Error in yii2 data display from a model

Question

This is the controller action list which should display users in a user table if a user role is not super

 public function actionList()
{
    $dataProvider = new ActiveDataProvider([
        'query'=>User::find()->where(['not', 'is_super']),
        'pagination' => [
            'pageSize' => 20,
        ],
    ]);

    return $this->render('list', ['dataProvider'=>$dataProvider]);
}

What could be the problem

So, what is that column name where all values are being saved. Like: is_super, is_no_super. There will be column name Right, where all values are being stored? — Nana Partykar
– Nana Partykar, Commented Jul 14, 2016 at 10:13

Nana Partykar · Accepted Answer · 2016-07-14 10:26:48Z

1

Change your query to

'query' => User::find()->where(['<>','is_super',1]),

Or,

<?php
$super = 1; 
$dataProvider = new ActiveDataProvider([
  'query' => User::find()->where(['<>','is_super',$super]),
  'pagination' => [
      'pageSize' => 20,
  ],
]);
?>

edited Jul 14, 2016 at 10:26

answered Jul 14, 2016 at 10:19

Nana Partykar

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2 Comments

Geoff Over a year ago

Still returns an error of ...[Only variables should be passed by reference]

Nana Partykar Over a year ago

Try, $super=1; 'query' => User::find()->where(['<>','is_super',$super]), @GEOFFREYMWANGI

Chinmay Waghmare · Accepted Answer · 2016-07-14 10:25:25Z

1

Try:

'query' => User::find()->where('is_super<>:is_super',[':is_super' => 1]),

answered Jul 14, 2016 at 10:25

Chinmay Waghmare

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1 Comment

Geoff Over a year ago

Thanks works, but the first answer above worked ...thanks

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