Question

我做了这个依赖的下拉列表，对于第一个下拉列表，它已经从mysql数据库加载了，但是对于第二个下拉列表，它无法被第一个值加载。

查看

<div class="form-group">
    <label class="sr-only" for="form-email">Country</label>
    <select name="country" id="country" class="form-control" onChange="get_state(this.value, 'ship');">
        <option value="" selected="selected">Pilih Voucher</option>
        <?php
        if ($results) {
        foreach ($results as $row) {
        echo "<option value=\"" . $row -> country_id . "\">" . $row -> country_name . "</option>";
        }
        }
        ?>
    </select>
</div>
<div class="form-group">
    <label class="sr-only" for="form-email">State Option</label>
    <select name="state_ship" id="state_ship" class="form-control">
        <option value="" selected="selected">State Option</option>
        <?php
        if ($result_state) {
            $str = 1;
            foreach ($result_state as $row) {
            while($str <= $row -> country ){
            echo "<option value=\"" . $str . "\">" . $str . "</option>";
            $str++;
            }
        }
        }
        ?>
    </select>
</div>

我的控制器

public function ship()
{
    $data=array(
       'results'=> $this->produk_model->getAllData('produk'),
        );
    $this->load->view('transaksi/header'); 
    $this->load->view('transaksi/ship', $data);   
    $this->load->view('transaksi/footer');  


public function get_state($country)
{

    $this->db->where('country_id', $country_id);
    $result_state = $this->db->get('country_name');   
}

我的模特

public function getAllData($table)
    {
        return $this->db->get($table)->result();
    }

JQUERY

 function get_state(value, type)
    {
            $('#state_' + type).html('Loading State...');
            $.ajax({
               dataType      : "json",
                type     : "POST",
                url      : "produk/get_state",
                data     : "data=" + value + "&type=" + type,
                success  : function(res){
                        $('#state_' + type).html('');
                        if(res.status == 200)
                        {
                            for(var i = 0; i < res.data.length; i++)
                            {

                                $('#state_' + type).append('<option value="' + res.data[i].id_country + '">' + res.data[i].nama_country + '</option>');
                            }

                        }
                        else
                        {
                            $('#state_' + type).append('<option> Data Unloaded </option>');
                        }
                    },
                error   : function(){ 
            },
                timeout: 50000          
            });             
        }
    }

感谢您的帮助

Answer 1

<%= f.select :start, Location.all.collect {|x [x.city_name, x.id]}, {:include_blank => 'Select One'} %>

并在你的观点

success: function(data)
{
    $('#state_ship').html('<option value="">-- Select State--</option>');

        $.each(data, function(i, data){
        $('#state_ship').append("<option value='"+data.state_id+"'>"+data.state_name+"</option>");
        });
}

codeigniter jquery依赖下拉列表

1 个答案: