How to insert filename in database using codeigniter upload lib

Question

It's ok to upload image in file using codeigniter upload library with this example.

However, I want to send filename in database. Can anybody pls give me example ? Thanks

<?php

class Upload extends Controller {

function Upload()
{
    parent::Controller();
    $this->load->helper(array('form', 'url'));
}

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

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);
    }
}   
}
?>

Or do we have any alternative?

Sarfraz · Accepted Answer · 2010-09-02 18:57:07Z

4

You can get it with data() method like this:

$upload_data = $this->upload->data();
echo $upload_data['file_name'];

Note that you can also get other upload data, just check what you have:

print_r($upload_data);

answered Sep 2, 2010 at 18:57

Sarfraz

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3 Comments

ktm Over a year ago

do i need to write model function ? like send_file_info_to_db() and write $this->db->insert('some_table') in controller ? any code help would be great. Thanks

ktm Over a year ago

I did $this->db->insert('upload_file', $data['upload_data']); but it's sending all the data arrays, I just want four of those fields going to database.How do i do that ? please help

Maartje Over a year ago

I know this is 4 years ago, but it really helped me out, Thank you!!

rabidmachine9 · Accepted Answer · 2010-09-02 21:28:14Z

2

Create an array with whatever you need to upload in your db...and then insert it in the db put something like this under else:

 //create array to load to database
                $insert_data = array(
                    'name' => $image_data['file_name'],
                    'path' => $image_data['full_path'],
                    'thumb_path'=> $image_data['file_path'] . 'thumbs/'. $image_data['file_name'],
                    'tag' => $tag
                     );

          $this->db->insert('photos', $insert_data);//load array to database

Important:the indexes of the array got to have same name with the columns of your table!

also try and use your model it will make your code a bit easier to maintain,read etc best of luck!

edited Sep 2, 2010 at 21:28

answered Sep 2, 2010 at 21:21

rabidmachine9

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2 Comments

ktm Over a year ago

Thank you very much, this is very close to what i want , howerver I am getting this error: Message: Undefined variable: image_data and Column 'file_name' cannot be null INSERT INTO upload_file (file_name, full_path, orig_name) VALUES (NULL, NULL, NULL) ......... my show create table is CREATE TABLE upload_file ( id int(11) NOT NULL, file_name varchar(80) NOT NULL, full_path varchar(80) NOT NULL, orig_name varchar(80) NOT NULL ) ENGINE=MyISAM DEFAULT CHARSET=latin1

rabidmachine9 Over a year ago

try and put this line of code before: $image_data = $this->upload->data();

Ahmed Almahlasi · Accepted Answer · 2013-03-05 07:19:27Z

0

you can use this way :

$data = array('upload' => $this->upload->data());
echo $data['upload']['file_name'];

answered Mar 5, 2013 at 7:19

Ahmed Almahlasi

1

Comments

Er Bhaskar Dutt Sharma · Accepted Answer · 2018-02-26 12:55:46Z

0

$config['upload_path'] = './assets/uploads/';  // Upload Path will be here
        $config['allowed_types'] = 'gif|jpg|png';
        $this->upload->initialize($config);

        if ($this->upload->do_upload('upload')) {
           $file_data = $this->upload->data();
           $file_name= $file_data['file_name']; //get file name of your uploaded file from here..

         }
         else { 

          echo "Error";
         } 

        $data = array(

        'image_name' =>  $file_data['file_name'], //insert your image name from here..
        );

        $this->add_model->image_upload($data); //that will go to the model function..
        }

answered Feb 26, 2018 at 12:55

Er Bhaskar Dutt Sharma

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