']    = $coregID.".png";
            if(file_exists($config['upload_path'].$config['file_name'])){
                unlink($config['upload_path'].$config['file_name']);
            }
            $this->load->library('upload', $config);

            if ( ! $this->upload->do_upload()){             
                $data['error'] = $this->upload->display_errors();   
            }   else    {
                $this->upload->data();
                // now move the image into the DB
                $fp = fopen($config['upload_path'].$config['file_name'], 'r');
                $data = fread($fp, filesize($config['upload_path'].$config['file_name']));

                $this->db->where('ID',$coregID);
                $this->db->update('COREG',array('IMAGE' =>$data));
                fclose($fp);
                // optionally delete the file from the HD after this step
                //unlink($config['upload_path'].$config['file_name']);
            }
        }

            $this->load->view("add_image", $data);  

    }

Answer 5

尝试此代码控制器：-

<?php
class Profile extends CI_Controller {
        public function __construct()
        {
                parent::__construct();
                $this->load->helper(array('form', 'url'));
        }

        public function index()
        {
                $this->load->view('profile', array('error' => ' ' ));
        }

        function do_upload(){
            $config['upload_path']          = './uploads/';
            $config['allowed_types']        = 'gif|jpg|png';
            $this->load->library('upload', $config);
            if ( ! $this->upload->do_upload())
            {
                    $error = array('error' => $this->upload->display_errors());

                    $this->load->view('profile', $error);
            }
            else
            {
                    $data = $this->upload->data();
                    $data['img']=base_url().'./uploads/'.$data['file_name'];
                    $image['profile_pic'] = $data['file_name'];
                    $this->db->insert('user_profile_pic', $image);
                    $this->load->view('profile', $data);    
            }
        }   
}

Answer 6

视图：

<?php echo form_open_multipart('profile/do_upload');?>
<input type="file" name="userfile" size="20" />
<br /><br /> 
<input type="submit" value="upload" />
<img src="<?php echo $img ?>" width="300px" height="300px">
</form>
?>

如何使用CodeIgniter在数据库中存储和检索图像

6 个答案: