javascript - 使用ajax时如何解决未定义索引？

Question

这个问题已经有答案了:

"Notice: Undefined variable", "Notice: Undefined index", "Warning: Undefined array key", and "Notice: Undefined offset" using PHP (29 个回答)

已关闭 5 年前。

我正在尝试建立一个依赖combobox使用ajax，但我不知道如何解决错误( Undefined index: faculty_id )我认为代码没有拼写错误或错误，并且当我在 sqlyog 上测试它时查询工作正常

这是我的ajax代码,,

$(document).ready(function(){

    $('#faculty').change(function(){
        var faculty_id = $(this).val();

        $.ajax({
            type:'POST',
            url:'option.php',
            data:'faculty_id='+faculty_id,
            success: function(response){
                    $('#department').html(response);
                }

            }); 



    });
});

这是我的 php 代码

<?php
include"../connection.php";
$faculty_id=$_POST['faculty_id'];
$query=mysql_query('select * from department where faculty_id=$faculty_id');
echo "<option value='' disabled selected>Choose One</option>";
while($departement=mysql_fetch_array($query))
{   
echo"<option values=".$departement[1].">".$departement[1]."</option>";  
}
?>

这是我选择的代码

<select name='faculty' id='faculty'>
<?php
include"../connection.php";
$query=mysql_query('select * from faculty');
while($faculty=mysql_fetch_array($query))
{
echo"<option value=".$faculty[0].">".$faculty</option>";                   
}
echo"<option value='' disabled selected>ChooseOne</option>";   
?>
</select>

<select name='departement' id="department">                  
</select>

当我更改数据时

data:faculty_id; or data:{'faculty_id':faculty_id}

仍然无法解决错误请帮助我T_T

Answer 1

您正在ajax中发送data:'faculty_id='+faculty_id，并且您正在php中接收$faculty_id=$_POST['faculty'];。改变

$faculty_id=$_POST['faculty'];

到

$faculty_id=$_POST['faculty_id'];