PHP file_get_contents() returns "failed to open stream: HTTP request failed!"


Question

I am having problems calling a url from PHP code. I need to call a service using a query string from my PHP code. If I type the url into a browser, it works ok, but if I use file-get-contents() to make the call, I get:

Warning: file-get-contents(http://.... ) failed to open stream: HTTP request failed! HTTP/1.1 202 Accepted in ...

The code I am using is:

$query=file_get_contents('http://###.##.##.##/mp/get?mpsrc=http://mybucket.s3.amazonaws.com/11111.mpg&mpaction=convert format=flv');
echo($query);

Like I said - call from the browser and it works fine. Any suggestions?

I have also tried with another url such as:

$query=file_get_contents('http://www.youtube.com/watch?v=XiFrfeJ8dKM');

This works fine... could it be that the url I need to call has a second http:// in it?

1
78
9/11/2016 9:29:25 AM

Accepted Answer

Try using cURL.

<?php

$curl_handle=curl_init();
curl_setopt($curl_handle, CURLOPT_URL,'http://###.##.##.##/mp/get?mpsrc=http://mybucket.s3.amazonaws.com/11111.mpg&mpaction=convert format=flv');
curl_setopt($curl_handle, CURLOPT_CONNECTTIMEOUT, 2);
curl_setopt($curl_handle, CURLOPT_RETURNTRANSFER, 1);
curl_setopt($curl_handle, CURLOPT_USERAGENT, 'Your application name');
$query = curl_exec($curl_handle);
curl_close($curl_handle);

?>
100
3/30/2009 3:19:59 PM


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