Find out if a website is on or off

Question

Is there any PHP function to find out if an IP address (meaning website) is in the air or not?

I found this function on the WEB but she did not satisfy me.... it takes too long to load.

function curl_info($url){
    $ch = curl_init();
    curl_setopt( $ch, CURLOPT_URL, $url );
    curl_setopt( $ch, CURLOPT_HEADER, 1);
    curl_setopt( $ch, CURLOPT_RETURNTRANSFER, 1 );
    curl_setopt( $ch, CURLOPT_CONNECTTIMEOUT, $timeout );
    curl_setopt( $ch, CURLOPT_FOLLOWLOCATION, 1 );

    $content = curl_exec( $ch );
    $info = curl_getinfo( $ch );

    return $info;
}

$site = 'http://www.locaweb.com.br';
$info = curl_info( $site );
if( $info['http_code']==200 ) {
    echo '<u>'.$site . '</u> - <strong>está no ar!!</strong><br />';
} else {
    echo '<u>'.$site . '</u> - está fora do ar<br />';
}

Answer 1

1) Using fsockopen, note the format of the URL

if( fsockopen( 'www.locaweb.com.br' , 80 , $errno , $errstr , 30 ) ){
    echo 'site online!';
} else {
    echo 'site offline.';
}

---

2) Using checkdnsrr: Checks DNS records that match a given host name or Internet IP address.

checkdnsrr( 'google.com' ) // TRUE
checkdnsrr( 'yaruuu.com' ) // FALSE

---

3) Using Curl

$ch = curl_init('http://www.locaweb.com.br');  
curl_setopt($ch, CURLOPT_TIMEOUT, 5);  
curl_setopt($ch, CURLOPT_CONNECTTIMEOUT, 5);  
curl_setopt($ch, CURLOPT_RETURNTRANSFER, true);  
$data = curl_exec($ch);  
$httpcode = curl_getinfo($ch, CURLINFO_HTTP_CODE);  
curl_close($ch);  

if( $httpcode >= 200 && $httpcode < 300 ){  
    echo 'site online!';
} else {
    echo 'site offline.';
}

In the case of CURL, the URL http://www.locaweb.com.br returns a status 301 redirect to the URL http://www.locaweb.com.br/default.html, we may accept the site as ONLINE. In the above example it does not consider redirect status as ONLINE, but you can change and accept these cases:

status redirect:

300 => 'Multiple Choices',
301 => 'Moved Permanently',
302 => 'Found', // 1.1
303 => 'See Other',
304 => 'Not Modified',
305 => 'Use Proxy',
// 306 is deprecated but reserved
307 => 'Temporary Redirect',

It is up to you to define the status that will be valid as ONLINE.

Answer 2

I had the same problem and changed the script.
The "responsibility" of getting data from the page passed to another function.

Function to check only if the given link returns any response:

//verifica se o link informado retorna alguma resposta
function isUrl($_sUrl){
    $bRet = false;

    $cl = curl_init($_sUrl);
    curl_setopt($cl,CURLOPT_VERBOSE, true);
    curl_setopt($cl,CURLOPT_CONNECTTIMEOUT,1);
    curl_setopt($cl,CURLOPT_HEADER,true);
    curl_setopt($cl,CURLOPT_NOBODY,true);
    curl_setopt($cl,CURLOPT_RETURNTRANSFER,true);
    curl_setopt($cl,CURLOPT_SSL_VERIFYPEER, false);
    $response = curl_exec($cl);
    curl_close($cl);
    if($response) $bRet = true;

    return $bRet;
}

In my case, I check if there is any video on the page:

function getVideo($_sUrl){
    $sVideo = '';
    //necessário a extensão php_openssl activa para ulr's com https
    $html = new DOMDocument();
    @$html->loadHTML(file_get_contents($_sUrl));
    $aMeta = $html->getElementsByTagName('meta');
    foreach($html->getElementsByTagName('meta') as $meta) {
        if($meta->getAttribute('property')=='og:video'){
            $sVideo = $meta->getAttribute('content');
            break;
        }
    }

    return $sVideo;
 }

To use the code above:

$sUrl = 'https://www.youtube.com/watch?v=OZ5gVkSjC8k';
if(isUrl($sUrl))
    $sVideo = getVideo($sUrl);

Answer 3

I’m no expert on the Web part, but I believe that if you run a Get at the address and return successfully, the address is "online"

var urlValida = '/echo/html/';
var urlInvalida =  '/NaoExiste/html/';

$.ajax({    
    type: 'GET',
    url: urlValida,
    success: function (data) {
        alert(urlValida + ' OnLine');
    }
}).error( function () {
        alert(urlValida + 'OffLine');
});

$.ajax({
    type: 'GET',
    url: urlInvalida,
    success: function (data) {
        alert(urlInvalida + ' OnLine');
    }
}).error( function () {
        alert(urlInvalida + 'OffLine');
});

The code can be tested in this example in jsFiddle