$var = "http://site.com/image.png";
if (file_exists($var))
echo 'yes';
else
echo 'no';
Why does this script always returns false?
-
1Because you're looking at a URL, not a physical path on your computer (or web server)Phill– Phill2011-02-03 08:14:26 +00:00Commented Feb 3, 2011 at 8:14
Add a comment
|
4 Answers
Because that's not actually a file, rather it's a URL.
Up until PHP5, the file_exists function was intended to detect whether a file or directory exists.
At PHP5, support was introduced to allow support for some URL wrappers, as per the note on this page:
Tip: As of PHP 5.0.0, this function can also be used with some URL wrappers. Refer to Supported Protocols and Wrappers to determine which wrappers support stat() family of functionality.
The wrappers are detailed here but, unfortunately, the http one does not support stat, a pre-requisite to allowing file_exists to work.
If you're running on the server, you could possibly convert it using $_SERVER['DOCUMENT_ROOT'] or find another way to locate it on the file system.
Comments
If you want to check the presence of remote files (actually "resources") then use:
$i = get_headers($url);
$ok = strpos($i[0], "200");
And check the HTTP status code.
Comments
Then most probably your server does not allow outside connections from within PHP.
You can check this by opening a local file and see if that works. If it does, your code is correct.
Make sure allow_url_fopen is set to ON.
1 Comment
berkes
Never mind, I jumped to conclusions. @paxdiablo has the correct answer.
The $var variable should describe a path on the system, not a URL. For example
$var = "/var/www/html/image.png";
