如何从部门中仅选择特定的名称

Question

这是我的代码

<div class="form-group">
                <label for="salary" class="control-label">Department:*
                </label>
                <select  class="form-control" id="selDepartment" 
                 name="selDepartment" autofocus="autofocus">
                    <option value="-1" onmouseout="" ="">Select 
                 Department</option>
                    <?php



            $query="SELECT * FROM department_master";
           $results=mysqli_query($con,$query);
              foreach ($results as $dm_department) {

            ?>

            <option value="<?php echo $dm_department["dm_department"];?>";?>
              <?php echo $a=$dm_department["dm_department"];?></option>
           <?php
           }
                   ?>

                  </select>
              </div>

              <div class="form-group">
                <label for="salary" class="control-label">Designation:*
                </label>
                <select  class="form-control" id="selDesignation" 
                name="selDesignation" autofocus="autofocus">
                    <option value="-1" onmouseout="" ="">Select 
                Designation</option>
                    <?php



           $query="SELECT * FROM designation_master";
           $results=mysqli_query($con,$query);
              foreach ($results as $dm_designation) {

         ?>

          <option value="<?php echo $dm_designation["dm_designation"];?>";?>
          <?php echo $dm_designation["dm_designation"];?></option>
            <?php
           }
                   ?>

                  </select>
              </div>

如何从部门中仅选择特定的名称，这是我的代码。

Answer 1

请在脚本

之前包含JS库

<select  class="form-control" id="selDepartment"  name="selDepartment" autofocus="autofocus" onchange='xyz(this.value)'>

<script>

function xyz(temp){
  $.ajax({
    type:'post',
url: "abc.php",
data: "department="+temp
success: function(result){
        $("#selDesignation").html(result);
    }});
</script>

在abc.php中

 $department=$_POST['department'];
  $html="<option value="-1" onmouseout="" ="">Select 
                    Designation</option>";

               $query="SELECT * FROM designation_master where department=$department";
               $results=mysqli_query($con,$query);
                  foreach ($results as $dm_designation) {
        $html=$html+" <option value=$dm_designation['dm_designation']> $dm_designation['dm_designation']</option>';

               }

           echo $html;