如何在PHP中创建异步GET请求？

Question

I wish to make a simple GET request to another script on a different server. How do I do this?

In one case, I just need to request an external script without the need for any output.

make_request('http://www.externalsite.com/script1.php?variable=45'); //example usage

In the second case, I need to get the text output.

$output = make_request('http://www.externalsite.com/script2.php?variable=45');
echo $output; //string output

To be honest, I do not want to mess around with CURL as this isn't really the job of CURL. I also do not want to make use of http_get as I do not have the PECL extensions.

Would fsockopen work? If so, how do I do this without reading in the contents of the file? Is there no other way?

Thanks all

Update

I should of added, in the first case, I do not want to wait for the script to return anything. As I understand file_get_contents() will wait for the page to load fully etc?

Answer 1

file_get_contents will do what you want

$output = file_get_contents('http://www.example.com/');
echo $output;

Edit: One way to fire off a GET request and return immediately.

Quoted from http://petewarden.typepad.com/searchbrowser/2008/06/how-to-post-an.html

function curl_post_async($url, $params)
{
    foreach ($params as $key => &$val) {
      if (is_array($val)) $val = implode(',', $val);
        $post_params[] = $key.'='.urlencode($val);
    }
    $post_string = implode('&', $post_params);

    $parts=parse_url($url);

    $fp = fsockopen($parts['host'],
        isset($parts['port'])?$parts['port']:80,
        $errno, $errstr, 30);

    $out = "POST ".$parts['path']." HTTP/1.1
";
    $out.= "Host: ".$parts['host']."
";
    $out.= "Content-Type: application/x-www-form-urlencoded
";
    $out.= "Content-Length: ".strlen($post_string)."
";
    $out.= "Connection: Close

";
    if (isset($post_string)) $out.= $post_string;

    fwrite($fp, $out);
    fclose($fp);
}

What this does is open a socket, fire off a get request, and immediately close the socket and return.