eays
(eays)
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注册 2005-8-1
来自 四川.中国
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[广告]: 代充Paypal帐号美元
Socket连接第二次发送请求时获得的还是第一次的数据????
我的Socket类:
<?php
class Socket_Client{
var $SocketClient;
var $Host;
var $Port;
var $TimeOut;
var $ErrNo;
var $ErrStr;
var $SendContent = '';
var $GetContent;
var $HttpPost;
var $HttpGet;
function Connect($Host , $Port = 80 , $TimeOut = 30 ){
$this->Port = $Port;
$this->TimeOut = $TimeOut;
$this->Host = $Host;
return $this->SocketClient = fsockopen(gethostbyname($this->Host),$this->Port,$this->ErrNo,$this->ErrStr,$this->TimeOut) or exit($this->ErrStr."--->".$this->ErrNo);
}
function SendPost($Posto,$PostFrom,$ContentArray){
if(!$this->SocketClient) return 'Socket Client Not Opened.';
$flag = 0;
foreach ($ContentArray as $key=>$value) {
if ($flag!=0) {
$this->SendContent .= "&";
$flag = 0;
}
$this->SendContent.= $key."="; $this->SendContent.= urlencode($value);
$flag = 1;
}
$this->HttpPost = "POST ".$Posto." HTTP/1.1rn";
$this->HttpPost .= "Host:".gethostbyname($this->Host)."rn";
$this->HttpPost .= "Referer:".$PostFrom."rn";
$this->HttpPost .= "Content-Type: application/x-www-form-urlencodedrn";
$this->HttpPost .= "Content-Length: ".strlen($this->SendContent)."rn";
$this->HttpPost .= "Connection: Closernrn";
$this->HttpPost .= $this->SendContent."rn";
return fputs($this->SocketClient,$this->HttpPost);
}
function SendGet($Geto,$GetForm){
$this->HttpGet = "GET ".$Geto." HTTP/1.1rn";
$this->HttpGet .= "Host: ".gethostbyname($this->Host)."rn";
$this->HttpGet .= "Accept: */*rn";
$this->HttpGet .= "Referer: ".$GetForm."rn";
$this->HttpGet .= "User-Agent: Mozilla/4.0 (compatible; MSIE 6.0; Windows NT 5.1; SV1)rn";
$this->HttpGet .= "Connection: Closernrn";
return fputs($this->SocketClient,$this->HttpGet);
}
function Get(){
if(!$this->SocketClient) return 'Socket Client Not Opened.';
while (!feof($this->SocketClient)) {
$this->GetContent .= fgets($this->SocketClient,1024);
}
return $this->GetContent;
}
function Close(){
fclose($this->SocketClient);
}
}
?> 使用类:
$Socket = new Socket_Client();
$Socket->SendPost("/cgi-bin/nEWpAYcENTER/Domain/DomainLogin","/domainmanage/management.htm",$ContentArray);
echo $Socket->Get(); #这里返回的是第一次请求POST返回的内容.
$Socket->SendGet($GetHttpURL,"/cgi-bin/nEWpAYcENTER/Domain/DomainLogin");
echo $Socket->Get(); #这里应该是请求GET时的内容,但还是第一次POST的内容.
如果建两个Socket连接则不会有这问题.请问有没有办法用一个Socket连接搞定~~~
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