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











Your Answer






StackExchange.ifUsing("editor", function () {
StackExchange.using("externalEditor", function () {
StackExchange.using("snippets", function () {
StackExchange.snippets.init();
});
});
}, "code-snippets");

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "1"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f54254401%2fcodeigniter-obtain-image-name-for-multiple-files%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























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


















draft saved

draft discarded




















































Thanks for contributing an answer to Stack Overflow!


  • Please be sure to answer the question. Provide details and share your research!

But avoid …



  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.


To learn more, see our tips on writing great answers.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f54254401%2fcodeigniter-obtain-image-name-for-multiple-files%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown





















































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown

































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown







-

Popular posts from this blog

Homophylophilia

Image
Haec pagina ad tempus protecta est ob maleficia perpetua. Quaesemus, emendationes suggere in pagina disputationis vel magistratum ad hanc deprotegendam pete. Homophylophilia Hyacinthus complexus ab amante suo Apolline, in pictura Mors Hyacinthi quam Jean Broc pinxit. Homophylophilia est sui sexus vel generis amor vel mores romantici vel sexuales, res venerea masculina in masculos, aut mulier in mulieres appetentes. Hoc ingenium sexuale est "exemplar permanens vel inclinatio ad experiendam sexualem, adfectionalem, vel romanticam vim attrahendi," praecipue vel solum ex hominibus eiusdem sexús; notio "etiam ad hominis iudicium identitatis privatae socialisque in illis attractionibus conditum refert, in moribus quoque qui illos exprimant, et in societate communitatis aliorum qui illos communicent." [1] [2] Per aevos alii aliter rem definiverunt, sed Occidentali nostrae aetatis cultu civili non tantum ad ipsa acta erotica spectat, sed etiam igitur
Read more

Updating UILabel text programmatically using a function

Image
0 so I am trying to update the UIlabel text using a func. I have seen few other questions like mine and I tried their method but it doesn't work. fun returns no value when used in the UIlabel. I have shown my code below.It has to be noted that when I use the function in viewdidload I can retrieve the data. lazy var name: UILabel = { let lb = UILabel() lb.text? = getthename() lb.lineBreakMode = NSLineBreakMode.byWordWrapping lb.textColor = .mainGreen lb.textAlignment = .left lb.font = UIFont.boldSystemFont(ofSize:12.0) return lb }() func getthename() -> String { var wins : String = "" let ud = (Auth.auth().currentUser?.uid) self.ref = Database.database().reference() self.ref.child("Users").child(ud!).observeSingleEvent(of: .value, with: {(snapshot) in
Read more

Cloud Functions - OpenCV Videocapture Read method fails for larger files from cloud storage

Image
0 I created a cloud function to read a video file from cloud storage and read frames and create an image for each frame. This function works for small files which is 1-2 MB, However, this method fails while reading larger file 130 MB. import cv2 from google.cloud import storage vidcap = cv2.VideoCapture("https://storage.googleapis.com/[bucker_name]/[filename.mp4]") if vidcap.isOpened(): vidcap.open("https://storage.googleapis.com/[bucker_name]/[filename.mp4]") success,image = vidcap.read() print(success) Expected the print will return true and an image will hold an image of a first frame. python-3.x function opencv cloud share | improve this question
Read more