php 判断访客是否为搜索引擎易做图的函数代码

答案：
复制代码 代码如下:
 
/** 
* 判断是否为搜索引擎易做图 
* 
* @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; 
} 
} 

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