php - yii2中如何显示关系数据

Question

我无法理解关系在 yii 2 中是如何工作的

我在 mysql 数据库中有 2 个表，author 和 book。

book 有一个名为 author 的列，它通过外键链接到 author 表的 id。

我已经使用 gii 生成了 CRUD，我希望作者姓名出现在 ListView 中，以及作者姓名的下拉列表出现在创建和更新 View 中。

但即使在 ListView 中，我似乎也无法使关系正常工作。

这是我的代码

书籍模型:

<?php

namespace app\models;

use Yii;
use app\models\Author;

/**
 * This is the model class for table "book".
 *
 * @property integer $id
 * @property string $name
 * @property integer $author
 */
class Book extends \yii\db\ActiveRecord
{
    /**
     * @inheritdoc
     */
    public static function tableName()
    {
        return 'book';
    }

    /**
     * @inheritdoc
     */
    public function rules()
    {
        return [
            [['name', 'author'], 'required'],
            [['author'], 'integer'],
            [['name'], 'string', 'max' => 11]
        ];
    }

    /**
     * @inheritdoc
     */
    public function attributeLabels()
    {
        return [
            'id' => 'ID',
            'name' => 'Name',
            'author' => 'Author',
        ];
    }

    public function getAuthor()
    {
        return $this->hasOne(Author::className(), ['id' => 'author']);
    }
}

图书搜索模型:

<?php

namespace app\models;

use Yii;
use yii\base\Model;
use yii\data\ActiveDataProvider;
use app\models\Book;

/**
 * BookSearch represents the model behind the search form about `app\models\Book`.
 */
class BookSearch extends Book
{
    /**
     * @inheritdoc
     */
    public function rules()
    {
        return [
            [['id', 'author'], 'integer'],
            [['name'], 'safe'],
        ];
    }

    /**
     * @inheritdoc
     */
    public function scenarios()
    {
        // bypass scenarios() implementation in the parent class
        return Model::scenarios();
    }

    /**
     * Creates data provider instance with search query applied
     *
     * @param array $params
     *
     * @return ActiveDataProvider
     */
    public function search($params)
    {
        $query = Book::find();
        $query->joinWith('author');

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

        var_dump($dataProvider);
        if (!$this->validate()) {
            // uncomment the following line if you do not want to return any records when validation fails
            // $query->where('0=1');
            return $dataProvider;
        }

        $query->andFilterWhere([
            'id' => $this->id,
            'author' => $this->author,
        ]);

        $query->andFilterWhere(['like', 'name', $this->name]);

        return $dataProvider;
    }
}

此外，这是 View 文件:

<?php

use yii\helpers\Html;
use yii\grid\GridView;

/* @var $this yii\web\View */
/* @var $searchModel app\models\BookSearch */
/* @var $dataProvider yii\data\ActiveDataProvider */

$this->title = 'Books';
$this->params['breadcrumbs'][] = $this->title;
?>
<div class="book-index">

    <h1><?= Html::encode($this->title) ?></h1>
    <?php // echo $this->render('_search', ['model' => $searchModel]); ?>

    <p>
        <?= Html::a('Create Book', ['create'], ['class' => 'btn btn-success']) ?>
    </p>

    <?= GridView::widget([
        'dataProvider' => $dataProvider,
        'filterModel' => $searchModel,
        'columns' => [
            ['class' => 'yii\grid\SerialColumn'],

            'id',
            'name',
            [
                'attribute' => 'author',
                'value'     => 'author.name',
            ],

            ['class' => 'yii\grid\ActionColumn'],
        ],
    ]); ?>

</div>

作者模型:

<?php

namespace app\models;

use Yii;

/**
 * This is the model class for table "author".
 *
 * @property integer $id
 * @property string $name
 */
class Author extends \yii\db\ActiveRecord
{
    /**
     * @inheritdoc
     */
    public static function tableName()
    {
        return 'author';
    }

    /**
     * @inheritdoc
     */
    public function rules()
    {
        return [
            [['name'], 'required'],
            [['name'], 'string', 'max' => 200]
        ];
    }

    /**
     * @inheritdoc
     */
    public function attributeLabels()
    {
        return [
            'id' => 'ID',
            'name' => 'Name',
        ];
    }


}

我想我可能必须在 author/authorSearch 模型的某处更改一些内容。

谁能帮忙

谢谢

Answer 1

您还可以使用来自匿名函数的值将列添加到 gridview，如此处所述 http://www.yiiframework.com/doc-2.0/yii-grid-datacolumn.html# $值(value)细节。例如，您可以在 gridview 中像这样显示作者姓名:

<?= GridView::widget([
'dataProvider'=>$dataProvider,
'filterModel'=>$searchModel,
'columns'=>[
    [
        'attribute'=>'author.name',
        'value'=>function ($model, $key, $index, $column) {
            return $model->author->name;
        },
    ],
    //...other columns
]);
?>

你也可以像这样返回一个 html 链接到作者的详细 View :

//...
'columns'=>[
    [
        'attribute'=>'author',
        'value'=>function ($model, $key, $index, $column) {
            return Html::a($model->author->name, ['/author/view', 'id'=>$model->author->id]);
        },
    ],
    //...
],
//...