doujichan1399 2014-10-04 05:33
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打开文件URL并在PHP中读取它?

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

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  • dqmhgz5848 2014-10-04 05:41
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    try file_get_contents()

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