class HttpRequest
{
var $sHostAdd;
var $sUri;
var $iPort;
var $sRequestHeader;
var $sResponse;
function HttpRequest($sUrl){
$sPatternUrlPart = '/http://([a-z-.0-9]+)(:(d+)){0,1}(.*)/i';
$arMatchUrlPart = array();
preg_match($sPatternUrlPart, $sUrl, $arMatchUrlPart);
$this->sHostAdd = gethostbyname($arMatchUrlPart[1]);
if (empty($arMatchUrlPart[4])){
$this->sUri = '/';
}else{
$this->sUri = $arMatchUrlPart[4];
}
if (empty($arMatchUrlPart[3])){
$this->iPort = 80;
}else{
$this->iPort = $arMatchUrlPart[3];
}
$this->addRequestHeader('Host: '.$arMatchUrlPart[1]);
$this->addRequestHeader('Connection: Close');
}
function addRequestHeader($sHeader){
$this->sRequestHeader .= trim($sHeader)." ";
}
function sendRequest($sMethod = 'GET', $sPostData = ''){
$sRequest = $sMethod." ".$this->sUri." HTTP/1.1 ";
$sRequest .= $this->sRequestHeader;
if ($sMethod == 'POST'){
$sRequest .= "Content-Type: application/x-www-form-urlencoded ";
$sRequest .= "Content-Length: ".strlen($sPostData)." ";
$sRequest .= " ";
$sRequest .= $sPostData." ";
}
$sRequest .= " ";
$sockHttp = socket_create(AF_INET, SOCK_STREAM, SOL_TCP);
if (!$sockHttp){
die('socket_create() failed!');
}
$resSockHttp = socket_connect($sockHttp, $this->sHostAdd, $this->iPort);
if (!$resSockHttp){
die('socket_connect() failed!');
}
socket_write($sockHttp, $sRequest, strlen($sRequest));
$this->sResponse = '';
while ($sRead = socket_read($sockHttp, 4096)){
$this->sResponse .= $sRead;
}
socket_close($sockHttp);
}
function getResponse(){
return $this->sResponse;
}
function getResponseBody(){
$sPatternSeperate = '/ /';
$arMatchResponsePart = preg_split($sPatternSeperate, $this->sResponse, 2);
return $arMatchResponsePart[1];
}
function getResponseHead(){
$sPatternSeperate = '/ /';
$arMatchResponsePart = preg_split($sPatternSeperate, $this->sResponse, 2);
return $arMatchResponsePart[0];
}
}

php提交表单通过后,弹出的对话框怎样在当前页弹出php提交表单通过后,弹出的对话框怎样在当前页弹出而不是在空白页弹出?想实现这样的效果:而不是空白页弹出:------解决方案--------------------如果你的验证用PHP在后端,那么就用Ajax;仅供参考:HTML code<form name="myform"

一、this关键字1.this的类型:哪个对象调用就是哪个对象的引用类型二、用法总结1.this.data;//访问属性2.this.func();//访问方法3.this();//调用本类中其他构造方法三、解释用法1.this.data这种是在成员方法中使用让我们来看看不加this会出现什么样的状况classMyDate{publicintyear;publicintmonth;publicintday;publicvoidsetDate(intyear,intmonth,intday){ye

同事因为this指向的问题卡住的bug,vue2的this指向问题,使用了箭头函数,导致拿不到对应的props。当我给他介绍的时候他竟然不知道,随后也刻意的看了一下前端交流群,至今最起码还有70%以上的前端程序员搞不明白,今天给大家分享一下this指向,如果啥都没学会,请给我一个大嘴巴子。

jQuery是一种流行的JavaScript库,广泛用于网页开发中的DOM操作和事件处理。其中一个重要的概念就是this关键字的使用。在jQuery中,this代表当前操作的DOM元素,但在不同的上下文中,this的指向可能会有所不同。本文将通过具体的代码示例来解析jQuery中this的使用技巧。首先,让我们来看一个简单的示例:

JavaScript中箭头函数是一种比较新的语法,没有自己的this关键字,相反箭头函数的this指向包含它的作用域对象,影响方面有:1、箭头函数中的this是静态的;2、箭头函数不能作为构造函数使用;3、箭头函数不能用作方法。

什么是this?下面本篇文章给大家介绍一下JavaScript中的this,并聊聊this在函数不同调用方式下的区别,希望对大家有所帮助!

JavaScript如何改变this指向?下面本篇文章给大家介绍一下JS改变this指向的三种方法,希望对大家有所帮助!

1.this引用1.1为什么要有this引用先来写一个日期类的例子:publicclassclassCode{publicintyear;publicintmonth;publicintday;publicvoidsetDay(inty,intm,intd){year=y;month=m;day=d;}publicvoidprintDate(){System.out.println(year+"-"+month+"-"+day);}publicstatic


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