我调用了一个AJAX脚本来将数据填充到我的选择框中,我成功地获取了值,但它们没有附加到选择框。
$('#choose_area').on('change',function(){
var stateID = $(this).val();
alert(stateID);
if(stateID){
var xz=$.ajax({
type:'post',
url:'../processRequest/locationRefine.php?stateID='+stateID,
data:{'stID':stateID},
success: function(html){
$('#choose_loc').append(html);
$('#choose_clg').html('<option value="">Please choose Location of College</option>');
}
});
console.log(xz);
}else{
$('#choose_loc').html('<option value="">Please choose state to apply</option>');
$('#choose_clg').html('<option value="">Please choose Location of College</option>');
}
});
<div class="input-field col s4">
<br>
<label for="area_selection">Choose State Applying for</label>
<?php
include '../config/db.php';
$sql = "SELECT * FROM college_info";
$run = mysqli_query($conn,$sql);
//$row = mysqli_num_rows($run)
?>
<select id="choose_area">
<option value="" disabled selected>Choose Your State</option>
<?php
while($row = mysqli_fetch_array($run)){
//extract($row);
//print_r ($row);
//exit;
echo '<option value="'.$row['id'].'">'.$row['state'].'</option>';
?>
<?php
}
?>
</select>
</div>
<div class="input-field col s4">
<br>
<label for="location_selection">Choose Location of College</label>
<select id="choose_loc">
<option value="">Please Choose Location</option>
</select>
</div>
<div class="input-field col s4">
<br>
<label for="college_selection">Choose College to apply</label>
<select id="choose_clg">
<option value="">Please Choose College</option>
</select>
</div>
我又试图根据查询无法使用$ .each函数追加选择值,有人能告诉我我在做什么错吗?