php 判断访客是否为搜索引擎蜘蛛的函数代码_php技巧

php 判断访客是否为搜索引擎蜘蛛的函数代码。需要的朋友可以参考下。

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;
}
}

网上倒是能搜到一大把，不过都是千篇一律的复制来复制去的，也不够全面，我这里整理了一份比较全面的代码：

复制代码 代码如下:

function is_spider(){
$robot = 0;
$USER_AGENT = strtolower($_SERVER['HTTP_USER_AGENT']);
if(strpos($USER_AGENT,"bot")) $robot = 1;
if(strpos($USER_AGENT,"spider")) $robot = 1;
if(strpos($USER_AGENT,"slurp")) $robot = 1;
if(strpos($USER_AGENT,"mediapartners-google")) $robot = 1;
if(strpos($USER_AGENT,"fast-webcrawler")) $robot = 1;
if(strpos($USER_AGENT,"altavista")) $robot = 1;
if(strpos($USER_AGENT,"ia_archiver")) $robot = 1;
if($robot == 1){
//do something
}
return '';
}