在sITeController中,我编写查询并将dataPRovider数组传递给index.PHP以显示包含在表格中.在index.PHP中,我想显示成员名而不是memberID.为此,我编写了内部查询,并成功在命令提示符下成功运行.这里我无法打印名字而不是’member id’

public function actionIndex()
      {              
     $query = new \yii\db\Query;
      $query->select(['member.FirstName','complaint.practiceCode','complaint.id','complaint.description','member.firstName'])
           ->From(['complaint'])
           ->innerJoin(['member','complaint.memberID = member.id'])
           ->groupBy(['complaint.id'])
           ->where(['complaint.deleted' => 'N']);
     $query->createCommand();

这里我通过创建$dataProvider4来传递数据,但是我无法设置firstName而不是memberID的值.

$dataProvider4= new ActiveDataProvider([
            'query' => $query,'pagination' => false,]);

        return $this->render('index',[
       'dataProvider4'=>$dataProvider4]);

<?= GridView::widget([
    'dataProvider'=>$dataProvider4,'summary'=>'total'.'&amp;nbsp<b>'.$complaintModel.'</b>&nbsp'.'Complaints','columns' => [
       ['class' => 'yii\grid\SerialColumn'],'practiceCode',// 'memberID','description','status',],]); ?>

我通过dataProvider传递数据.

public function actionIndex()
    {              
    $query = Member::find()->
        ->select(['firstName',complaint.practiceCode',complaint.id','complaint.description'])
        ->groupBy(['complaint.id'])
        ->joinWith('complaints')//Tells Yii to use the complains relation that we define below. By default it is an inner join
        ->where(['complaint.deleted' => 'N']);

$dataProvider= new ActiveDataProvider([
    'query' => $query,]);

return $this->render('index',[
    'dataProvider4'=>$dataProvider]);

public function getComplaints(){
    return $this->hasmany(Complaints::classname(),'memberID' => 'id');
}