Question

我遇到的问题是ajax只是将变量附加到当前URL的末尾。我需要它指向ProHost.php。

我正在使用此处的教程tutorial

我使用它的方式是来自表单的onSubmit="return validateInput();验证我的输入，然后调用function ajaxFunction()将数据发送到我的php文件ProHost.php

但是现在ajax就是这样做的

NewHost.php?FName=d&LName=d&email=samesc%40yahoo.com&BirthDate=2014-06-8&Pass=s&PassRe=s

什么时候看起来像这样

ProHost.php?FName=d&LName=d&email=samesc%40yahoo.com&BirthDate=2014-06-18&Pass=s&PassRe=s

如果我自己编辑URL（ProHost.php?FName=d&LName=...），PHP文件就会在空白页面上以纯文本形式报告。

以下是HTML表单：

<form  action=""  method ="get" onSubmit="return validateInput(); " >
 <p id="demo" style="text-align: center">&hellip;</p> 
First Name: <input Id="FName" name="FName" type="text" required placeholder="john" size="35"  value="d" >

Last Name: <input Id="LName" name="LName" type="text" required placeholder="Smith" size="35" value="d">

Email: <input Id="email" name="email" type="email" required placeholder="John@Smith.com" size="35"  >

Birth Date: <input Id="BirthDate" name="BirthDate" type="date" required  min="1900-01-01"  >

Password: <<input Id="Pass" name="Pass" type="password" required size="35"  placeholder="Secret" >

Retype Password: <input Id="PassRe" name="PassRe" type="password" required size="35" placeholder="Secret" >
<input type="reset"  class="Button" value="Clear" name="ResetBtn">
<input type="submit" class="Button" value="Submit">
<input id="clickMe" type="button" value="clickme" onclick="doFunction();" />
</form>

这是脚本：

<script language="javascript" type="text/javascript">
function ajaxFunction(){


var ajaxRequest;  // The variable that makes Ajax possible!

 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! Please use a current web browser");
         return false;
      }
   }
 }
 // Create a function that will receive data 
 // sent from the server and will update
 // div section in the same page.
 ajaxRequest.onreadystatechange = function(){
   if(ajaxRequest.readyState == 4){
      var ajaxDisplay = document.getElementById('demo');
     ajaxDisplay.innerHTML = ajaxRequest.responseText;
   }
 }
 // Now get the value from user and pass it to 
 // server script.

//Mine
var Fname = document.getElementById("FName").value;
var Lname = document.getElementById("LName").value;              
var BirthDate = document.getElementById("BirthDate").value;
var email = document.getElementById("email").value;
var Pass = document.getElementById("Pass").value;

 var queryString = "?Fname=" + Fname + "&Lanme="  + LName +
                   "&BirthDate=" + BirthDate + "&EMail=" + email + "&Pass=" + Pass;
 ajaxRequest.open("GET", "ProHost.php" + queryString, true);
 ajaxRequest.send(null);  
return false
}

Answer 1

尝试更改此行：

ajaxRequest.open("GET", "ProHost.php" + queryString, true);

对此：

ajaxRequest.open("GET", "/ProHost.php" + queryString, true);

使用/

Ajax附加到当前URL的末尾而不是提供的文件名

1 个答案: