rename() help

[RobertSubnet]

Hello all. Hopefully this is an easy question.

 

I am having a problem with rename().

 

I have successfully uploaded a jpg to an upload directory on my web server-IIS. After the upload, I want to change the name of the uploaded file. The name change goes like this:

 

timestamp + p1 + uploaded_filename.jpg

 

I am trying to use rename() to rename the jpg file allready on the web server.

 

PHP keeps putting out the following error:

 

PHP Warning: rename() expects parameter 3 to be resource, string given in C:\Inetpub\wwwroot\kaydoo\upload.php on line 38

 

What sort of "resource" is PHP looking for?

 

I am using PHP v.5.2.3

 

Any help would be much appreciated.

 

Thanks,

~Robert

 

The code and output is below:

 

<?php

//beta kaydoo picture uploader

 

$p1 = 'p1';

$name = $_FILES['uploaded']['name'];

$time = time();

$newname = $time.$p1.$name;

 

//These two lines set the upload directory location, don't delete either line

$target = "upload/";

$target = $target.basename($_FILES['uploaded']['name']);

 

 

if(move_uploaded_file($_FILES['uploaded']['tmp_name'], $target)) {

  echo "The file ".basename($_FILES['uploaded']['name']). " has been uploaded to the Kaydoo server";

 

}else{

 

  echo "File upload, she no work man";

}

 

echo "<br>";

echo "<br>";

echo "The original name of the uploaded file is $name";

echo "<br>";

echo "<br>";

echo "The renamed file is $newname";

echo "<br>";

echo "<br>";

 

$directory = "upload/";

rename($name, $newname, $directory);

 

?>

-----------------

Script output:

 

The file picacho.jpg has been uploaded to the Kaydoo server

 

The original name of the uploaded file is picacho.jpg

 

The renamed file is 1192259718p1picacho.jpg

 

PHP Warning: rename() expects parameter 3 to be resource, string given in C:\Inetpub\wwwroot\kaydoo\upload.php on line 38

 

 

 

 

[derwert]

The 3rd parameter is to be used for context options when using streams; it's not for a directory.

[JJohnsenDK]

you need to point out where the file is located in the first parameter:

 

$name = "images\$myfile";

 

dont use the 3. parameter.