【发布时间】:2015-03-08 02:49:36
【问题描述】:
我想制作一个脚本来检查我所有的网站是否启动。 该代码适用于一个站点,但是当我尝试检查例如一次 10 个站点,代码停止工作。
<?php
function checkStatus($url){
$agent = "Mozilla/5.0 (Macintosh; U; Intel Mac OS X 10_5_8; pt-pt) AppleWebKit/533.20.25 (KHTML, like Gecko) Version/5.0.4 Safari/533.20.27";
$ch=curl_init();
curl_setopt ($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);
curl_exec($ch);
$httpcode = curl_getinfo($ch, CURLINFO_HTTP_CODE);
curl_close($ch);
if($httpcode>=200 && $httpcode<300)
return true;
else
return false;
}
$do = array();
$n = 0;
$myfile = fopen("domens.txt", "r") or die("Unable to open file!");
while(!feof($myfile)) {
$do = fgets($myfile);
$n = $n + 1;
}
fclose($myfile);
echo '<br><br>';
$trimmed = file('domens.txt', FILE_SKIP_EMPTY_LINES);
for($x=0;$x<$n;$x++){
if(checkStatus($trimmed[$x]))
echo " <br>Website is up " . $trimmed[$x];
else
echo " <br> Website is down ". $trimmed[$x];
}
?>
【问题讨论】:
-
您能重新格式化您的代码吗?不知道那里发生了什么..