TypeError Yii2 Framework

Question

yii\base\View::{closure}(): Argument #2 ($model) must be of type Libro, app\models\Libro given

in C:\xampp\htdocs\biblioteca\views\libro\index.php

'filterModel' => $searchModel,
'columns' => [
    ['class' => 'yii\grid\SerialColumn'],

    'id',
    'titulo',
    'imagen',
    [
        'class' => ActionColumn::className(),
     *   'urlCreator' => function ($action, Libro $model, $key, $index, $column) {*
            return Url::toRoute([$action, 'id' => $model->id]);
         }
    ],
],

]); ?>

Error message says "must be of type Libro, app\models\Libro given". So, I guess you are in "app\models" namespace, so use \Libro $model or use Libro; after your namespace statement. — Syscall
– Syscall, Commented Jan 7, 2022 at 20:34

Top-Master · Accepted Answer · 2022-01-07 20:49:29Z

2

Change your closure from:

'urlCreator' => function ($action, Libro $model, $key, $index, $column) {*
    return Url::toRoute([$action, 'id' => $model->id]);
}

Into:

'urlCreator' => function ($action, \app\models\Libro $model, $key, $index, $column) {*
    return Url::toRoute([$action, 'id' => $model->id]);
}

In other words, use right class as type (\app\models\Libro instead of Libro).

answered Jan 7, 2022 at 20:49

Top-Master

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Comments

Senthil · Accepted Answer · 2022-05-25 14:25:12Z

0

Include below line at the beginning of the script along with other use statements:

use app\models\Libro;

answered May 25, 2022 at 14:25

Senthil

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