捕获真实的谷歌搜索引擎术语

发布于 2024-12-10 11:09:18 字数 840 浏览 2 评论 0原文

我使用以下行来捕获谷歌搜索引荐来源网址，但它不再有效用于捕获真实的人查询

$referrer= $_SERVER['HTTP_REFERER'];
if(preg_match('/(www\.google.*)|(go\.google\.com)/i',$referrer)) {
   $pattern = "/^.*q=([^&]+)&?.*\$/i";
   $query = preg_replace($pattern, '$1', $referrer);
}

谷歌抓取页面，就像访问者来自搜索引擎一样。我在 5 分钟内获得了 22000 个类似搜索引荐来源网址的关注。

http://www.google.co.in/m/search?sa=2&q=xxxx&site=mobile
http://www.google.ca/search?sourceid=navclient&ie=UTF-8&rlz=1T4ACAW_enCA305CA305&q=xxxxx
http://www.google.com/url?sa=t&source=web&cd=7&ved=0CEwQFjAG&urlhttp://xxxxxxx&rct=j&q=query&ei=Sq2YTpyCMYKOiAKD74jnCQ&usg=AFQjCNEziZmOdqSBoIM_IBo8mZXrmiwkUg&sig2=cno3znIQoQ_wGt_tyWEyqA

知道如何捕获真实的用户搜索查询吗？

编辑

谷歌分析显示了正确的搜索引荐来源网址数量，想知道他们如何评估搜索引荐来源网址？

原文

I use following line for capturing google search referrers but its no longer valid for capturing real people queries

$referrer= $_SERVER['HTTP_REFERER'];
if(preg_match('/(www\.google.*)|(go\.google\.com)/i',$referrer)) {
   $pattern = "/^.*q=([^&]+)&?.*\$/i";
   $query = preg_replace($pattern, '$1', $referrer);
}

Google crawls pages like visitors are coming from search engines. I got 22000 following similar search referrers in less than 5 mins

http://www.google.co.in/m/search?sa=2&q=xxxx&site=mobile
http://www.google.ca/search?sourceid=navclient&ie=UTF-8&rlz=1T4ACAW_enCA305CA305&q=xxxxx
http://www.google.com/url?sa=t&source=web&cd=7&ved=0CEwQFjAG&urlhttp://xxxxxxx&rct=j&q=query&ei=Sq2YTpyCMYKOiAKD74jnCQ&usg=AFQjCNEziZmOdqSBoIM_IBo8mZXrmiwkUg&sig2=cno3znIQoQ_wGt_tyWEyqA

Any idea how to capture real user search queries?