我在工作的项目,使用一个动态或相关的下拉列表codeigniter。这是我的noob逻辑,视图中的国家值传递给模型,然后模型从视图中获取具有相同国家值的状态数据,然后控制器从该模型中获取状态数据并再次传递给视图。问题是状态下拉列表什么都不显示。对不起,我是新来的。
//This is the Controllers (Pkl.php)
$data['all_country'] = $this->Server_Model->get_country_model();
$data['all_state'] = $this->Server_Model->get_state_model();
$this->load->view('contents/page_dashboard', $data);
//This is the Models (Server_Model.php)
function get_country_model(){
$db_jarlap = $this->load->database('gis_bali', TRUE);
$db_jarlap->select("*");
$db_jarlap->from("country");
$que = $db_jarlap->get();
return $que->result();
}
function get_state_model(){
$kakakoko = filter_input(INPUT_POST, 'country_id');
$db_jarlap = $this->load->database('gis_bali', TRUE);
$db_jarlap->select("*");
$db_jarlap->from("state");
$db_jarlap->where(" id_country = $kakakoko ");
$que = $db_jarlap->get();
return $que->result();
}
//This is the Views Content (page_dashboard.php)
<script src="<?php echo base_url() ?>resources/js/jquery-3.3.1.min.js"</script>
<script>
$(document).ready(function(){
$('#formCountry').change(function(){
var country_id = $(this).val();
$.ajax({
url: "<?php echo base_url() ? >application/models/Server_Model.php",
method: "POST",
data: {country_id:country_id},
success: function(data) {
$('#formState').html(data);
}
});
});
});
</script>
<select name="formCountry" id="formCountry">
<?php foreach($all_country as $semua_country): ?>
<option value="<?php echo $semua_country->id_country; ?>"><?php echo $semua_country->nama_country; ?></option>
<?php endforeach; ?>
</select>
<select name="formState" id="formState" >
<?php foreach($all_state as $semua_state): ?>
<option value="<?php echo $semua_state->id_state; ?>"><?php echo $semua_state->nama_state; ?></option>
<?php endforeach; ?>
</select>
谢谢你的帮助。
1条答案
按热度按时间c90pui9n1#
您不应该直接调用model,而应该调用
Pkl
控制员,我假设你用的是index()
上的方法Pkl
控制器: