从属下拉选择代码点火器Ajax

Question

我试图在第二个下拉列表中从第一个下拉列表中选择的相应房屋中填补房屋。 在第二个下拉菜单中，我没有任何结果。

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AJAX：

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

}

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;
}

Answer 1

您必须在php函数fill_houses()返回该值。

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;
}

在javascript ajax函数get_houses()您还需要获取变量。 和几个编辑。

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

}