我的 mysql 表中有三个文件,它们是:id、url、status
如何检查列 url 中的所有 url 并将 1(可用)或 0(不可用)写入状态列?
要在没有 mysql 的 php 中手动检查 url,我可以使用这个:
<?php
function Visit($url){
$agent = "Mozilla/4.0 (compatible; MSIE 5.01; Windows NT 5.0)";
$ch = curl_init();
curl_setopt ($ch, CURLOPT_URL,$url );
curl_setopt($ch, CURLOPT_USERAGENT, $agent);
curl_setopt ($ch, CURLOPT_RETURNTRANSFER, 1);
curl_setopt ($ch,CURLOPT_VERBOSE,false);
curl_setopt($ch, CURLOPT_TIMEOUT, 5);
$page=curl_exec($ch);
//echo curl_error($ch);
$httpcode = curl_getinfo($ch, CURLINFO_HTTP_CODE);
curl_close($ch);
if($httpcode >= 200 && $httpcode < 300){
return true;
}
else {
return false;
}
}
if(Visit("http://www.google.com")){//maybe make it a variable from a result of a mysql select, but how to process it one by one?
echo "Website OK"; //maybe somesql here to wtite '1'
}
else{
echo "Website DOWN";//maybe somesql here to wtite '0'
}
?>