Question

我无法显示保存在MYSQL数据库中的图像 以下是我上传文件的视图代码

<?php echo form_open_multipart('Welcome/do_upload');?>
<input name="file" size="40" type="file" />
<input type="submit" value="Upload" />

这是我的控制器文件

public function do_upload()
    {
        $this->output->enable_profiler(TRUE);
        $config['upload_path'] = './uploads/';
        $config['allowed_types'] = 'gif|jpg|png';
        $config['max_size'] = '3000';
        $config['max_width']  = '1024';
        $config['max_height']  = '768';
        $this->load->library('upload', $config);

        if ( ! $this->upload->do_upload('file'))
        {
            $error = array('error' => $this->upload->display_errors());
            echo "<pre>";
            print_r($this->input->post());
            print_r($error);
        }
        else
        {
            $data = array('upload_data' => $this->upload->data('file'));
            echo "Upload";
            $this->Blog->insert_images($this->upload->data());
             $data = array('upload_data' => $this->upload->data());

        }
    }

这是我的模型文件 此文件首先插入图像，然后显示数据库中的所有图像

public function insert_images($image_data = array()){
      $data = array(
          'i_id' => "NULL",
          'i_s' => $image_data['full_path']
      );
        $img=$this->db->get('images')->result();
        foreach($img as $i){
            ?>
        <img src="<?php echo $i->i_s?>">

            <?php
            }

        }

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Answer 1

问题是保存数据库中的路径

$data = array(
          'i_id' => "NULL",
          'i_s' => $image_data['full_path']
      );

通过此

替换此代码

$data = array(
          'i_id' => "NULL",
          'i_s' => $image_data['file_name']
      );

并显示来自数据库的图像编写此代码

<?php foreach($im as $imd){?>
<img src="<?php echo base_url('uploads/'.$imd->i_s);?>">
<?php }?>

Answer 2

You are not supposed to store file absolute path. You should store relative path of image in database. Relative to

    private void button2_Click(object sender, EventArgs e)
        {
            SqlCommand rcmd = new SqlCommand("SELECT ID, Column1, Column2 FROM [TEST].[dbo].[Table_1] where ID=@ID", connection);

            rcmd.Parameters.AddWithValue("@ID", textBox3.Text);
            connection.Open();
            SqlDataReader reader = rcmd.ExecuteReader();


            while (reader.Read())
            {

                textBox4.Text = reader["Column1"].ToString();
                textBox5.Text = reader["Column2"].ToString();

            }
            connection.Close();
       }

For example: if your settings are

FCPATH

, stored string in DB should be something like

$config['base_url'] = 'http://localhost/site/';

When you want to see it in browser, you would call

img.jpg

图像库使用CodeIgniter

2 个答案: