Izvlačenje google backlinksaKonkretno moja pretraga bi bila tipa link:domena.com
skripta bi trebala iz toga ispisati sve url-ove koje pretraga nađe. Znači ispisala bi sve site-ove koji prema google-u imaju backlink na domena.com.
Pokusavao sam nesto sam napisati, ali nije islo, pa sam kopao po netu ne bi li nasao nesto slicno. Uglavnom najblize onome sto trebam je ovo:
Code:
<?php
function google_search_api($args, $referer = 'http://localhost/testing/', $endpoint = 'web')
{
$url = "http://ajax.googleapis.com/ajax/services/search/".$endpoint;
if ( !array_key_exists('v', $args) )
$args['v'] = '1.0';
//$args['key']="ABQIAAAArMTuM-CBxyWL0PYBLc7SuhT2yXp_ZAY8_ufC3CFXhHIE1NvwkxT-uD75NXlWUsDRBw-8aVAlQ29oCg";
//$args['userip']=$_SERVER['REMOTE_ADDR'];
$args['rsz']='8';
$url .= '?'.http_build_query($args, '', '&');
$ch = curl_init();
curl_setopt($ch, CURLOPT_URL, $url);
curl_setopt($ch, CURLOPT_RETURNTRANSFER, 1);
// note that the referer *must* be set
curl_setopt($ch, CURLOPT_REFERER, $referer);
$body = curl_exec($ch);
curl_close($ch);
//decode and return the response
return json_decode($body,true);
}
$query=urldecode(isset($_GET['q'])?$_GET['q']:"none");
echo "Results for: $query<br />-----<br />";
$res = google_search_api(array('q' => $query));
$pages=$res['responseData']['cursor']['pages'];
$nres=0;
for($i=0;$i<count($pages);$i++)
{
$res = google_search_api(array('q' => $query,'start'=>$rez['responseData']['cursor']['pages'][$i]['start']));
for($j=0;$j<count($res['responseData']['results']); $j++)
{
$nres++;
echo urldecode($res['responseData']['results'][$j]['url'])."<br />";
}
}
echo "<br />---<br />Total number of reuslts: $nres";
?>
<?php
function google_search_api($args, $referer = 'http://localhost/testing/', $endpoint = 'web')
{
$url = "http://ajax.googleapis.com/ajax/services/search/".$endpoint;
if ( !array_key_exists('v', $args) )
$args['v'] = '1.0';
//$args['key']="ABQIAAAArMTuM-CBxyWL0PYBLc7SuhT2yXp_ZAY8_ufC3CFXhHIE1NvwkxT-uD75NXlWUsDRBw-8aVAlQ29oCg";
//$args['userip']=$_SERVER['REMOTE_ADDR'];
$args['rsz']='8';
$url .= '?'.http_build_query($args, '', '&');
$ch = curl_init();
curl_setopt($ch, CURLOPT_URL, $url);
curl_setopt($ch, CURLOPT_RETURNTRANSFER, 1);
// note that the referer *must* be set
curl_setopt($ch, CURLOPT_REFERER, $referer);
$body = curl_exec($ch);
curl_close($ch);
//decode and return the response
return json_decode($body,true);
}
$query=urldecode(isset($_GET['q'])?$_GET['q']:"none");
echo "Results for: $query<br />-----<br />";
$res = google_search_api(array('q' => $query));
$pages=$res['responseData']['cursor']['pages'];
$nres=0;
for($i=0;$i<count($pages);$i++)
{
$res = google_search_api(array('q' => $query,'start'=>$rez['responseData']['cursor']['pages'][$i]['start']));
for($j=0;$j<count($res['responseData']['results']); $j++)
{
$nres++;
echo urldecode($res['responseData']['results'][$j]['url'])."<br />";
}
}
echo "<br />---<br />Total number of reuslts: $nres";
?>
Problem kod ove skripte je sto je limit 64 linka, a to mi ne odgovara jer stranica moze imati puno vise backlinkova.
Jel mi moze tko pomoci, kako da editiram ovu skriptu (nisam cak siguran da je to i moguce, jer koristi google API, a mislim da on definira taj limit od 64). Mozete li mi napisati neki primjer funkcije koja bi radila ovo sto mi treba, vec dva dana se patim sa ovim i ne mogu naci rijesenje.
Hvala unaprijed svima,
Lijep pozdrav
