动态下拉菜单[PHP]

时间:2016-07-12 16:48:29

标签: javascript php jquery ajax twitter-bootstrap

我试图创建一个动态的两个下拉列表,通过选择一个选项,它会改变另一个,就像我们现在在网站上看到的国家和城市示例一样,但在我的情况下,我想通过选择商店,它显示所有的空话。我不知道它为什么不起作用。

  

以下是代码:

<div class="row wow fadeInRight">
        <div class="col-sm-3 col-sm-offset-2">
            <form action="#">
                <div class="form-group">
                <label id="drops-labels"><b id="centraliza-txt">Store</b></label>
                <span data-toggle="tooltip" data-placement="top" title="Choose a Store">
                        <?php echo "<select class='form-control input-sm' name='store' id='store-list' onChange='pickFranch(this.value)'>";
                                  echo "<option value='0'>---Select---</option>";
                                        while($dadosRedes = mysql_fetch_array($buscarDadosSuperv)) {
                                            echo "<option value='".$dadosRedes['Store']."'>".$dadosRedes['Store']."</option>";
                                                }
                                echo "</select>";
                    ?>
                        </span>
                </div>
            </form>
        </div>
<div class="col-sm-3">
        <label id="drops-labels"><b id="centraliza-txt">Franchise</b></label>
            <select name="franch" id="franch-list" class='form-control input-sm'">
                <option value="0">---Select---</option>
            </select>   
</div>


<script>
        function pickFranch(val) {
            $.ajax({
            type: "POST",
            url: "ajaxPickStore.php",
            data: // i dont know what to put here
            success: function(data){
            $("#lojas-list").html(data);
                        }
            });
            }
</script>
  

AjaxPickStore.php

<?php session_start();

require_once("connect.php");
$db_handle = new DBController();
if(!empty($_POST["store_id"])) {
    $query ="SELECT Loja FROM indicadores_rv_m1 WHERE store = '".$_POST['store_id']."' and Supervisor = '".$_SESSION['usuarioNome']."'";
    $results = $db_handle->runQuery($query);
?>
    <option value="0">---Select---</option>
<?php
    foreach($results as $franch) {
?>
    <option value="<?php echo $franch["franch"]; ?>"><?php echo $franch["franch"];?></option>
<?php
    }
}

?>

我几乎不了解Ajax,所以如果你能解释一下synthax,它会帮助我很多!

0 个答案:

没有答案