ajax代码无法从php访问json响应

这是我的ajax电话

<script src = "js/jquery.js" type = "text/javascript"></script>
<script type="text/javascript">

$(document).ready(function(){

    $('#DepartmentNameDropdown').change(function(){

        $('#DepartmentName').attr('value', document.getElementById("DepartmentNameDropdown").options[document.getElementById("DepartmentNameDropdown").selectedIndex].text);                                 

    $.ajax({
        type: "POST",
        dataType: "json",
        url: "http://localhost:8081/crownregency/getInfo.php",
        data: {id : $('#DepartmentNameDropdown').val()}, //data: {status: status, name: name},
        success:function(data){
            $('#DepartmentDescription').attr('value', data.Desc);
        }           
    });
});                    
});
$('#DepartmentName').attr('value', $('#DepartmentNameDropdown').text());    

</script>

这是我的getinfo.php

<?php
mysql_connect("localhost", "root", "");
mysql_select_db("jkings");
$value = $_POST['id'];
$query = mysql_query("SELECT * FROM department WHERE dept_ID = $value");
$row = mysql_fetch_array($query);
$desc = $row['dept_desc'];
$mission = $row['dept_mission'];
$struct = array('Desc' => '$desc', 'Mission' => '$mission');
session_start();
$_SESSION['id'] = $value;

echo json_encode($struct);
//echo $row['dept_desc'];
?>

---

我不知道我在代码中哪里犯了错误，但似乎我无法访问json响应。我已经尝试了不同的方法，我仍然在学习ajax和json，所以我很难看出我在哪里犯了错误，我不完全确定，但我想你可能必须改变

进入

而且

$struct=array'Desc'=>'$Desc'，'Mission'=>'$Mission'

应改为

$struct=数组'Desc'=>$Desc'，任务'=>$Mission

---

您的javascript似乎很好，我建议您这样做，以检测问题所在

 $.ajax({
    type: "POST",
    dataType: "json",
    url: "http://localhost:8081/crownregency/getInfo.php",
    data: {id : $('#DepartmentNameDropdown').val()}, //data: {status: status, name: name},
    success:function(data){
         alert(data.Desc)// should alert [object object] if all is well with your php code
        $('#DepartmentDescription').attr('value', data.Desc);
    }           
});

我真的很怀疑，但可能是一个案件敏感的问题，所以可能会改变

dataType:'json' to dataType:'JSON'

另外，如果您正在使用chrome使用开发者工具查看是否存在javascript错误，请为safari开发

首先更正此行$struct=array'Desc'=>$Desc'，Mission'=>$Mission；

 $.ajax({
    type: "POST",
    dataType: "json",
    url: "http://localhost:8081/crownregency/getInfo.php",
    data: {id : $('#DepartmentNameDropdown').val()}, //data: {status: status, name: name},
    success:function(data){
         alert(data.Desc)// should alert [object object] if all is well with your php code
        $('#DepartmentDescription').attr('value', data.Desc);
    }           
});

dataType:'json' to dataType:'JSON'