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I'm triying to fetch the name of a file uploaded and set it as the value of a hidden input on the same form. I've tried this and it uploads the file but dont send the file name.

   if(isset($_FILES['image'])){
  $errors= array();
  $file_name = $_FILES['image']['name'];
  $file_size = $_FILES['image']['size'];
  $file_tmp = $_FILES['image']['tmp_name'];
  $file_type = $_FILES['image']['type'];
  $file_ext=strtolower(end(explode('.',$_FILES['image']['name'])));

  $expensions= array("jpeg","jpg","png");

  if(in_array($file_ext,$expensions)=== false){
     $errors[]="extension not allowed, please choose a JPEG or PNG file.";
  }

  if($file_size > 2097152) {
     $errors[]='File size must be excately 2 MB';
  }

  if(empty($errors)==true) {
     move_uploaded_file($file_tmp,"../img/plates/".$file_name);
     header("Location: backend.php");
  }else{
     print_r($errors);
  }

}

After that i try to inset the $file_name in to the same form as the file input. Something like this

<input name="image" type="file" placeholder="Select photo">
<input type="hidden" value="<?php echo $file_name ?>" name="photo" />
<button type="submit" name="add">ADD NEW ITEM</button>
Daniel Acevedo
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    You're redirecting the user after a successful upload, which would make `$file_name` empty after that. Can you please explain in more details what you're trying to do here? What problem are you trying to solve etc? – M. Eriksson Sep 15 '18 at 05:04
  • @MagnusEriksson I see what you're saying now. Let me read about it and how to solve it. This form insert the data into a json file, i want to save the name of the uploaded file to a field named "photo". All the rest of the script works fine. Thanks for your comment and sorry for the noob question. – Daniel Acevedo Sep 15 '18 at 05:09
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    No worries. We've all been there and asking questions is the right way of learning (and it's what this site is for :-)). Why would you need that, though? You already have the name in the `$_FILES` array? – M. Eriksson Sep 15 '18 at 05:14
  • @MagnusEriksson something like $_SESSION['NameOfFile'] = $_POST[$file_name]; /// and then inserting it like =$_SESSION['NameOfFile']?> ? Is this solution better than using javascript as Danial suggest? – Daniel Acevedo Sep 15 '18 at 05:20
  • I don't think I understand what you're actually trying to do here. Why would you need to save the file name in a separate input field when you already get the name in the `$_FILES`-array when you post the form? My suggestion about sessions was to repopulate the form _after_ the post. The JS solution would populate the name _before_ the post. Totally different things. But since you haven't explain why or when you need this, I can't say which is better. Two different use cases. – M. Eriksson Sep 15 '18 at 05:26
  • Im Sorry @MagnusEriksson i think is a language barrier. Even so.. I use your suggestion and got it to work! I insert the variable in to a session and then use it to populate the field. So, thanks again! – Daniel Acevedo Sep 15 '18 at 05:39

1 Answers1

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You cant do it just by php, because php update your page after request and response if you like to get the file name and file in a request you can use java script like this :

$("#imagefile").change(function(){
  $("#imagename").val(this.files[0].name);
});

and change your code like this :

<input id="imagefile" name="image" type="file" placeholder="Select photo">
<input id="imagename" type="hidden" value="<?php echo $file_name ?>" name="photo" />
<button type="submit" name="add">ADD NEW ITEM</button>

you can read more at this link :

How to display file name for custom styled input file using jquery?

Danial Jalalnouri
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