Javascript 在选择第一个下拉列表时未加载从属下拉列表

我有两个下拉列表。我使用Jquery加载第二个下拉列表。没有jqyery，我的php代码运行良好。但当我使用jquery时，第二个下拉列表在选择第一个下拉列表时变为空

首次辍学教育

$sqleducation = "select * from education order by education_id asc ";
$reseducation = mysqli_query($conn,$sqleducation);

<select name="education" id="education">
<option value="-1">Please Select</option>
<?php while($roweducation=mysqli_fetch_array($reseducation)){ ?>
<option value="<?php echo $roweducation['education_id']?>">
<?php echo $roweducation['education_name']?>
 </option>
<?php }?>
</select>

二次降阶

<select name="degree" id="degree" >
<option value="-1">Please Select</option>

 <?php if(isset($_POST["education_id"]) && !empty($_POST["education_id"])){

$sqldegree = "SELECT * FROM degree WHERE education_id = ".$_POST['education_id']." ";
 $resdegree = mysqli_query($conn,$sqldegree);
while($rowdegree=mysqli_fetch_array($resdegree))
    { ?>

              <option value="<?php echo $rowdegree['degree_id']?>">
              <?php echo $rowdegree['degree_name']?>
              </option>
           <?php } }?>
  </select>

第二个下拉菜单是使用juery加载第一个下拉菜单的选择

<script src="https://code.jquery.com/jquery-3.2.1.min.js"
 integrity="sha256-hwg4gsxgFZhOsEEamdOYGBf13FyQuiTwlAQgxVSNgt4="
  crossorigin="anonymous"></script>

 <script type="text/javascript">
  $(document).ready(function(){
   $('#education').on('change',function(){
    var educationID = $(this).val();
    if(educationID){
        $.ajax({
            type:'POST',
            url:'education-career.php',
            data:'education_id='+educationID,
            success:function(html){
                $('#degree').html(html);

            }
        }); 
    }else{
        $('#degree').html('<option value="">Select Education first</option>');

    }
});});</script>

---

在ajax中，您正在使用

data:'education_id='+educationID,

用于发布数据。变量名将显示在这里

在您的第二页中，您试图获得：

isset($_POST["education"]) 

仅此而已。所以，在第二页中，您必须用教育id替换教育。

尝试在下面的行中进行更改

data:'education_id='+educationID,

到

---

试试这个。第二个选择标记必须放在第一页才能使用$'degree'.html

第一个下拉列表

$sqleducation = "select * from education order by education_id asc ";
$reseducation = mysqli_query($conn,$sqleducation);

<select name="education" id="education">
<option value="-1">Please Select</option>
<?php while($roweducation=mysqli_fetch_array($reseducation)){ ?>
<option value="<?php echo $roweducation['education_id']?>">
<?php echo $roweducation['education_name']?>
 </option>
<?php }?>
</select>
<select name="degree" id="degree" >
   <option value="-1">Please Select</option>
</select>

第二个下拉列表

<option value="-1">Please Select</option>

 <?php if(isset($_POST["education_id"]) && !empty($_POST["education_id"])){

$sqldegree = "SELECT * FROM degree WHERE education_id = ".$_POST['education_id']." ";
 $resdegree = mysqli_query($conn,$sqldegree);
while($rowdegree=mysqli_fetch_array($resdegree))
    { ?>
      <option value="<?php echo $rowdegree['degree_id']?>">
         <?php echo $rowdegree['degree_name']?>
      </option>
   <?php } }?>

更改：

$('#education').on('change',function(){

致：

---

我试过了，如果设置$\u POST[education\u id]&&！空的$_POST[education\u id]但没有问题你也必须在这里更改：数据：'education\u id='+educationID，使用这个：数据：{'education\u id'；educationID}，你做了什么更改？@SaritaSharma，第二个选择标签必须放在第一页才能使用$'degree'。html…@Star。在第二个下拉列表中检查您的where条件。@CodeLღ弗尔，非常感谢。我修好了。@Star\u人嗯，很好，现在很好

$('#education').on('change',function(){

$('select#education').change(function(){