php url 检查可用

Question

我想检查我的数据库中是否有 url。我选择fopen，但是我从我的数据库中测试了 30 行，这将花费将近 20 秒。有什么办法可以提高效率吗？谢谢。

<?php
$start_t = microtime(true); 
//connect database and select query
while ($row = mysql_fetch_array($result)){
//$url = 'http://www.google.com'; //not test from database, but a google.com, one url will cost 0.49 seconds.
$url = $row['url'];
$res = @fopen($url, "r "); 
if($res){
    echo $row['url'].' yes<br />';
}else{
    echo $row['url']. ' no<br />';
}   
}
$end_t = microtime(true);
$totaltime = $end_t-$start_t;
echo "<br />".$totaltime." s";
?>

Answer 1

尝试使用比fopen更快的fsockopen

<?php

$t = microtime(true);

$valid = @fsockopen("www.google.com", 80, $errno, $errstr, 30);

echo (microtime(true)-$t);

if (!$valid) {
   echo "Failure";
} else {
   echo "Success";
}
?>

输出:

0.0013298988342285