Question

我试图在我的控制器中测试一个方法.... 我可以通过浏览器提供输入，使用REQUEST类型来获取我的值.... 是否可以提供文件路径以及其他键值对... 我试图给出文件路径，但它没有存储... 继承我的代码......

$profilepic = $_FILES['logo']['name'];
$logintype=$_REQUEST['logintype'];

if(!empty($profilepic))
{

    $imagepath = pathinfo($profilepic);
    $pic=date('YmdHis').'.'.$imagepath['extension'];    
    $config['upload_path'] = './user_images/';
    $config['allowed_types'] = 'gif|jpg|jpeg|png';
    $config['max_size'] = '0';
    $config['max_width']  = '0';
    $config['max_height']  = '0';
    $config['file_name'] = $pic; 
    $this->load->library('upload', $config);

if ($this->upload->do_upload('logo'))
{
  $upload_data = $this->upload->data();
  $profilepicture=$upload_data['file_name'];
  $this->load->model('user_model');             
  $results = $this->user_model->NewSignUp($logintype,$profilepicture);
}           
}
else
{

  $message = array('message'=>"provide values","logintype"=>$logintype,"image"=>"empty image");
  echo json_encode($message);

我尝试将值传递为

url： baseurl / folder / controller / method？logintype = 0＆amp; logo = C：\ Users \ php-04 \ Pictures \ tulips.jpg

我总是得到回应：

{"message":"provide values","logintype":"0","image":"empty image"}

Answer 1

你的意思是？

$profilepicturePath = $upload_data['full_path'];

如何在codeigniter中将文件路径传递给我的控制器以获取fileupload

1 个答案: