如何从数据库中获取数据并在 yii2 中设置为子菜单？

Question

我在 yii2 中有一个小部件菜单：

<?= \yii\widgets\Menu::widget([
        'encodeLabels' => false,
        'options' => ['id' => 'dock'],
        'items' => [

            ['label' => 'ab...',
                'template' => '<i class="fa fa-dashboard"></i><a href="{url}">{label}</a>',
                'options' => ['class' => 'launcher dropdown hover'],
                'submenuTemplate' => "\n<ul class='dropdown-menu'>\n{items}\n</ul>\n",
                'items' => [

                    ['label' => 'a',
                        'url' => ['users/..'],
                        'visible' => Yii::$app->user->isGuest
                    ],
                    ['label' => 'b',
                        'url' => ['users/..'],
                        'visible' => Yii::$app->user->isGuest
                    ],
                    ...
                ],
            ],

]);

我想从 database.That 中获取子菜单项，这意味着项目的数量可能会有所不同。我无法手动输入项目。例如：

'items' => [
                    $query="select title from book";
                    foreach($query as $items){

                    ['label' => $items['title'],
                        'url' => ['users/..'],
                        'visible' => !Yii::$app->user->isGuest
                    ],
               }
           ],

此代码不正确。我应该使用 foreach 循环吗？或者这个小部件有这种可能性吗？你有示例代码吗？

Answer 1

您可以在小部件外部构建子菜单并分配给它

              $models=YouBookModel::find()->select( 'title')->findAll();
              $subMenu = '';
              foreach($models as $items){

                 $subMenu .= "['label' => $items['title'],
                      'url' => ['users/..'],
                      'visible' => !Yii::$app->user->isGuest
                  ],";
              }

然后

       'items' => [

        ['label' => 'ab...',
            'template' => '<i class="fa fa-dashboard"></i><a href="{url}">{label}</a>',
            'options' => ['class' => 'launcher dropdown hover'],
            'submenuTemplate' => "\n<ul class='dropdown-menu'>\n{items}\n</ul>\n",
            'items' => $subMenu
            ],
        ],

Answer 2

这样：

// If you don't need object representation, you can use an array to speed up the action (and preserve memory)
$models = Model::find()->asArray()->all();

$items = [];
foreach ($models as $m) {
    $items[] = [
        'label' => $m['title'], 
        'url' => ['users/..'], 
        'visible' => !Yii::$app -> user -> isGuest
    ];
}
?>

然后打印输出

echo \yii\widgets\Menu::widget([
        'encodeLabels' => false,
        'options' => ['id' => 'dock'],
        'items' => [
            ['label' => 'ab...',
                'template' => '<i class="fa fa-dashboard"></i><a href="{url}">{label}</a>',
                'options' => ['class' => 'launcher dropdown hover'],
                'submenuTemplate' => "\n<ul class='dropdown-menu'>\n{items}\n</ul>\n",
                'items' => $items 
            ],
        ]
]);

如何从数据库中获取数据并在 yii2 中设置为子菜单？

How to fetch data from database and set as submenus in yii2?

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