php 判断访客是否为搜索引擎蜘蛛的函数代码_PHP教程
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2022-06-01 11:49:10
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复制代码 代码如下:
/**
* 判断是否为搜索引擎蜘蛛
*
* @author Eddy
* @return bool
*/
function isCrawler() {
$agent= strtolower($_SERVER['HTTP_USER_AGENT']);
if (!empty($agent)) {
$spiderSite= array(
"TencentTraveler",
"Baiduspider+",
"BaiduGame",
"Googlebot",
"msnbot",
"Sosospider+",
"Sogou web spider",
"ia_archiver",
"Yahoo! Slurp",
"YoudaoBot",
"Yahoo Slurp",
"MSNBot",
"Java (Often spam bot)",
"BaiDuSpider",
"Voila",
"Yandex bot",
"BSpider",
"twiceler",
"Sogou Spider",
"Speedy Spider",
"Google AdSense",
"Heritrix",
"Python-urllib",
"Alexa (IA Archiver)",
"Ask",
"Exabot",
"Custo",
"OutfoxBot/YodaoBot",
"yacy",
"SurveyBot",
"legs",
"lwp-trivial",
"Nutch",
"StackRambler",
"The web archive (IA Archiver)",
"Perl tool",
"MJ12bot",
"Netcraft",
"MSIECrawler",
"WGet tools",
"larbin",
"Fish search",
);
foreach($spiderSite as $val) {
$str = strtolower($val);
if (strpos($agent, $str) !== false) {
return true;
}
}
} else {
return false;
}
}