Variable inside an if statement giving me an error even though if condition not met

[the_loserkid]

I’m trying to upload an image to the database. I’m using getimagesize to check if it is an image. It is stored in a variable called $image_size. Even though this variable is inside the if statement, I keep getting an error that says:

Undefined variable: image_size in C:\xampp1\htdocs\admin.php on line 10

I believe the page is not supposed to worry about that variable unless we have $_POST [‘image’] being sent to the server. I do not know how I can fix this.

$status = '';

if (isset($_POST['image'])) {
    $image = file_get_contents ($_FILES ['image']['tmp_name']);
    $image_size = getimagesize ($_FILES ['image']['tmp_name']);
}
        
if ($image_size == 0) {
    $status = "That is not an image. Please select an image.";
} else {
    $sql = "UPDATE company_info SET 
    co_image = :co_image
    WHERE id = 1";
    $stmt = $pdo->prepare($sql);
    $stmt->bindParam(':co_image', $image);
    $stmt->execute();
    $status = "Profile updated successfully!";
}

[fretburner]

Hi the_loserkid,

The problem is that if no image is submitted, the variable $image_size is never created, so your second IF statement results in the ‘undefined variable’ error.

One way to fix this would be to change the second IF statement to this:

if ( empty($image_size) ) {

The empty() function will return true if the variable is not set, or has a zero value.

[WebMachine]

Why don’t you nest the second conditional inside the first one, so it doesn’t kick in unless there is a $image and $image_size set?

[the_loserkid]

Yes, I realised it was a silly error. I meant to nest it inside the first one and thought I had done that but was still getting the error. Don’t know how I managed that but thanks to you both for pointing it out.