Codeigniter obtain image name for multiple files












0















I have some code that i am using to upload multiple images i a form. I am uploading and renaming the images like this



//Obtain Picture



        $config['upload_path']          = './uploads/';

        $config['allowed_types']        = 'gif|jpg|png|pdf';

        $config['max_size']             = 5000;

        $config['max_width']            = 2000;

        $config['max_height']           = 2000;

        $rand_name = 'vehicles'.rand(321,8999).'_'.time();

        $config['file_name'] = $rand_name;

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

        $this->upload->do_upload('logbook');

        $this->upload->do_upload('inrep');

        $this->upload->do_upload('insurance');

        $this->upload->do_upload('vp');

        $data = array('upload_data' => $this->upload->data());


the form names for the multiple uploads which are unrelated are as follows logbook,inrep,insurance and vp



The code above uploads the files and renames them but i would like to obtain the new names of the files given the form name of the fields.



Right now, the files are uploaded and renamed but i cant they are from which field name.






php codeigniter







share|improve this question























  • Why you are not using loop to upload multiple images?

    – Danish Ali
    Jan 18 at 12:55











  • forms fields are unrelated.

    – Gandalf
    Jan 18 at 12:58
















0















I have some code that i am using to upload multiple images i a form. I am uploading and renaming the images like this



//Obtain Picture



        $config['upload_path']          = './uploads/';

        $config['allowed_types']        = 'gif|jpg|png|pdf';

        $config['max_size']             = 5000;

        $config['max_width']            = 2000;

        $config['max_height']           = 2000;

        $rand_name = 'vehicles'.rand(321,8999).'_'.time();

        $config['file_name'] = $rand_name;

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

        $this->upload->do_upload('logbook');

        $this->upload->do_upload('inrep');

        $this->upload->do_upload('insurance');

        $this->upload->do_upload('vp');

        $data = array('upload_data' => $this->upload->data());


the form names for the multiple uploads which are unrelated are as follows logbook,inrep,insurance and vp



The code above uploads the files and renames them but i would like to obtain the new names of the files given the form name of the fields.



Right now, the files are uploaded and renamed but i cant they are from which field name.






php codeigniter







share|improve this question























  • Why you are not using loop to upload multiple images?

    – Danish Ali
    Jan 18 at 12:55











  • forms fields are unrelated.

    – Gandalf
    Jan 18 at 12:58














0












0








0








I have some code that i am using to upload multiple images i a form. I am uploading and renaming the images like this



//Obtain Picture



        $config['upload_path']          = './uploads/';

        $config['allowed_types']        = 'gif|jpg|png|pdf';

        $config['max_size']             = 5000;

        $config['max_width']            = 2000;

        $config['max_height']           = 2000;

        $rand_name = 'vehicles'.rand(321,8999).'_'.time();

        $config['file_name'] = $rand_name;

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

        $this->upload->do_upload('logbook');

        $this->upload->do_upload('inrep');

        $this->upload->do_upload('insurance');

        $this->upload->do_upload('vp');

        $data = array('upload_data' => $this->upload->data());


the form names for the multiple uploads which are unrelated are as follows logbook,inrep,insurance and vp



The code above uploads the files and renames them but i would like to obtain the new names of the files given the form name of the fields.



Right now, the files are uploaded and renamed but i cant they are from which field name.






php codeigniter







share|improve this question














I have some code that i am using to upload multiple images i a form. I am uploading and renaming the images like this



//Obtain Picture



        $config['upload_path']          = './uploads/';

        $config['allowed_types']        = 'gif|jpg|png|pdf';

        $config['max_size']             = 5000;

        $config['max_width']            = 2000;

        $config['max_height']           = 2000;

        $rand_name = 'vehicles'.rand(321,8999).'_'.time();

        $config['file_name'] = $rand_name;

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

        $this->upload->do_upload('logbook');

        $this->upload->do_upload('inrep');

        $this->upload->do_upload('insurance');

        $this->upload->do_upload('vp');

        $data = array('upload_data' => $this->upload->data());


the form names for the multiple uploads which are unrelated are as follows logbook,inrep,insurance and vp



The code above uploads the files and renames them but i would like to obtain the new names of the files given the form name of the fields.



Right now, the files are uploaded and renamed but i cant they are from which field name.






php codeigniter




php codeigniter






share|improve this question













share|improve this question











share|improve this question




share|improve this question










asked Jan 18 at 12:51









GandalfGandalf

5,4892269121




5,4892269121













  • Why you are not using loop to upload multiple images?

    – Danish Ali
    Jan 18 at 12:55











  • forms fields are unrelated.

    – Gandalf
    Jan 18 at 12:58



















  • Why you are not using loop to upload multiple images?

    – Danish Ali
    Jan 18 at 12:55











  • forms fields are unrelated.

    – Gandalf
    Jan 18 at 12:58

















Why you are not using loop to upload multiple images?

– Danish Ali
Jan 18 at 12:55





Why you are not using loop to upload multiple images?

– Danish Ali
Jan 18 at 12:55













forms fields are unrelated.

– Gandalf
Jan 18 at 12:58





forms fields are unrelated.

– Gandalf
Jan 18 at 12:58












1 Answer
1






active

oldest

votes


















1














You can get new file name using 



$this->upload->data();


This will return details of uploaded files.
So you can do it like



$config['upload_path']          = './uploads/';

$config['allowed_types']        = 'gif|jpg|png|pdf';

$config['max_size']             = 5000;

$config['max_width']            = 2000;

$config['max_height']           = 2000;

$rand_name = 'vehicles'.rand(321,8999).'_'.time();

$config['file_name'] = $rand_name;

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



$this->upload->do_upload('logbook');

$logbook = $this->upload->data();

$logbook_new_filename = $logbook['file_name'];



$this->upload->do_upload('inrep');

$inrep = $this->upload->data();

$inrep_new_filename = $inrep['file_name'];



$this->upload->do_upload('insurance');

$insurance= $this->upload->data();

$insurance_new_filename = $insurance['file_name'];



$this->upload->do_upload('vp');

$vp= $this->upload->data();

$vp_new_filename = $vp['file_name'];


I haven't tested it. But I use like this to get uploaded filename.Hope it will help.






share|improve this answer
























  • That worked but may i ask $logbook_new_filename = $logbook['file_name']; what this does

    – Gandalf
    Jan 18 at 13:42











  • " $logbook_new_filename " will be name of your new file.

    – Cppi Khatri
    Jan 18 at 13:43






  • 1





    And your final array will be $upload_data = array('logbook'=>$logbook_new_filename,'inrep'=>$inrep_new_filename,'insurance'=>$insurance_new_filename,'vp'=>$vp_new_filename); Now you can use this array in your insert query.

    – Cppi Khatri
    Jan 18 at 13:50











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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









1














You can get new file name using 



$this->upload->data();


This will return details of uploaded files.
So you can do it like



$config['upload_path']          = './uploads/';

$config['allowed_types']        = 'gif|jpg|png|pdf';

$config['max_size']             = 5000;

$config['max_width']            = 2000;

$config['max_height']           = 2000;

$rand_name = 'vehicles'.rand(321,8999).'_'.time();

$config['file_name'] = $rand_name;

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



$this->upload->do_upload('logbook');

$logbook = $this->upload->data();

$logbook_new_filename = $logbook['file_name'];



$this->upload->do_upload('inrep');

$inrep = $this->upload->data();

$inrep_new_filename = $inrep['file_name'];



$this->upload->do_upload('insurance');

$insurance= $this->upload->data();

$insurance_new_filename = $insurance['file_name'];



$this->upload->do_upload('vp');

$vp= $this->upload->data();

$vp_new_filename = $vp['file_name'];


I haven't tested it. But I use like this to get uploaded filename.Hope it will help.






share|improve this answer
























  • That worked but may i ask $logbook_new_filename = $logbook['file_name']; what this does

    – Gandalf
    Jan 18 at 13:42











  • " $logbook_new_filename " will be name of your new file.

    – Cppi Khatri
    Jan 18 at 13:43






  • 1





    And your final array will be $upload_data = array('logbook'=>$logbook_new_filename,'inrep'=>$inrep_new_filename,'insurance'=>$insurance_new_filename,'vp'=>$vp_new_filename); Now you can use this array in your insert query.

    – Cppi Khatri
    Jan 18 at 13:50
















1














You can get new file name using 



$this->upload->data();


This will return details of uploaded files.
So you can do it like



$config['upload_path']          = './uploads/';

$config['allowed_types']        = 'gif|jpg|png|pdf';

$config['max_size']             = 5000;

$config['max_width']            = 2000;

$config['max_height']           = 2000;

$rand_name = 'vehicles'.rand(321,8999).'_'.time();

$config['file_name'] = $rand_name;

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



$this->upload->do_upload('logbook');

$logbook = $this->upload->data();

$logbook_new_filename = $logbook['file_name'];



$this->upload->do_upload('inrep');

$inrep = $this->upload->data();

$inrep_new_filename = $inrep['file_name'];



$this->upload->do_upload('insurance');

$insurance= $this->upload->data();

$insurance_new_filename = $insurance['file_name'];



$this->upload->do_upload('vp');

$vp= $this->upload->data();

$vp_new_filename = $vp['file_name'];


I haven't tested it. But I use like this to get uploaded filename.Hope it will help.






share|improve this answer
























  • That worked but may i ask $logbook_new_filename = $logbook['file_name']; what this does

    – Gandalf
    Jan 18 at 13:42











  • " $logbook_new_filename " will be name of your new file.

    – Cppi Khatri
    Jan 18 at 13:43






  • 1





    And your final array will be $upload_data = array('logbook'=>$logbook_new_filename,'inrep'=>$inrep_new_filename,'insurance'=>$insurance_new_filename,'vp'=>$vp_new_filename); Now you can use this array in your insert query.

    – Cppi Khatri
    Jan 18 at 13:50














1












1








1







You can get new file name using 



$this->upload->data();


This will return details of uploaded files.
So you can do it like



$config['upload_path']          = './uploads/';

$config['allowed_types']        = 'gif|jpg|png|pdf';

$config['max_size']             = 5000;

$config['max_width']            = 2000;

$config['max_height']           = 2000;

$rand_name = 'vehicles'.rand(321,8999).'_'.time();

$config['file_name'] = $rand_name;

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



$this->upload->do_upload('logbook');

$logbook = $this->upload->data();

$logbook_new_filename = $logbook['file_name'];



$this->upload->do_upload('inrep');

$inrep = $this->upload->data();

$inrep_new_filename = $inrep['file_name'];



$this->upload->do_upload('insurance');

$insurance= $this->upload->data();

$insurance_new_filename = $insurance['file_name'];



$this->upload->do_upload('vp');

$vp= $this->upload->data();

$vp_new_filename = $vp['file_name'];


I haven't tested it. But I use like this to get uploaded filename.Hope it will help.






share|improve this answer













You can get new file name using 



$this->upload->data();


This will return details of uploaded files.
So you can do it like



$config['upload_path']          = './uploads/';

$config['allowed_types']        = 'gif|jpg|png|pdf';

$config['max_size']             = 5000;

$config['max_width']            = 2000;

$config['max_height']           = 2000;

$rand_name = 'vehicles'.rand(321,8999).'_'.time();

$config['file_name'] = $rand_name;

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



$this->upload->do_upload('logbook');

$logbook = $this->upload->data();

$logbook_new_filename = $logbook['file_name'];



$this->upload->do_upload('inrep');

$inrep = $this->upload->data();

$inrep_new_filename = $inrep['file_name'];



$this->upload->do_upload('insurance');

$insurance= $this->upload->data();

$insurance_new_filename = $insurance['file_name'];



$this->upload->do_upload('vp');

$vp= $this->upload->data();

$vp_new_filename = $vp['file_name'];


I haven't tested it. But I use like this to get uploaded filename.Hope it will help.







share|improve this answer












share|improve this answer



share|improve this answer










answered Jan 18 at 13:10









Cppi KhatriCppi Khatri

526




526













  • That worked but may i ask $logbook_new_filename = $logbook['file_name']; what this does

    – Gandalf
    Jan 18 at 13:42











  • " $logbook_new_filename " will be name of your new file.

    – Cppi Khatri
    Jan 18 at 13:43






  • 1





    And your final array will be $upload_data = array('logbook'=>$logbook_new_filename,'inrep'=>$inrep_new_filename,'insurance'=>$insurance_new_filename,'vp'=>$vp_new_filename); Now you can use this array in your insert query.

    – Cppi Khatri
    Jan 18 at 13:50



















  • That worked but may i ask $logbook_new_filename = $logbook['file_name']; what this does

    – Gandalf
    Jan 18 at 13:42











  • " $logbook_new_filename " will be name of your new file.

    – Cppi Khatri
    Jan 18 at 13:43






  • 1





    And your final array will be $upload_data = array('logbook'=>$logbook_new_filename,'inrep'=>$inrep_new_filename,'insurance'=>$insurance_new_filename,'vp'=>$vp_new_filename); Now you can use this array in your insert query.

    – Cppi Khatri
    Jan 18 at 13:50

















That worked but may i ask $logbook_new_filename = $logbook['file_name']; what this does

– Gandalf
Jan 18 at 13:42





That worked but may i ask $logbook_new_filename = $logbook['file_name']; what this does

– Gandalf
Jan 18 at 13:42













" $logbook_new_filename " will be name of your new file.

– Cppi Khatri
Jan 18 at 13:43





" $logbook_new_filename " will be name of your new file.

– Cppi Khatri
Jan 18 at 13:43




1




1





And your final array will be $upload_data = array('logbook'=>$logbook_new_filename,'inrep'=>$inrep_new_filename,'insurance'=>$insurance_new_filename,'vp'=>$vp_new_filename); Now you can use this array in your insert query.

– Cppi Khatri
Jan 18 at 13:50





And your final array will be $upload_data = array('logbook'=>$logbook_new_filename,'inrep'=>$inrep_new_filename,'insurance'=>$insurance_new_filename,'vp'=>$vp_new_filename); Now you can use this array in your insert query.

– Cppi Khatri
Jan 18 at 13:50


















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