codeigniter upload image - php

Question

Hello all im working on a admin system that can create news with a image but i cant find out how to send the image name from my model file to my controller,
this is my model file:
function uploadImg()
{
$config = array(
'allowed_types' => 'jpg|jpeg|gif|png',
'upload_path' => $this->gallery_path,
'max_size' => 2000,
'encrypt_name' => true
);
$this->load->library('upload', $config);
$this->upload->do_upload();
$image_data = $this->upload->data();
$config = array(
'source_image' => $image_data['full_path'],
'new_image' => $this->gallery_path . '/thumbs',
'maintain_ration' => true,
'width' => 200,
'height' => 200,
'encrypt_name' => true,
'max_size' => 2000
);
$this->load->library('image_lib', $config);
$this->image_lib->resize();
# Ret profil billed navn #
$file_array = $this->upload->data('file_name');
return $billed_sti['billed_sti'] = $file_array['file_name'];
//$this->db->where('username', $this->input->post('username'));
//$this->db->update('users', $profilBilledNavn);
}
This is my controller:
function opret() {
$this->form_validation->set_rules('overskrift', 'overskrift', 'required');
$this->form_validation->set_rules('description', 'description', 'required');
$this->form_validation->set_rules('indhold', 'indhold', 'required');
if($this->form_validation->run() == true)
{
$this->load->model('admin/nyheder_model');
$billed_sti = $this->nyheder_model->uploadImg();
$data = array(
'overskrift' => $this->input->post('overskrift'),
'description' => $this->input->post('description'),
'indhold' => $this->input->post('indhold'),
'billed_sti' => $billed_sti,
'brugernavn' => $this->session->userdata('username'),
'godkendt' => 'ja'
);
$this->db->insert('nyheder', $data);
redirect('admin/nyheder/index');
} else {
$this->index();
}
}

Answer 1

I do the image processing in the controller rather than the model.
"Models are PHP classes that are designed to work with information in your database."
from: http://codeigniter.com/user_guide/general/models.html

Answer 2

What you need to do is move the code for uploading the image to the controler.
function do_upload()
{
$config['upload_path'] = './uploads/';
$config['allowed_types'] = 'gif|jpg|png';
$config['max_size'] = '100';
$config['max_width'] = '1024';
$config['max_height'] = '768';
$this->load->library('upload', $config);
if ( ! $this->upload->do_upload())
{
$error = array('error' => $this->upload->display_errors());
$this->load->view('upload_form', $error);
}
else
{
$data = array('upload_data' => $this->upload->data());
$this->load->view('upload_success', $data);
}
}
Once you did that,
You can insert the name of the file from the $data variable created in this line:
$data = array('upload_data' => $this->upload->data());
and you can get the value like this:
$data['file_name']
The file will upload the the folder you configured, and you will insert the filename to the DB From the controller.
I hope it helps.

Answer 3

Please use the upload function in your controller as the model classes are used to handle the database information. Please check the code below
//Controller Class
function upload_image()
{
//Check for the submit
// Submit Name refers to the name attribute on the submit input tag.
// $filename refers to the name attribute of the file input tag.
if($_SERVER['REQUEST_METHOD'] == "POST")
{
$submit = $this->input->post('submit');
if($submit == "Submit Name")
{
//Load the relevant classes and libraries
$this->load->library('upload');
$this->load->model('admin/nyheder_model','nmodel');
$filename = "image_file";
//Define the config array
$config = array();
$config['upload_path'] = $this->gallery_path;
$config['allowed_types'] = "jpg|gif|png";
$config['max_size'] = 0; //0 is for no limit
$this->upload->initalize($config);
if(!$this->upload->do_upload("$filename"))
{
echo $this->upload->display_errors();
}
else
{
$file_data = $this->upload->data();
$filename_1 = $file_data['file_name'];
$insert_array = array('filename'=>"$filename_1");
$this->nmodel->insert_data($insert_array);
} // end of the else statement
} // end of the isset statement
} // end of the outer conditional statement
Now you have the value of the filename in the $filename_1 variable which you can pass to the model class and can store the value in the database.
Thanks
J