Question

关于AJAX中POST方法后台接收的问题！

首先声明：.NET中手写AJAX代码！非ASP.NET AJAX控件！ 找了很多的例子，都说Request.Form可以接收POST方法传递的参数。但我这个程序死活接收不到！ 请高手指教 QQ289543823 function doRequestUsingPOST(){ createXMLHttpRequest(); xmlHttp.onreadystatechange=handleStateChange; xmlHttp.open("POST","getAndPostExample2.aspx?Date="+new Date().getTime(),true); xmlHttp.send("firstName="+document.getElementById("firstName").value+"&middleName="+document.getElementById("middleName").value+"&birthday="+document.getElementById("birthday").value); alert(document.getElementById("firstName").value+document.getElementById("middleName").value+document.getElementById("birthday").value); } .CS if (Request.Form["firstName"] != null && Request.Form["middleName"]!= null & Request.Form["birthday"] != null) { string firstname = Request.Form["firstName"].ToString(); string middlename = Request.Form["middleName"].ToString(); Response.Write(firstname); Response.End(); }

Answer 1

使用post方法要加两个参数 xmlHttp.open("POST","getAndPostExample2.aspx?Date="+new Date().getTime(),true); xmlHttp.setRequestHeader("content-length",param.length);//post提交设置项 xmlHttp.setRequestHeader("content-type","application/x-www-form-urlencoded");//post提交设置项