Javascript 依赖下拉选择codeigniter Ajax

我试图让房子在第二个下拉列表中从第一个下拉列表中选择的各自的房地产中填充。我在第二个下拉列表中没有得到结果

这是HTML：

                           <div class="col-sm-4 col-xs-12">
                                <div class="form-group drop-custum">
                                    <select id="estate" name="estate" data-live-search="true" class="form-control show-tick"
                                    onchange="get_houses(this.value)">
                                        <option value="">-- Estate --</option>
                                        <?php
                                        $sql = $this->db->query("select * from estates ORDER BY estate_name asc");
                                        $result = $sql->result();
                                        foreach ($result as $estates):
                                            ?>
                                            <option class="text-uppercase"
                                                    value="<?= $estates->estate_name?>"> <?= $estates->estate_name?></option>
                                        <?php endforeach; ?>
                                    </select>
                                </div>
                            </div>
                            <div class="col-sm-4 col-xs-12">
                                <div class="form-group drop-custum">

                                    <select name="house" data-live-search="true" id="house"
                                            class="form-control show-tick">
                                        <option value="">-- House --</option>
                                    </select>
                                </div>
                            </div>

这是PHP：

public function fill_houses()
{

    $query = "select * from houses where estate_name='" . $_POST["estate_name"] . "' order by house_number asc ";
    $result = $query->result();
    foreach ($result as $estates):
        '<option class="text-uppercase" value="'.$estates->house_id.'"> '.$estates->house_number.'</option>';
    endforeach;
}

public function fill_house（）
{
$query=“从房地产名称=”的房屋中选择*”$\u POST[“房地产名称”]。““按房屋编号asc排序”；
$result=$query->result（）；
foreach（结果为$estates）：
“.$estates->house_number.”；
endforeach；
}

必须在php函数

fill_house（）

中返回值

---

因为你忘记了

echo

在

fill\u house

？

$query

是一个字符串，你不能在它上面调用

result

方法。js中的

get\u house

对

val

变量一无所知。完成了

echo

@u\u mulder…仍然是nothing@PeterMader你能帮我更正一下吗，非常感谢。欢迎@Lexieheyy再次光临我的网站，为什么它没有发布到

public function fill_houses()
{

    $query = "select * from houses where estate_name='" . $_POST["estate_name"] . "' order by house_number asc ";
    $result = $query->result();
    foreach ($result as $estates):
        '<option class="text-uppercase" value="'.$estates->house_id.'"> '.$estates->house_number.'</option>';
    endforeach;
}

public function fill_houses()
{

    $query = $this->db->query("select * from houses where estate_name='" . $_POST["estate_name"] . "' order by house_number asc ");
    $result = $query->result();
    foreach ($result as $estates):
      return  '<option class="text-uppercase" value="'.$estates->house_id.'"> '.$estates->house_number.'</option>';
    endforeach;
}

function get_houses(val)
 {
$.ajax({
    url:"fill_houses/",
    type:"POST",
    data: { 'estate_name': val} ,
    success:function(data)
    {
        $("#house").html(data);
        alert('success');
    }
});

}