my variable will get empty value in my array

Question

I want to upload file and store name of file into my database but here in this code my file is stored at my desired location but in database i am not getting value probably my $filename is not getting inserted into my data_insert array. input->post('pic_file');

//$pic_file1 = str_replace( "\\", '/', $pic_file1);
// $filename = time().basename($pic_file1);
//$filename = "";

$config['upload_path']   = './uploads/';
$config['allowed_types'] = 'gif|jpg|png';
$config['max_size']      = 1000;
$config['encrypt_name']  = true;
// $config['overwrite'] = FALSE;
//$config['file_name'] =  $filename;

$this->load->library('upload', $config);
$this->upload->initialize($config);

if (!$this->upload->do_upload('pic_file')) {
    $error = array('error' => $this->upload->display_errors());
    print_r($error);
} else {
    $data     = array('upload_data' => $this->upload->data());
    $filename = $data['upload_data']['file_name'];

    // print_r($data);
    // $first_names = array_column($data, 'file_name');
    //print_r($first_names);
    //$file_name =  implode(" ",$first_names);

} //die();
echo $filename;

//die();
$data_insert = array(
    'setpassword' => $setpassword,
    'conpassword' => $conpassword,
    'city'        => $city,
    'products'    => $products,
    'bank_type'   => $bank_type,
    'bank_name'   => $bank_name,
    'dsa_code'    => $dsa_code,
    'pic_file'    => $filename,
);

$this->db->where('id', $dataId);
$this->db->update('tbl_reg_dsa', $data_insert);

do 1 thing. print_r($this->upload->data()); check you are getting file_name in array — aman
– aman, Commented May 16, 2019 at 11:48
i am getting filename in $filename variable but whenever i pass that $filename into my $data_insert array at that time i get blank value into my database — user3636036
– user3636036, Commented May 16, 2019 at 11:51
$dataId is not defined in your code snippet, so we don't know what it's value? — Vickel
– Vickel, Commented May 16, 2019 at 15:29
use it like that $upload_data = $this->upload->data(); $filename = $upload_data['file_name']; — aman
– aman, Commented May 21, 2019 at 8:06

Rahul · Accepted Answer · 2019-05-16 14:00:03Z

1

You can fetch that using

Method 1

$temp = $this->upload->data();
echo $temp['orig_name']; // or file_name

Method 2

$this->upload->data('pic_file'); // should return name of file

Here is more details about CI DOC.

answered May 16, 2019 at 14:00

Rahul

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