PHP file upload count set file

Question

I want to count set file upload. Here is my using code. Are there any better method to do this. Thanks.

<form action="index.php" method="post" enctype="multipart/form-data">
    <input name="new_image[]"  type="file" />
    <input name="new_image[]" type="file" />
    <input name="new_image[]" type="file" />
    <input name="new_image[]" type="file" />
    <input name="new_image[]" type="file" />
<button name="submit" type="submit">Upload</button>
</form>
<?php

$img_error = '0';
$fill_img_count = '0';
if(isset($_POST['submit']))
{
    $img_count = count($_FILES['new_image']);
    echo "Total : ".$img_count."<br>";
    for ($i=0 ; $i<=$img_count ; $i++)
    {
        if (isset($_FILES['new_image']) && !empty($_FILES['new_image']['name'][$i]))
        {
            $fill_img_count++;
        }
    }
    echo "Set : ".$fill_img_count."<br>";
}
?>

Anders R. Bystrup · Accepted Answer · 2013-01-11 21:57:50Z

2

$count_files = 0;
foreach ($_FILES['picture']['error'] as $item) {
    if ($item != 4) {
        $count_files++;
    }
}
echo $count_files;

edited Jan 11, 2013 at 21:57

Anders R. Bystrup

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answered Jan 11, 2013 at 21:38

Mike

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Comments

Álvaro González · Accepted Answer · 2011-07-15 12:48:13Z

1

I'd recommend testing each ['error'] key against UPLOAD_ERR_OK.

answered Jul 15, 2011 at 12:48

Álvaro González

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Comments

Community · Accepted Answer · 2017-05-23 10:31:26Z

0

You don't require to have name="new_image[]" as the name ... just new_image will suffice. If you post 1 or many, on the PHP side, you'll see $_FILES[]

Some useful links for you:

Some code:

  if (empty($_FILES)) { echo "0 files uploaded"; } 
  else { echo count($_FILES) . " files uploaded"; }

Edit based on comment:

How to handle multiple file upload using PHP

From that post:

  echo count($_FILES['file']['tmp_name']);

edited May 23, 2017 at 10:31

CommunityBot

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answered Jul 15, 2011 at 13:02

sdolgy

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1 Comment

user188654 Over a year ago

This code will fail in case empty file fields are sent which is something OP is explicitly looking to avoid.

user188654 · Accepted Answer · 2011-07-15 13:33:34Z

0

<?php
$count = 0;
foreach($_FILES['new_image']['error'] as $status){
    if($status === UPLOAD_ERR_OK) {
        $count++;
    }
}
var_dump($count);
?>
<form action="test.php" method="post" enctype="multipart/form-data">
    <input name="new_image[]"  type="file" />
    <input name="new_image[]" type="file" />
    <input name="new_image[]" type="file" />
    <input name="new_image[]" type="file" />
    <input name="new_image[]" type="file" />
<button name="submit" type="submit">Upload</button>
</form>

answered Jul 15, 2011 at 13:33

user188654

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