我知道这个问题已被多次询问,但我发现的所有其他问题都与我想要的不同。
我想将文件名和文件路径上传到名为'factoryimages'的表。
这样做的最佳方式是什么?
我的控制器功能:
function do_upload()
{
$config['upload_path'] = './assets/uploads/';
$config['allowed_types'] = 'gif|jpg|png';
$config['max_size'] = '1000';
$config['max_width'] = '';
$config['max_height'] = '';
$config['overwrite'] = TRUE;
$config['remove_spaces'] = TRUE;
$this->load->library('upload', $config);
if ( ! $this->upload->do_upload())
{
$error = array('error' => $this->upload->display_errors());
$this->load->view('members/header');
$this->load->view('members/upload_form', $error);
$this->load->view('members/footer');
}
else
{
$data = array('upload_data' => $this->upload->data());
$this->load->view('members/header');
$this->load->view('members/upload_success', $data);
$this->load->view('members/footer');
}
}
我在我的模型中试过这个,但它不起作用:
function insert_images()
{
$insert_data = array(
'filename' => $image_data['file_name'],
'fullpath' => $image_data['full_path']
);
$this->db->insert('bedrijfimages', $insert_data);
}
我的观点:
<?php echo $error;?>
<br/>
<?php echo form_open_multipart('upload/do_upload');?>
<input type="file" name="userfile" size="20" multiple="true" />
<br /><br />
<input type="submit" value="upload" />
</form>
我收到服务器错误,但我看不出它是什么。
我的控制器功能:
function do_upload()
{
$config['upload_path'] = './assets/uploads/';
$config['allowed_types'] = 'gif|jpg|png';
$config['max_size'] = '1000';
$config['max_width'] = '';
$config['max_height'] = '';
$config['overwrite'] = TRUE;
$config['remove_spaces'] = TRUE;
$this->load->library('upload', $config);
if ( ! $this->upload->do_upload())
{
$error = array('error' => $this->upload->display_errors());
$this->upload_model->insert_images($this->upload->data());
$this->load->view('members/header');
$this->load->view('members/upload_form', $error);
$this->load->view('members/footer');
}
else
{
$data = array('upload_data' => $this->upload->data());
$this->load->view('members/header');
$this->load->view('members/upload_success', $data);
$this->load->view('members/footer');
}
}
我的模特功能:
function insert_images($data = array())
{
$data = array(
'filename' => $image_data['file_name'],
'fullpath' => $image_data['full_path']
);
$this->db->insert('bedrijfimages', $data);
}
可能是什么问题?
答案 0 :(得分:3)
您必须将此数据作为数组$this->upload->data()
在您的控制器中,您必须在验证完成时进行设置
else
{
$this->MODELNAME->insert_images($this->upload->data());
$data = array('upload_data' => $this->upload->data());
$this->load->view('members/header');
$this->load->view('members/upload_success', $data);
$this->load->view('members/footer');
}
在你的模特:
function insert_images($image_data = array()){
$data = array(
'filename' => $image_data['file_name'],
'fullpath' => $image_data['full_path']
);
$this->db->insert('bedrijfimages', $data);
}
您可以在CI文档页面查看此数组中的信息:Codeigniter upload library