$_FILES array is not empty upon uploading nothing

Question

Here is the form

form action="index.php" method="POST" enctype="multipart/form-data" >
        <input type="file" name="image[]" multiple="multiple">
        <input type="submit" value="upload">
    </form>

I am trying to only run my code when only when if(!empty($_FILES['image'])){ but for some reason the array is not empty upon submitting no files and only clicking submit.

Here is the rest of the code if that will help, thanks.

<html>

Image Upload

    <form action="index.php" method="POST" enctype="multipart/form-data" >
        <input type="file" name="image[]" multiple="multiple">
        <input type="submit" value="upload">
    </form> 

    <?php

    include 'connect.php';

    if(!empty($_FILES['image'])){
    echo $_FILES['image']['error'];
        $allowed = array('jpg', 'gif', 'png', 'jpeg');
        $count = 0;

        foreach($_FILES['image']['name'] as $key => $name){
        $image_name = $name;
        $tmp = explode('.', $image_name);
        $image_extn = strtolower(end($tmp)); //can only reference file
        $image_temp = $_FILES['image']['tmp_name'][$count];
        $filesize = filesize($_FILES['image']['tmp_name'][$count]);
        $count = $count +1;

        if(count($_FILES['image']['tmp_name']) > 5){
            echo "You can upload a maximum of five files.";
            break;
        }

        else if(in_array($image_extn, $allowed) === false){
            echo $name." is not an allowed file type<p></p>";
        }

        else if($filesize > 1024*1024*0.3){
            echo $name." is too big, can be a maximum of 3MB";
        }

        else{

            $image_path = 'images/' . substr(md5($name), 0, 10) . '.' . $image_extn;


            move_uploaded_file($image_temp, $image_path);

            mysql_query("INSERT INTO store VALUES ('', '$image_name', '$image_path')") or die(mysql_error());

            $lastid = mysql_insert_id();
            $image_link = mysql_query("SELECT * FROM store WHERE id = $lastid");
            $image_link = mysql_fetch_assoc($image_link);
            $image_link = $image_link['image'];
            echo "Image uploaded.<p></p> Your image: <p></p><a href = $image_link>$image_path</a>";
            }

        }
        }

    else{
            echo "Please select an image.";
        }

    ?>

I don't think it will ever be empty because it holds all the data for the upload. For example, it will tell you that the file size is 0. It will be a string of some sort - use print_r($_FILES) to check the array and see what you want to use to check if an image is not uploaded. — Oliver Tappin
– Oliver Tappin, Commented Sep 21, 2012 at 22:01
What do you mean array is not empty ?? It should not be empty — Baba
– Baba, Commented Sep 21, 2012 at 22:03

Community · Accepted Answer · 2017-05-23 12:17:21Z

19

This is how $_FILES array looks like when nothing uploaded

Array ( [image] => Array ( [name] => [type] => [tmp_name] => [error] => 4 [size] => 0 ) )

So it's never empty.

The error code 4 [error] => 4 indicates no file was uploaded and error code 0 indicates no error and file was uploaded so you can check

if($_FILES['image']['error']==0) {
    // file uploaded, process code here
}

Here is another answer on SO.

edited May 23, 2017 at 12:17

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answered Sep 21, 2012 at 22:13

The Alpha

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1 Comment

hakre Over a year ago

You can get different errors, I also see UPLOAD_ERR_PARTIAL (3) next to UPLOAD_ERR_NO_FILE (4).

Oliver Tappin · Accepted Answer · 2012-09-21 22:03:30Z

8

Use the is_uploaded_file PHP function instead:

if(is_uploaded_file($_FILES['image']['tmp_name'])) {
  //code here
}

http://php.net/manual/en/function.is-uploaded-file.php

answered Sep 21, 2012 at 22:03

Oliver Tappin

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7 Comments

user1672267 Over a year ago

This worked when I put it under if(!empty($_FILES['image'])){ and used it as if(is_uploaded_file($_FILES['image']['tmp_name'][$count])) {.

hakre Over a year ago

Well, there are a thousand of other different things you could to as well, but this does not mean that it is actually the right way to do nor that you understand what you do. You need to check for the error code, see as well: php.net/manual/en/features.file-upload.errors.php - this become especially important with the $count you use here, because, well, you should better know with which kind of data you're working. See: php.net/manual/en/features.file-upload.post-method.php (uploading array of files)

user1672267 Over a year ago

This way worked fine. The only problem with I was having with the original code was when I submitted a blank form this code: ` else if(in_array($image_extn, $allowed) === false){ echo $name." is not an allowed file type<p></p>"; }` would still run.

hakre Over a year ago

Yes, sure. It works until it breaks. Question is if you're able to actually understand why something works and why not. I can only try to help you in the understanding. I added an answer that shows you what is actually going on, maybe it helps you to clarify things: stackoverflow.com/a/12538983/367456

user1672267 Over a year ago

I do understand what happened now. I was only checking if(!empty($_FILES['image'])){ to make sure the rest of the code involving the '$_FILES['image']' array was set otherwise an empty submitted form would return an array which would make this if(in_array($image_extn, $allowed) === false){ execute.

|

hakre · Accepted Answer · 2012-09-21 23:03:25Z

4

You should first of all take a look into the PHP manual because - you're not the first one with that problem - the solution has been written in there:

If no file is selected for upload in your form, PHP will return $_FILES['userfile']['size'] as 0, and $_FILES['userfile']['tmp_name'] as none.

So if you actually want to find out if any file for the image element has has been submitted, check for it:

 $noFile = $_FILES['image']['size'][0] === 0 
           && $_FILES['image']['tmp_name'][0] === '';

Yes, that simple it is.

The test you used:

empty($_FILE);

will only tell you if the whole form has been submitted or not. So an example in full:

$submitted = empty($_FILE);
if ($submitted) {
    $noFile    = $_FILES['image']['size'][0] === 0 
                 && $_FILES['image']['tmp_name'][0] === '';
    if ($noFile) {
        ...
    }
}

edited Sep 21, 2012 at 23:03

answered Sep 21, 2012 at 22:30

hakre

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2 Comments

Oliver Tappin Over a year ago

You save a few bytes by using if(is_uploaded_file($_FILES['image']['tmp_name'])) resulting in a faster page load.. lol.

hakre Over a year ago

Sidenote: There might not be an error but the filesize is 0 bytes. That's a replay of the upload form w/o actually uploading any file-data. Can happen with browsers, I've seen it with Chromium.

A.L · Accepted Answer · 2015-06-02 11:05:46Z

2

Check value is not null:

in_array(!null,$_FILES['field_name']['name'])

edited Jun 2, 2015 at 11:05

A.L

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answered Apr 24, 2015 at 12:12

Pavnish Yadav

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Comments

Adam Michalik · Accepted Answer · 2015-12-16 20:13:23Z

2

if($_FILES['image']['error'] === UPLOAD_ERR_OK) {
    // Success code Goes here .. 
}

UPLOAD_ERR_OK returns value 0 if there is no error, the file was uploaded successfully.

edited Dec 16, 2015 at 20:13

Adam Michalik

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answered Dec 16, 2015 at 19:37

rarenicks

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Comments

jimp · Accepted Answer · 2012-09-21 22:15:49Z

1

http://php.net/manual/en/features.file-upload.post-method.php

If no file is selected for upload in your form, PHP will return $_FILES['userfile']['size'] as 0, and $_FILES['userfile']['tmp_name'] as none.

You get an array entry per "file" upload field, even if the user didn't select a file to upload.

answered Sep 21, 2012 at 22:15

jimp

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Comments

Calin Rusu · Accepted Answer · 2014-10-16 09:35:27Z

1

I have confronted this issue with a multiple file input. What I found to be working for checking if any file has been selected is:

<?php
$size_sum = array_sum($_FILES['img']['size']);
if ($size_sum > 0) {
 // at least one file has been uploaded
} else {
 // no file has been uploaded
}
?>

answered Oct 16, 2014 at 9:35

Calin Rusu

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Comments

Eng Hany Atwa · Accepted Answer · 2013-06-28 23:20:03Z

0

if(!empty($_FILES['image']['name'][0]) || !empty($_FILES['image']['name'][1]) ||  
!empty($_FILES['image']['name'][2]))

or

for($1=0;$i<count($_FILES['image']);$i++){
if(!empty($_FILES['image']['name'][$i])){
   // do something
}
}

answered Jun 28, 2013 at 23:20

Eng Hany Atwa

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Comments

Badshah Sahib · Accepted Answer · 2015-04-21 16:49:53Z

-1

if(isset($_FILES['file'])){//do some thing here}

answered Apr 21, 2015 at 16:49

Badshah Sahib

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3 Comments

A.L Over a year ago

How is this code different from the other 6 answers? What is its goal? Please explain it in your answer.

Badshah Sahib Over a year ago

i just shared the simple code that can be easily understand. i shared for beginners

A.L Over a year ago

You should add an explanation: how this code should be used?

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