Question

php判断一个网站是否能够打开

Answer 1

<?php
$url = '';
$ch = curl_init();
curl_setopt($ch, CURLOPT_URL, $url);
curl_setopt($ch, CURLOPT_TIMEOUT, 30);
curl_setopt($ch, CURLOPT_RETURNTRANSFER, 1);
curl_exec($ch); // $resp = curl_exec($ch);
$curl_code = curl_getinfo($ch, CURLINFO_HTTP_CODE);
curl_close($ch);
if ($curl_code == 200) {
echo '连接成功，状态码：' . $curl_code;
} else {
echo '连接失败，状态码：' . $curl_code;
}
如果对于302这样的跳转也算作访问成功的话，你也可以加入到判断里面。
if ($curl_code == 200 || $curl_code == 302) {
echo '连接成功，状态码：' . $curl_code;
} else {
echo '连接失败，状态码：' . $curl_code;
}