CODEWITHSUNDEEP
php
Toni
Toni

Reputation: 41

$_FILES and form data

Trying to read if user had uploaded file or not.

HTML:

<input type="file" name="img" id="img">

PHP:

if(isset($_FILES['img'])) {
    $upload = 1;
    echo "upload 1";
}
else{
    $upload = 0;
    echo "upload 0";
}

I want to see if user had selected file to be uploaded or not. In my code program goes every time to $upload = 1; even there is no file included.

Upvotes: 0

Views: 54

Answers (3)

&#193;lvaro Gonz&#225;lez
&#193;lvaro Gonz&#225;lez

Reputation: 146350

As per the docs:

if (isset($_FILES['img']) && $_FILES['img']['error'] === UPLOAD_ERR_OK) {
    $upload = true;
    echo "upload true";
} else {
    $upload = false;
    echo "upload false";
}

My example also uses actual booleans. In fact you can also do:

$upload = isset($_FILES['img']) && $_FILES['img']['error'] === UPLOAD_ERR_OK;

Upvotes: 0

pstryk
pstryk

Reputation: 2111

Don't check only isset($_FILES['img']), check what's the value of $_FILES['img']['error']. If it's 0, there was no error, and the file should be there - but this should also be verified with is_uploaded_file($_FILES['img']['tmp_name']). You can check list of potential upload errors

More details about file upload handling

Upvotes: 0

Pupil
Pupil

Reputation: 23948

You can use file_exists and is_uploaded_file functions of PHP to check if file is uploaded.

if (!file_exists($_FILES['img']['tmp_name']) || !is_uploaded_file($_FILES['img']['tmp_name'])) {
  $upload = 1;
  echo "upload 1";
} else {
  $upload = 0;
  echo "upload 0";
}

Upvotes: 1

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