doubi2145 2015-11-02 23:54
浏览 30
已采纳

空查询结果

in a web form there are two drop-down lists. The second list items should change dynamically depending on the value selected on the first drop-down list.

This is how am I trying to do it:

index.php:

...

<script>
function getClient(val) {

    $.ajax({
    type: "POST",
    url: "get_contacts.php",
    data:'client_id='+val,
    success: function(data){
        $("#contacts-list").html(data);
    }
    });
}
</script>

...

 <div class="form-group">
        <label for="mto_client" class="col-sm-2 control-label">MTO Client</label>
        <div class="col-sm-10">
  <select name="mto_client" id="clients_list" onChange="getClient(this.value)">
  <option value="">Select a Client</option>
            <?php 
do {  
?>
            <option value="<?php echo $row_RSClients['id_client']?>" ><?php echo $row_RSClients['client_name']?></option>
            <?php
} while ($row_RSClients = mysql_fetch_assoc($RSClients));
?>
          </select>
      </div>
      </div>

       <div class="form-group">
        <label for="mto_client_contact" class="col-sm-2 control-label">MTO Client Contact</label>
        <div class="col-sm-10">

                  <select name="state" id="contacts-list">
            <option value="">Select Client Contact</option>
            </select>


      </div>
      </div>

get_contacts.php

<?php
require_once("dbcontroller.php");
$db_handle = new DBController();

if(!empty($_POST["client_id"])) {
    $query ="SELECT * FROM tb_client_contacts WHERE contact_client_id = '" . $_POST["client_id"] . "'";
    $results = $db_handle->runQuery($query);
?>
    <option value="">Select Client Contact</option>
<?php
    foreach($results as $state) {
?>
    <option value="<?php echo $state["id_client_contact"]; ?>"><?php echo $state["contact_name"]; ?></option>
<?php
    }
}
?>

There are objects on the table tb_clients_contact that meet the condition, but the second drop-down list doesn't show any objects.

Any help is welcome.

  • 写回答

1条回答 默认 最新

  • douxianji6104 2015-11-03 00:34
    关注

    Instead of

    $("#contacts-list").html(data);
    

    It should be

    $('#contacts-list').empty().append(data);
    

    empty() will clear first the options inside the contacts-list select field, then append() will insert the options from the result of your AJAX.

    You can also look at the console log for errors. If you are using Google Chrome, hit F12 to display the console log.

    本回答被题主选为最佳回答 , 对您是否有帮助呢?
    评论

报告相同问题?

悬赏问题

  • ¥15 随身WiFi网络灯亮但是没有网络,如何解决?
  • ¥15 gdf格式的脑电数据如何处理matlab
  • ¥20 重新写的代码替换了之后运行hbuliderx就这样了
  • ¥100 监控抖音用户作品更新可以微信公众号提醒
  • ¥15 UE5 如何可以不渲染HDRIBackdrop背景
  • ¥70 2048小游戏毕设项目
  • ¥20 mysql架构,按照姓名分表
  • ¥15 MATLAB实现区间[a,b]上的Gauss-Legendre积分
  • ¥15 delphi webbrowser组件网页下拉菜单自动选择问题
  • ¥15 linux驱动,linux应用,多线程