php include to external url

Question

I am currently trying to use the php function 'include' to include an external url. This is so that whenever the webpage is updated it will automatically update mine. The trouble I'm having however is that I keep getting an error saying the following...

Warning: require() [function.require]: http:// wrapper is disabled in the server configuration by allow_url_include=0 in C:\wamp\www\starterpack\starterpack2\header.php on line 48

I have tried to find a way to fix this error or find a way around it but cannot find one. Does anyone have any ideas?

P.S I am building the site using wampserver, could permissions of the wampserver be causing this error?

diolemo · Accepted Answer · 2012-02-08 15:35:08Z

11

You would be better using echo file_get_contents($url) as the include statement could execute any PHP code returned by the other site.

answered Feb 8, 2012 at 15:35

diolemo

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Barry Chapman · Accepted Answer · 2012-02-08 15:26:23Z

8

Look at your php.ini and make sure allow_url_include is set to 1. Restart HTTPD, done.

answered Feb 8, 2012 at 15:26

Barry Chapman

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Prashanth · Accepted Answer · 2016-03-17 19:34:32Z

5

function getter($url) {
    $ch = curl_init();
    curl_setopt($ch, CURLOPT_URL, $url);
    curl_setopt($ch, CURLOPT_SSL_VERIFYPEER, false);
    curl_setopt($ch, CURLOPT_RETURNTRANSFER, true);
    curl_setopt($ch, CURLOPT_HEADER, 0);
    $data = curl_exec($ch);
    curl_close($ch);
    return $data;
}

echo getter('http://yourdomain.com/externalfile.php');

And you're done :)

edited Mar 17, 2016 at 19:34

Prashanth

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answered Feb 8, 2012 at 15:34

Nicholas Valbusa

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Wayne Nort · Accepted Answer · 2012-07-25 12:43:32Z

0

This will load an external website and also gives external links an absolute website link address

$ch = curl_init();
curl_setopt($ch, CURLOPT_URL, "http://www.your_external_website.com");
curl_setopt($ch, CURLOPT_HEADER, 0);
curl_setopt($ch, CURLOPT_RETURNTRANSFER, TRUE);
$result = curl_exec($ch);
curl_close($ch);
$result = preg_replace("#(<\s*a\s+[^>]*href\s*=\s*[\"'])(?!http)([^\"'>]+)([\"'>]+)#",'$1http://www.your_external_website.com/$2$3', $result);
echo $result

answered Jul 25, 2012 at 12:43

Wayne Nort

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1 Comment

Tom Over a year ago

Your answer is identical @squallstars.

Prashanth · Accepted Answer · 2016-03-17 18:44:49Z

0

Look at your php.ini and make sure allow_url_include is set to 1

Otherwise use following ...

function getter($url) 
{
$ch = curl_init();
curl_setopt($ch, CURLOPT_URL, $url);
curl_setopt($ch, CURLOPT_SSL_VERIFYPEER, false);
curl_setopt($ch, CURLOPT_RETURNTRANSFER, true);
curl_setopt($ch, CURLOPT_HEADER, 0);
$data = curl_exec($ch);
curl_close($ch);
return $data;}
echo getter('http://yourdomain.com/externalfile.php');

answered Mar 17, 2016 at 18:44

Prashanth

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