我有模式弹出窗口,单击按钮即可打开。我有一个具有用户类型的选择下拉列表,根据该用户类型,它从mysql表中获取相应的数据并填充到另一个选择下拉列表中。我正在使用.on(change)
函数,这是模式弹出菜单的select下拉列表:
<label for="inputEmail3" class="col-sm-4 col-form-label">User Type</label>
<select name="user_type" id="user_type" class="form-control form-control-sm" style="width: 100%;" required="">
<option selected="selected" value="">Select user type </option>
<option value="admin">Admin </option>
<option value="manager">Manager </option>
<option value="executive">Executive </option>
</select>
<label for="inputEmail3" class="col-sm-4 col-form-label">Employee/User</label>
<select name="user" id="user" class="form-control form-control-sm" style="width: 100%;" required="" disabled>
<option selected="selected" value="">Select Employee </option>
</select>
我最初拥有的Javascipt:
<script type="text/javascript">
$(document).ready(function(){
$('#user_type').on(function() {
var usertype = $(this).val();
$('#user').prop( "disabled", false );
$.ajax({
type:'POST',
url:'T_userData.php',
data:'usertype='+usertype,
success:function(html){
//alert(html);
$('#sub_prod').html(html);
//$('#city').html('<option value="">Select state first</option>');
}
});
});
});
</script>
然后我将其更改为
<script type="text/javascript">
$(document).ready(function(){
$('#trgtmodal').on("change","#user_type",function() {
var usertype = $(this).val();
$('#user').prop( "disabled", false );
$.ajax({
type:'POST',
url:'T_userData.php',
data:'usertype='+usertype,
success:function(html){
//alert(html);
$('#sub_prod').html(html);
//$('#city').html('<option value="">Select state first</option>');
}
});
});
});
</script>
仍然无法正常工作。如果需要,我可以发布整个Modal Popup HTML标记。请指教