Uploading multiple files using html5 and php

Question

I have a file upload form set up with the HTML5 multiple attribute.

However, the form still only uploads a single file. Do i need to create some sort of a looping function in the php or is there another way of doing this?

Here's my code...

form:

<form action="<?php $_SERVER['PHP_SELF'] ?>" method="post" enctype="multipart/form-data">        
       <input type="file" multiple="multiple" name="file[]" id="file" />
       <input name="submit" type="submit" value="Submit" />    
</form>

php:

<?php
if(isset($_POST['submit'])) {           
foreach($_FILES['newsImage'] as $file){ 
if ((($_FILES["file"]["type"] == "image/jpeg")
|| ($_FILES["file"]["type"] == "image/pjpeg")))
  {
  if ($_FILES["file"]["error"] > 0)
    {
    echo "Return Code: " . $_FILES["file"]["error"] . "<br />";
    }
  else
    {


    if (file_exists("upload/" . $_FILES["file"]["name"]))
      {
      echo $_FILES["file"]["name"] . " already exists. ";
      }
    else
      {
      move_uploaded_file($_FILES["file"]["tmp_name"],
      "upload/" . $_FILES["file"]["name"]);
      echo "Stored in: " . "upload/" . $_FILES["file"]["name"];
      }
    }
  }
else
  {
  echo "Invalid file";
  }
}
}
?>

post your form code also. it should contain a multi-part specification. — Kris
– Kris, Commented Jan 4, 2012 at 11:02
Your code seems to be processing only one file - are you sure the problem is on the upload side of things? — Pekka
– Pekka, Commented Jan 4, 2012 at 11:02
It is worth noting that IE (and Opera I think) does not support choosing multiple files through the <input type="file" /> element. — Christofer Eliasson
– Christofer Eliasson, Commented Jan 4, 2012 at 11:09

JJJ · Accepted Answer · 2012-01-04 11:05:14Z

3

I believe your field should be <input type="file" multiple="multiple" name="files[]" />

And then in PHP:

<?php
   foreach($_FILES['files'] as $file){
       // Handle one of the uploads
   }
?>

edited Jan 4, 2012 at 11:05

JJJ

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answered Jan 4, 2012 at 11:03

Shai Mishali

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2 Comments

Tom Over a year ago

I've amended my code based on your reply, it doesn't seem to upload anything now though... any ideas... thanks

Shai Mishali Over a year ago

Inside your loop you are still calling $_FILES['file']['type'] etc... you should call $file['type'] (the specific file inside the loop over your files array).

Tom · Accepted Answer · 2012-01-18 17:29:30Z

3

for ($i = 0; $i < count($_FILES['newsImage']['name']); $i++) {
    // handle upload
}

answered Jan 18, 2012 at 17:29

Tom

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3 Comments

Shai Mishali Over a year ago

Why did you accept your own answer which was answered two weeks after mine (which is the same solution ... ?) Seems a bit weird I have to admit (:

Erik Over a year ago

The foreach won't work due to the structure of the file array when using a multiple file input. You need to define each file info part as $_FILES['inputName']['name'][0] for the first filename, $_FILES['inputName']['name'][1] for the second, etc.

Martavis P. Over a year ago

Although this is an old post, I've come across this issue as well. Indeed, uploading multiple files does make the array structure more complicated, thus using a regular for loop like this one really helps. I'm sure there is a way to use foreach, but this is simpler.

Neysor · Accepted Answer · 2012-03-24 11:01:48Z

2

I believe this code could serve the purpose. It loops through the $_FILES array and creates an array with key => value pair of all the attributes for for each file.

$temp = array();
foreach ($_FILES['file'] as $key => $value) {
    foreach($value as $index => $val){
        $temp[$index][$key] = $val;
    }
}

edited Mar 24, 2012 at 11:01

Neysor

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answered Mar 18, 2012 at 6:39

Shikhar Kapoor

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Comments

iBug · Accepted Answer · 2019-01-14 12:31:58Z

-1

<?php   

    include 'db.php';

      extract($_POST);

        extract($_POST);
        if(isset($submit))
        {

          $count = count($_FILES['image']['name']);

           for($i=0;$i<$count;$i++)
           {
              $fname = $_FILES['image']['name'][$i];
              $file_tmp = $_FILES['image']['tmp_name'][$i];
              $file_size =  $_FILES['image']['size'][$i];
               $file_type=$_FILES['image']['type'][$i];
               echo $file_size,$file_type;         
               $target = "img/".$fname;
               move_uploaded_file($file_tmp,$target);
                echo "uploaded succ !"."<br>";

           }

        }

?>

edited Jan 14, 2019 at 12:31

iBug

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answered Jan 14, 2019 at 12:31

user8518571

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