如何在php中将动态表单值插入数据库

how to insert form values(dynamic checkbox,input) dynamically into database This my view file:

<form action="<?php base_url('controller/insert'); ?>"> 
    <table>
        <thead>
            <tr> 
                <th>Menu Id</th> 
                <th>Menu Name</th> 
                <th>Yes/No</th> 
            </tr> 
        </thead>
        <tbody>
            <?php foreach($result as $res) { ?>
                <tr> 
                    <td><input type="text" name="menu_id[]" value="<?= $res->menu_id ?>"></td> 
                    <td><input type="text" name="menu_name[]" value="<?= $res->menu_name ?>"></td> 
                    <td><input type="checkbox" name="yes[]" value="<?= $res->menu_id ?>"></td> 
                </tr> 
            <?php } ?>
        </tbody>
    </table>
</form>

this my controller function:

<?php
    $fields = array(
        'menu_id'   => $this->input->post('menu_id'),
        'menu_name' =>  $this->input->post('menu_name'),
        'yes'       =>$this->input->post('yes')
    );
    $this->db->insert('menu_table',$fields);
?>

when i print this array $fields it displays as:

Array ( [0] => 2 [1] => 3 ) Array ( [0] => Plant [1] => Line ) Array ( [0] => on [1] => on ).

this my form:the menuid and menu desc and checkbox displaying dynamically.i need to insert into table as same as the form below displays.

You need to loop through the arrays. You can use the menu id as the key as it is sequential e.g. the first of menu_id corresponds with the first of menu_name:

$menu_id = $this->input->post('menu_id');
$menu_name = $this->input->post('menu_name');
$yes = $this->input->post('yes');

$yes = is_null($yes) ? array() : array_flip($yes);

if (is_null($menu_id) || is_null($menu_name)) {
    show_error('Parameters missing');
}

// should really be in a model
foreach ($menu_id as $key => $value) {
    $data['menu_id'] = $value;
    $data['menu_name'] = $menu_name[$key];
    $data['yes'] = isset($yes[$value]) ? 'true' : 'false';
    $this->db->insert('menu_table',$data);
}