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我正在尝试使用jQuery在Codeigniter中创建动态下拉列表，但我对JQuery非常陌生，无法正常工作。用户首先必须选择地区，之后他必须从该地区的下拉菜单中选择。无法使用jQuery在CodeIgniter中创建动态依赖下拉列表
我的看法是：
<html>
<head>
<script src="//code.jquery.com/jquery-1.11.2.min.js"></script>
<script>
$('#school').hide();
$('#region').change(function(){
var region = $('#region').val();
if (region != ""){
var post_url = "/index.php/home/register/" + region;
$.ajax({
type: "POST",
url: post_url,
success: function(school)
{
$('#school').empty();
$('#school').show();
$.each(school,function(school_id,school)
{
var opt = $('<option />');
opt.val(school_id);
opt.text(school);
$('#school').append(opt);
});
}
});
} else {
$('#school').empty();
$('#school').hide();
}
});
</script>
</head>
<?php
echo "<body>";
echo validation_errors();
echo "<div class='container'>";
echo form_open('home/register');
echo "<h3>Register:</h3><br/>";
echo "<table border = '0' >";
echo "<tr><td> Username:* </td><td>";
$data=array(
'name' => 'username',
'class' => form_error('username') ? 'error' : '',
'value' => set_value('username')
);
echo form_input($data);
echo "</td></tr>";
//come code register form -passwords and so on..
echo "<tr><td> Region:* </td><td>";
echo "<select name = 'region' id='region' >";
foreach($regions as $row)
{
echo "<option value= '$row->region '>$row->region</option>";
}
echo "</select>";
echo "</td></tr>";
echo "<tr><td> School:* </td><td>";
echo "<select name = 'school' id='school'>";
foreach($school_show as $row)
{
echo "<option value= '$row->school_name '>$row->school_name</option>";
}
echo "</select>";
echo "</td></tr>";
echo "<tr ><td> Class:* </td><td>";
$options=array(
'8' => '8',
'9' => '9',
'class' => form_error('class[]') ? 'error' : ''
);
echo form_dropdown('class[]',$options);
echo "</td></tr>";
echo "</table><br/>";
$data=array(
'class' => 'btn btn-success ',
'id' => 'reg',
'value' => 'Register'
);
echo form_submit($data);
echo "</form>";
echo "</div>";
echo "</body>";
echo "</html>";
我的控制器：
<?php
class Home extends CI_Controller {
public function __construct() {
parent::__construct();
$this->load->library('table');
$this->load->library('session');
$this->load->helper('url');
$this->load->database();
}
public function register()
{
if ($this->form_validation->run()==FALSE)
{
$this->signup();
}
else
{
if($this->user_model->register())
{
$data['dynamic_view'] = 'success_reg';
$this->load->view('templates/main',$data);
}
else
{
$this->load->model('user_model');
$data['dynamic_view'] = 'register_form';
$data['regions'] = $this->user_model->regions_show();
$data['school_show'] = $this->user_model->school_show();
$this->load->view('templates/main',$data);
}
}
}
我的模式是：
public function regions_show() {
$this->db->select('region');
$this->db->distinct('region');
$this->db->from('schools');
$result=$this->db->get();
return $result->result();
}
public function school_show() {
$this->db->select('school_name');
$this->db->from('schools');
$result=$this->db->get();
return $result->result();
}
jQuery的不 “CI中” *做任何事情*。 JavaScript（jQuery）不关心你的框架，PHP或任何服务器端代码。 JavaScript在浏览器中运行，所以重要的是浏览器源代码中呈现的内容。向我们展示呈现的HTML标记。 – Sparky 2015-03-31 23:42:40
我的HTML标记位于JavaScript之后的视图中。 – 2015-04-01 06:57:38
我认为我的评论足够清晰。视图中的HTML不是***“***呈现*** HTML标记”。渲染标记是浏览器源代码中的输出，不包含任何PHP。 – Sparky 2015-04-01 13:44:26