Heim  >  Artikel  >  php教程  >  php 模拟POST提交的2种方法详解

php 模拟POST提交的2种方法详解

WBOY
WBOYOriginal
2016-06-06 20:30:231117Durchsuche

本篇文章是对php模拟POST提交的2种方法进行了详细的分析介绍,需要的朋友参考下

一、通过curl函数

复制代码 代码如下:


$post_data = array();
$post_data['clientname'] = "test08";
$post_data['clientpasswd'] = "test08";
$post_data['submit'] = "submit";
$url='http://xxx.xxx.xxx.xx/xx/xxx/top.php';
$o="";
foreach ($post_data as $k=>$v)
{
$o.= "$k=".urlencode($v)."&";
}
$post_data=substr($o,0,-1);
$ch = curl_init();
curl_setopt($ch, CURLOPT_POST, 1);
curl_setopt($ch, CURLOPT_HEADER, 0);
curl_setopt($ch, CURLOPT_URL,$url);
//为了支持cookie
curl_setopt($ch, CURLOPT_COOKIEJAR, 'cookie.txt');
curl_setopt($ch, CURLOPT_POSTFIELDS, $post_data);
$result = curl_exec($ch);


二、通过fsockopen

复制代码 代码如下:


$URL=‘';
$post_data['clientname'] = "test08";
$post_data['clientpasswd'] = "test08";
$post_data['submit'] = "ログイン";
$referrer="";
// parsing the given URL
$URL_Info=parse_url($URL);
// Building referrer
if($referrer=="") // if not given use this script as referrer
$referrer=

{1}


SERVER["SCRIPT_URI"]; // making string from $dataforeach($post_data as $key=>$value)$values[]="$key=".urlencode($value); $data_string=implode("&",$values);// Find out which port is needed - if not given use standard (=80)if(!isset($URL_Info["port"]))$URL_Info["port"]=80;//
building POST-request:$request.="POST ".$URL_Info["path"]." HTTP/1.1\n";$request.="Host: ".$URL_Info["host"]."\n";$request.="Referer: $referrer\n";$request.="Content-type: application/x-www-form-urlencoded\n";$request.="Content-length: ".strlen($data_string)."\n";$request.="Connection:
close\n";$request.="\n";$request.=$data_string."\n";$fp = fsockopen($URL_Info["host"],$URL_Info["port"]);fputs($fp, $request);while(!feof($fp)) { $result .= fgets($fp, 128);}fclose($fp);





Snoopy 类(2)

sourceforge.net/projects/snoopy/


?wp-includes/class-snoopy.php.htm


HTTP类(1,2)



PEAR HTTP_Request



Popularity: 70%



,美国空间,香港服务器,香港空间
Stellungnahme:
Der Inhalt dieses Artikels wird freiwillig von Internetnutzern beigesteuert und das Urheberrecht liegt beim ursprünglichen Autor. Diese Website übernimmt keine entsprechende rechtliche Verantwortung. Wenn Sie Inhalte finden, bei denen der Verdacht eines Plagiats oder einer Rechtsverletzung besteht, wenden Sie sich bitte an admin@php.cn