Question

我想在codeigniter中创建多个链式下拉列表，在第一个下拉列表中它正常工作，但我想在第二个组合框后选择另一个下拉列表。如何编写第二个下拉列表的ajax，这样我就可以得到第三个下拉结果。

我的模特

 function loadcaryear() {
    $query = $this->db->query("SELECT DISTINCT(year) FROM drv_master_car_brand");

    return $query->result();

}

function loadcarmake($year) {
    $query = $this->db->query("SELECT DISTINCT(make) FROM drv_master_car_brand where year = '{$year}' ");

    return $query->result();
}


function loadmodelfrombrand($make) {

    $query = $this->db->query("SELECT model FROM drv_master_car_brand WHERE make = '{$make}'");

    return $query->result();
}

我的控制器。

function carbyyear() {
    $arrCaryear = $this->modelRegister->loadcaryear();

    foreach ($arrCaryear as $caryears) {
        $arrcaryear[$caryears->year] = $caryears->year;
    }

    $data['year'] = $arrcaryear;

    $this->load->view('car',$data); 
}

function ajax_car_make() {

   if (isset($_POST) && isset($_POST['year'])) {

        $year = $_POST['year'];
        $arrMakes = $this->modelRegister->loadcarmake($year);

        //print_r($arrModels);
        foreach ($arrMakes as $makes) {
            $arrmakes[$makes->make] = $makes->make;
        }

        $data['make'] = $arrmakes;

        print form_dropdown('make',$arrmakes);


    } else {
        redirect('site');
    }   
}

我的观点。

<html>
<head>

    <script type="text/javascript" src="http://code.jquery.com/jquery-latest.min.js"></script>
    <script type="text/javascript">
        $(document).ready(function () { 
            $('#yearcombox select').change(function () {
                var selYear = $(this).val();
                console.log(selYear);
                $.ajax({   
                    url: "ajax_car_make", 
                    async: false,
                    type: "POST", 
                    data: "year="+selYear, 
                    dataType: "html", 

                    success: function(data) {

                        $('#makecombox').html(data);

                    },
                })
            });
        });

    </script>

</head>

<body>
    <div id="mydoubts">
        <div id="yearcombox"><?php echo form_dropdown('year',$year, 'onChange="showState(this);"'); ?></div>
        <div id="makecombox"></div>
    </div>
</body>
</html>

Answer 1

只需将您的jquery AJAX部分的数据更改为

即可

data: {"year":selYear},

希望它能奏效！

谢谢！

使用codeigniter进行多链式下拉列表

1 个答案: