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

Question

I have a widget menu in yii2:

<?= \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
                    ],
                    ...
                ],
            ],

]);

I want to fetch submenu items from database.That's mean the number of items may vary .I can not enter items manually. such as :

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

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

This code not true. Should I use the foreach loop? OR There is such a possibility for this widget? Do you hava a sample code?

Answer 1

You can build you submenu outside the widget and the assign to it

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

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

then

       '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

In this way:

// 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
    ];
}
?>

and then print the output

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 
            ],
        ]
]);

Answer 3

you can do small trick :-)

open the tag in the labelTemplet and close it in the submenuTemplate

echo Menu::widget([
            'encodeLabels' => true,
            'activateParents' => true,
            'labelTemplate'=>"<li class=''>
            <a href='javascript:;' class='has-arrow' aria-expanded='false'>
              <div class='parent-icon'>
                <ion-icon name='home-sharp' role='img' class='md hydrated' aria-label='home sharp'></ion-icon>
              </div>
              <div class='menu-title'>{label}</div>
            </a>
           ",
            'submenuTemplate' => " <ul class='mm-collapse' style='height: 0px;'>{items}</ul>
            </li>
              ",
            'items' => require '_menu_items.php',
          ]);

as you can see i started the <li> in labelTemplate and close it in the submenuTemplate it workes for me