【发布时间】：2015-10-10 09:45:51
【问题描述】：
controller
car.php
<?php
class Car extends CI_Controller {
public function __construct()
{
parent::__construct();
$this->load->database();
$this->load->helper('url');
$this->load->helper('form');
$this->load->model('company_model');
}
public function index()
{
//starts by running the query for the countries
//dropdown
$data['companydrop'] = $this->company_model->company();
//loads up the view with the query results
$this->load->view('car_view', $data);
}
//call to fill the second dropdown with the cities
public function car_model()
{
//set selected country id from POST
echo $company_id = $this->input->post('company_id',TRUE);
//run the query for the cities we specified earlier
$cardata['cardrop']=$this->company_model->car($company_id);
print_r($cardata);
$output = null;
foreach ($cardata['cardrop'] as $row)
{
//here we build a dropdown item line for each
// query result
$output .= "<option value='".$row->car_model."'>".$row->car_model."</option>";
}
echo $output;
}
}
?>
model
company_model
<?php
class Company_model extends CI_Model {
public function __construct()
{
$this->load->database();
}
//fill your contry dropdown
public function company()
{
$this->db->select('company_id,company_name');
$this->db->from('company');
$query = $this->db->get();
// the query mean select cat_id,category from
//category
foreach($query->result_array() as $row){
$data[$row['company_id']]=$row['company_name'];
}
// the fetching data from database is return
return $data;
}
//fill your cities dropdown depending on the selected city
public function car($company_id=string)
{
$this->db->select('car_id,car_model');
$this->db->from('car');
$this->db->where('company',$company_id);
$query = $this->db->get();
return $query->result();
}
}
?>
view
car_view
<html>
<head>
<title>car dealers</title>
<script type="text/javascript" src="http://ajax.googleapis.com/ajax/libs/jquery/1.4/jquery.min.js"></script>
<script type="text/javascript" src="http://ajax.googleapis.com/ajax/libs/jqueryui/1.8/jquery-ui.min.js"></script>
<script type="text/javascript">
$(document).ready(function() {
$("#companydrop").change(function(){
/*dropdown post *///
$.ajax({
url:"<?php echo base_url();?>index.php/car/car_model",
data: {id:$(this).val()},
type: "POST",
success:function(data){
$("#cardrop").html(data);
alert(data);
}
});
});
});
</script>
<style>
body{
no-repeat;
background:url(../../../video-fallback-background.jpg)
}
</style>
</head>
<body>
<!--company dropdown-->
<?php echo form_dropdown('companydrop',$companydrop,'','class="required" id="companydrop"'); ?>
<br />
<br />
<!--car dropdown-->
<select name="cardrop" id="cardrop">
<option value="">Select</option>
</select>
<br />
</body>
</html>
动态下拉菜单不能作为第一个选择，它是 公司名称正在工作，因为它是从数据库中获取的，但汽车型号不起作用，它没有提取到下拉列表中。我需要从数据库中获取汽车公司型号，然后在选择公司后，该指定公司的型号必须是在第二个下拉列表中列出。我在 phpmyadmin 中创建了数据库，并在公司 copany_id 和 company_name 中创建了两个表 car 和 company，其中 car 有 car_id、car_name 和 company_id
【问题讨论】：
-
在 ajax 调用后你是否获得了数据的价值？如果是这样，请使用 $("#cardrop").append(data);并尝试。还从 car_model 函数中删除所有回声和打印，除了 echo $data
-
如何判断ajax响应是否成功
-
$data 的输出是什么？
-
$data s ouptput is 'company name' from the table 'company' in database
-
请粘贴console.log($data)的输出。你也能看到警告框吗？
标签： codeigniter dynamic