php防止伪造数据从地址栏URL提交的方法

网络编程 2024/11/20 佚名

3 2 1

针对伪造的数据从URL提交的情况，首先是一个检查前一页来源的如下代码：

<"HTTP_REFERER"]; 
$sub_len=strlen($servername); 
$checkfrom=substr($sub_from,7,$sub_len); 
if($checkfrom!=$servername)die("警告！你正在从外部提交数据！请立即终止！"); 
}
"htmlcode">

/*创建XHR对象*/
function createXHR()
{
if (window.XMLHttpRequest){
var oHttp = new XMLHttpRequest();
return oHttp;
} 
else if (window.ActiveXObject){
var versions = ["MSXML2.XmlHttp.6.0","MSXML2.XmlHttp.3.0"];
for (var i = 0; i < versions.length; i++){
try {
var oHttp = new ActiveXObject(versions[i]);
return oHttp;
} catch (error) {}
}
}
throw new Error("你的浏览器不支持AJAX！");
}
/*用AJAX向page页面传递数据*/
function ajaxPost(url,query_string='')
{
var xhr;
xhr = createXHR();
xhr.open('POST',url,false);
xhr.setRequestHeader("Content-Type", "application/x-www-form-urlencoded; charset=gb2312");
xhr.onreadystatechange = function(){if (xhr.readyState == 4)if (xhr.status != 200)return;}
xhr.send(query_string);
}

伪造数据,URL提交