打开文件URL并在PHP中读取它？

Question

Given a URL that points to a file, how can I open it and read the contents in PHP?

$file = fopen("http://www.alinktomyfile.com/kasfafa", "r");

This line has an error. The docs say to use stream_get_contents() for a URL, however I get an error:

$file_contents = stream_get_contents($file);
fclose($file);

Error:

Warning: stream_get_contents() expects parameter 1 to be resource, null given in /Users/donald/Desktop/php_boilerplate.php on line 24

Answer 1

try file_get_contents()

$file = file_get_contents("http://www.alinktomyfile.com/kasfafa");