0

嗨，我需要使用 Ajax 将 url 加载到 div 标签中。这是我使用的代码。

function loadXMLDoc()
{

var xmlhttp;
if (window.XMLHttpRequest)
  {// code for IE7+, Firefox, Chrome, Opera, Safari
  xmlhttp=new XMLHttpRequest();

  }
else
  {// code for IE6, IE5
  xmlhttp=new ActiveXObject("Microsoft.XMLHTTP");

  }
xmlhttp.onreadystatechange=function()
  {
  if (xmlhttp.readyState==4 && xmlhttp.status==200)
    {

    document.getElementById("Loading_Page").innerHTML=xmlhttp.responseText;

}

  }

  xmlhttp.open("POST","UserTypesPHP.php",true);

xmlhttp.send();
}

这就是我的 UserTypesPHP.php 的样子。

<?php

// Create connection
include('connectionPHP.php');


// Check connection
if (mysqli_connect_errno($con))
  {
  echo "Failed to connect to MySQL: " . mysqli_connect_error();
  }
 //Adding Student records
 if (!empty($_POST['UserTypeSubmit'])){ 


  $sqlstu1="INSERT INTO user_type(User_ID ,User_Name) VALUES('$_POST[UserTypeID]','$_POST[UserType]')";



if (!mysqli_query($con,$sqlstu1))
  {
die("<script>alert( \"Error: ". mysqli_error($con)."\");window.location.href='AdminUser.php';</script>");


  }

 else
 echo "<script>alert ('The lecture was recorded successfully');
 window.location.href=' AdminUser.php';
 </script>";


 }



mysqli_close($con);
?>

我试图将 AdminUser.php 重定向到 Loading_Page div。但这并不像预期的那样工作。请有人告诉我原因。

先感谢您。

4

2 回答 2

0 
function loadXMLDoc()
{

var xmlhttp;
if (window.XMLHttpRequest)
  {// code for IE7+, Firefox, Chrome, Opera, Safari
  xmlhttp=new XMLHttpRequest();

  }
else
  {// code for IE6, IE5
  xmlhttp=new ActiveXObject("Microsoft.XMLHTTP");

  }
xmlhttp.onreadystatechange=function()
  {
  if (xmlhttp.readyState==4 && xmlhttp.status==200)
    {

    if(xmlhttp.responseText == 'success'){
      response(); 
    }
    //document.getElementById("Loading_Page").innerHTML=xmlhttp.responseText;

    }

  }

  xmlhttp.open("POST","UserTypesPHP.php",true);

xmlhttp.send();
}
function response(){
     alert ('The lecture was recorded successfully');
     window.location.href=' AdminUser.php';
}

在 php 文件上：

   <?php

// Create connection
include('connectionPHP.php');


// Check connection
if (mysqli_connect_errno($con))
  {
  echo "Failed to connect to MySQL: " . mysqli_connect_error();
  }
 //Adding Student records
 if (!empty($_POST['UserTypeSubmit'])){ 


  $sqlstu1="INSERT INTO user_type(User_ID ,User_Name) VALUES('$_POST[UserTypeID]','$_POST[UserType]')";



if (!mysqli_query($con,$sqlstu1))
  {
die("<script>alert( \"Error: ". mysqli_error($con)."\");window.location.href='AdminUser.php';</script>");


  }

 else
 echo "success";


 }



mysqli_close($con);
?>
于 2013-10-25T09:54:23.903 回答
0

这是一个简单的解决方案。它假定如果响应包含<script>，则整个响应是一个脚本——它不会尝试将脚本从其他 HTML 中解析出来。

xmlhttp.onreadystatechange=function() {
    if (xmlhttp.readyState==4 && xmlhttp.status==200) {
        if (/<script>/.test(xmlhttp.responseText)) {
            var script = document.createElement('script');
            script.appendChild(document.createTextNode(xmlhttp.responseText.replace(/<\/?script.*?>/g, '')));
            document.getElementByTagName("head").item(0).appendChild(script);
        } else {
            document.getElementById("Loading_Page").innerHTML = xmlhttp.responseText;
        }
    }
}
于 2013-10-25T09:55:01.067 回答