Question

我有一张这样的桌子表名：菜单

Id menu submenu      childmenu
1. Oil. Edible.      Coconut oil
2. Oil. Edible.      Sunflowe oil
3. Oil. Non edible.  Deepa oil
4. Oil  non edible.  Coconut oil

我的预期输出是

Oil 
 Edible
   Coconut oil
   Sunflower oil
 Non edible
   Deepa oil
   Coconut oil

如何创建这样的菜单。谢谢你的回答

Answer 1

这应该让你开始。如果您已阅读文档（您应该阅读），其余部分则不言自明。

$array = array(
    array(
        'id' => '1',
        'menu' => 'Oil',
        'submenu' => 'editable',
        'childmenu' => 'coco oil'
    ),
    array(
        'id' => '2',
        'menu' => 'Oil',
        'submenu' => 'editable',
        'childmenu' => 'sunflower'
    ),
    array(
        'id' => '3',
        'menu' => 'Oil2',
        'submenu' => 'noneditable',
        'childmenu' => 'deepa'
    ),
);
// CONTROLLER
$_data = array();
// $array will come from model result_array()
foreach ($array as $item) {
    $_data[$item['menu']][$item['submenu']][] = $item['childmenu'];
}
// send $_data to view
// VIEW
foreach ($_data as $menu => $submenu) {
    echo '<i>' . $menu . '</i><br>';
    foreach ($submenu as $name => $childmenu) {
        echo '<strong>' . $name . '</strong><br>';
        foreach ($childmenu as $item) {
            echo $item . '<br>';
        }
    }
}

基本上我重新排列数组以便更友好地使用。这样就可以防止我们以后在视图中检查foreach中的特定子菜单。

如何从单个表

1 个答案: