jQuery函数在回调时不返回任何内容

我有一个jQuery函数，它采用表单选择元素的选定选项并将其发送到php文件，该文件应该返回html以填充第二个选择元素。 不幸的是，jQuery正在触发，但返回值为空。 有什么想法吗？

表单元素：

if ($result) {
            echo '<label>*Team: <select name="team" class="team" style=''width: 150; font-size: 16px;'' autocomplete="off" tabindex="1">';
            echo '<option value="">Select</option>';
            while($row = mysql_fetch_array($result, MYSQL_ASSOC)) {
                    echo '<option value="'.$row['TeamID'].'">'.$row['CaptainLast']."  ".date('m-y',strtotime($row['ArrDate'])).'</option>';
            }
            echo '</select></label>';
} else {
            echo "0 results";
}

?>
<label>*Team Member: <select name="member" class="member" style='width: 150; font-size: 16px;' autocomplete="off" tabindex="2">
<option value="">Select</option>
</select></label>

jQuery：

$(document).ready(function(){
    $(document).on('change','.team', function(){
      var id=$(".team option:selected").val();
      var dataString = 'tmid='+ id;
      console.log(dataString);
      $.ajax({
         type: "POST",
         url: "scripts/memfltpop.php",
         data: dataString,
         dataType: 'html',
         success: function(html){
             $(".member").html(html);
         } 
      });
    });
});

和服务器文件：

<?php
$team = $_POST['tmid'];
echo $team;
$con = mysqli_connect('**********', '****', '******', '*******');
if ($con) {
        $sql = "SELECT MemberAdmin.IndID, MemberAdmin.First, MemberAdmin.Last ". "FROM MemberAdmin ". "LEFT JOIN Members ON MemberAdmin.IndID = Members.IndID ". "WHERE Members.TeamID='" .$team. "'";
        $result = mysqli_query($con,$sql) or die(mysqli_error());
        if ($result) {
            while($row = mysql_fetch_array($result, MYSQL_ASSOC)) {
                $id=$row['IndID'];
                $data=$row['First']." ".$row['Last'];
                echo '<option value="">Select</option>';
                echo '<option value="'.$id.'">'.$data.'</option>';
            }
        } else {
            echo "0 results";
        }
        }
        else {
            echo'Not connected to database';
        }