就在两天前,以下代码用于从 google 获取搜索查询:
$refer = parse_url($_SERVER['HTTP_REFERER']);
$host = parse_url($_SERVER['HTTP_REFERER'], PHP_URL_HOST);
$query = parse_url($_SERVER['HTTP_REFERER'], PHP_URL_QUERY);
if(strstr($host,'www.google.com'))
{
//do google stuff
$qstart = strpos($query, 'q=') +2;
$qend = strpos($query, '&', $qstart);
$qlength = $qend - $qstart;
$querystring = substr($query, $qstart, $qlength);
$querystring = str_replace('q=','',$querystring);
$keywords = explode('%20',$querystring);
$keywords = implode(' ', $keywords);
return $keywords;
}
但是,现在没有了。我使用 echo($query) 对其进行了测试,看来谷歌处理引荐来源网址查询请求的方式已经改变。以前包含 $query
"q=term1%20term2%20term3%20...
但是,现在,当回显 $query 时,我得到以下输出:
sa=t&rct=j&q=&esrc=s&source=web&cd=2&ved=0CCsQFjAB&url=http%3A%2F%2Fexample.com%2F&ei=vDA-UNnxHuOjyAHlloGYCA&usg=AFQjCNEvzNXHULR0OvoPMPSWxIlB9-fmpg&sig2=iPinsBaCFuhCLGFf0JHAsQ
有没有办法解决这个问题?