The second argument to copy() function cannot be a directory - php

Question

Anyone know why this:
<?PHP
$title = trim($_POST['title']);
$description = trim($_POST['description']);
// Array of allowed image file formats
$allowedExtensions = array('jpeg', 'jpg', 'jfif', 'png', 'gif', 'bmp');
foreach ($_FILES as $file) {
if ($file['tmp_name'] > '') {
if (!in_array(end(explode(".",
strtolower($file['name']))),
$allowedExtensions)) {
echo '<div class="error">Invalid file type.</div>';
}
}
}
if (strlen($title) < 3)
echo '<div class="error">Too short title</div>';
else if (strlen($description) > 70)
echo '<div class="error">Too long desccription.</div>';
else {
move_uploaded_file($_FILES['userfile']['tmp_name'], 'c:\wamp\www\uploads\images/');
}
Gives:
Warning: move_uploaded_file() [function.move-uploaded-file]: The second argument to copy() function cannot be a directory in C:\wamp\www\upload.php on line 41
Warning: move_uploaded_file() [function.move-uploaded-file]: Unable to move 'C:\wamp\tmp\php1AB.tmp' to 'c:\wamp\www\uploads\images/' in C:\wamp\www\upload.php on line 41

Answer 1

It's because you're moving a file and it thinks you're trying to rename that file to the second parameter (in this case a directory).
it should be:
move_uploaded_file($_FILES['userfile']['tmp_name'], 'c:/wamp/www/uploads/images/'.$file['name']);

Answer 2

You are specifying to move a file to a directory; neither PHP's move_uploaded_file nor its copy is as smart as a shell's copy -- you have to specify a filename, not a directory, for the destination.
So, one simple solution would be to take the basename of the source file and append that to the destination directory.

Answer 3

It sounds like the second argument to move_uploaded_file should be a full file name instead of just the directory name. Also, probably only a style issue, but you should use consistent slashes in 'c:\wamp\www\uploads\images/'

Answer 4

Because PHP is not a shell. You're trying to copy the file into the c:\wamp\www\uploads\images directory, but PHP doesn't know you mean that when you execute (within the move_uploaded_file function):
copy($_FILES['userfile']['tmp_name'], 'c:\wamp\www\uploads\images/');
This command tells it to rename the file to c:\wamp\www\uploads\images/, which it can't do because that's the name of an existing directory.
Instead, do this:
move_uploaded_file($_FILES['userfile']['tmp_name'],
'c:\wamp\www\uploads\images/' . basename($_FILES['userfile']['tmp_name']));

Answer 5

if you want just copy the file in two Dir different, try this :
if (move_uploaded_file($_FILES['photo']['tmp_name'], $target.$pic1))
{
copy("C:/Program Files (x86)/EasyPHP-5.3.9/www/.../images/".$pic1, "C:/Program Files (x86)/EasyPHP-5.3.9/www/.../images/thumbs/".$pic1)
}
You should write the complete path "C:/..."

Answer 6

Try adding the extension to the name file.
$filename = $_FILES['userfile']['tmp_name'].".jpg";

Answer 7

It will be like below in phalcon 3.42
if ($this->request->hasFiles() == true) {
// Print the real file names and sizes
foreach ($this->request->getUploadedFiles() as $file) {
//Print file details
echo $file->getName(), " ", $file->getSize(), "\n";
//Move the file into the application
$file->moveTo($this->config->application->uploadsDir.$file->getName());
}
}

Answer 8

FIRSTLY FIND YOUR CURRENT DIRECTORY echo getcwd(); IF YOU FIND THE CURRENT DIRECTORY SO MOVE THEM ANOTHER DIRECTORY
FOR EXAMPLE = echo getcwd(); these output is "your live directory name" like that -> /opt/lampp/htdocs/product and create new directory anywhere but your current directory help to go another directory ***
plz read carefully very thinking answer