wget return downloaded filename

Question

I'm using wget in a php script and need to get the name of the file downloaded.

For example, if I try

<?php
  system('/usr/bin/wget -q --directory-prefix="./downloads/" http://www.google.com/');
?>

I will get a file called index.html in the downloads directory.

EDIT: The page will not always be google though, the target may be an image or stylesheet, so I need to find out the name of the file that was downloaded.

I'd like to have something like this:

<?php
  //Does not work:
  $filename = system('/usr/bin/wget -q --directory-prefix="./downloads/" http://www.google.com/');
  //$filename should contain "index.html"
?>

Answer 1

Maybe that's some kind of cheating, but why not :

• decide yourself the name of the file that wget should create
• indicate to wget that the download should be made to that file
• when the download is finished, use that file -- as you already know the name.

Check out the -O option of wget ;-)

For example, running this from the command-line :

wget 'http://www.google.com/' -O my-output-file.html

Will create a file called my-output-file.html.

Answer 2

if your requirement is simple like just getting google.com, then do it within PHP

$data=file_get_contents('http://www.google.com/');
file_put_contents($data,"./downloads/output.html");

Answer 3

On Linux like systems you can do:

system('/usr/bin/wget -q --directory-prefix="./downloads/" http://www.google.com/');
$filename = system('ls -tr ./downloads'); // $filename is now index.html

This works if there is no other process creating file in the ./downloads directory.