Question

如何在不加载页面的情况下从视图的代码点火器获取值到控制器。
我在控制器中有2个类，index和create_data。
我需要打印$test dinamically
如何使我的代码工作？

查看

<?php echo form_open("example/create_data", "class='form-horizontal' row-border")?>

<select name="example" onchange()="this.form.submit()">
    <option value="1">1</option>
    <option value="2">2</option>
</select>
<button type="submit">Tambah</button>
<?php form_close()?>

控制器

function index(){
    $data['combo'] = comboData();
    $test = $this->input->post();
    $data['test'] = $test;
    echo $test;
    if ($test==1) { 
        doSomething(); 
    } else{ 
        doNothing(); 
    }
    $this->template->display('example/index',$data);
}

Answer 1

||图

<?php echo form_open("example/create_data", "class='form-horizontal' row-border")?>

<select name="example" onchange="change_option()">
    <option value="1">1</option>
    <option value="2">2</option>
</select>
<select id="your_second_dropdown_id_name">
    <option value="0">no data</option>
</select>
<button type="submit">Tambah</button>
<?php form_close()?>

||的javascript / jquery的

var category_id =  $(this).val();
    $.ajax({
        type: "POST",
        data: category_id,
        url: "<?= base_url() ?>learning/dependent_dropdown",
        success: function(data) {
            $.each(data, function(i, data) {
                var opt = $('<option />'); 
                opt.val(data.id_type);
                opt.text(data.name);
                $('#your_second_dropdown_id_name').append(opt);
            });
        }
    });

从Combo Cox获取价值（查看）并发送到控制器而不加载页面

1 个答案: