Question

我的代码有问题了。

在html文件中：

  <form id='changeAccountForm' action='usermenu.php'  method='post'>
      Name:<input type="text" name="name"/>
      Surname:<input type="text" name="surname"/>
      Password:<input type="password" name="password"/>
      Phone:<input type="text" name="phone"/>
      Paid mail:<input type="text" name="paidMail"/>
      Payment method:<input type="radio" name="paymentMethod" value="0"/>paypal 
                    <input type="radio" name="paymentMethod" value="1"/>payza
      <div id="changeAccountInformation" >change</div>
    </form>

在Javascript文件中我使用Javascript和jQuery。我使用带有ajax的javascript和名为：

的函数

function ajaxInit(){
  var ajaxRequest;

 try{
   // Opera 8.0+, Firefox, Safari
   ajaxRequest = new XMLHttpRequest();
 }catch (e){
   // Internet Explorer Browsers
   try{
      ajaxRequest = new ActiveXObject("Msxml2.XMLHTTP");
   }catch (e) {
      try{
         ajaxRequest = new ActiveXObject("Microsoft.XMLHTTP");
      }catch (e){
         // Something went wrong
         alert("Your browser broke!");
         return false;
      }
   }
 }
 return ajaxRequest;
}//end function ajaxInit
function updateUserDataWithAjax(){
 var ajaxRequest = ajaxInit(); 
 // Create a function that will receive data 
 // sent from the server and will update

 ajaxRequest.onreadystatechange = function(){ 
   if(ajaxRequest.readyState == 4){ 
     // find all form elements and add the responseText to placeholder attribute
     var childNodes = $("#changeAccountForm").children(),i,j=0;
     var inputAjaxRequest = ajaxRequest.responseText;
     //inputAjaxRequest = inputAjaxRequest.plit(",");//ERROR: 
     for(i=0; i<childNodes.length; i++)
       if(childNodes[i]!= "input"){
         childNodes[i].setAttribute("placeholder",inputAjaxRequest[j++]);    
       }

   }
 }

 var namesArray = "userId,password,name,phone,";
   namesArray+= "surname,paidMail,payMethod,";
   namesArray+= "score";
 namesArray = namesArray.split(',');    
 //gets all the childnodes and find these with attribute  
 var childNodes = $("#changeAccountForm").children(),values,names,i;
 var queryString = '';     
 var first = true;
 for(i=0; i<childNodes.length; i++)
   for(j=0;j<namesArray.length;j++)
   if(childNodes[i].getAttribute("name")==namesArray[j]){
     values = childNodes[i].value;
     names =  childNodes[i].getAttribute("name");
     if (first){
       queryString += names+"="+values;  
       first = false;
     }else
       queryString += "&"+names+"="+values; 

   }
 ajaxRequest.open("POST", "usermenu.php", true);
 ajaxRequest.setRequestHeader("Content-type", "application/x-www-form-urlencoded");
 ajaxRequest.setRequestHeader("Content-length", queryString .length);
 ajaxRequest.setRequestHeader("Connection", "close");
 ajaxRequest.send(queryString); 

}
//and i call this function 
$(document).ready( function(){
 $("#changeAccountInformation").click(function(){
   var user = updateUserDataWithAjax();
 });

});

在php文件中：当我使用该方法发布错误并打印未定义的索引：[键值名称] in。来自ajax的响应文本是一个名为html文件的数组。

include "userMenu.html";
isset($_POST['name']);
 echo $_POST['name'];

任何信息都将不胜感激，谢谢。

方法post没有来自ajax请求的值

0 个答案: