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Editorial Team
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Asked: 2026-06-11T17:00:22+00:00

Here is the form form action=index.php method=POST enctype=multipart/form-data > <input type=file name=image[] multiple=multiple> <input

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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.";
        }

    ?>

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1 Answer

  1. Editorial Team

    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

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