php – Yii2,如何使用另一个模型在视图文件中显示

我很抱歉我的英语.所以,我有3个表,有很多关系.

并且此代码在视图文件中显示属性:

    <?= DetailView::widget([
    'model' => $model,
    'attributes' => [
        'scientist_id',
        'scientist_name',
        'scientist_surname',
        'scientist_patronymic',
        'scientist_birthdate',
        'scientist_email:email',
        'scientist_phone',
        'scientist_photo',
        'scientist_status',
        'scientist_job:ntext',
        'scientist_additional_information:ntext',
        'field_id', //display field but no data
    ],
]) ?>

所以我需要在“field_id”中显示来自SUMMARY_FIELD表的相应“scientist_id”.我怎么能这样做?

《php – Yii2,如何使用另一个模型在视图文件中显示》

科学家模型中与表格的关系:

    public function getSummaryFields()
{
    return $this->hasMany(SummaryField::className(), ['scientist_id' => 'scientist_id']);
}
public function getFields()
{
    return $this->hasMany(Field::className(), ['field_id' => 'field_id'])->viaTable('summary_field', ['scientist_id' => 'scientist_id']);
}

SummaryField模型中的关系:

    public function getField()
{
    return $this->hasOne(Field::className(), ['field_id' => 'field_id']);
}

/**
 * @return \yii\db\ActiveQuery
 */
public function getScientist()
{
    return $this->hasOne(Scientist::className(), ['scientist_id' => 'scientist_id']);
}

最佳答案 在Model中创建函数:

function getFieldId($model)
{
   $string = '';
   foreach ($model->summaryFields as $cat) {
      $string .= $cat->field_id . " ";
   }
return $string;
}

并在视图中访问使用$model-> functionName():

<?= DetailView::widget([
   'model' => $model,
   'attributes' => [ 
     [
     'attribute'=>'field_id',
     'value' => $model->getFieldId($data),
     ],
   ],
]); ?>
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