php - 不符合条件 无法在 Yii2 中工作

Question

我把它清理了一下，所以问题是为什么当我说不的时候 searchmodel 返回 id=30 的房间
搜索模型代码

public function search($params)
{
   $ids = [];
   $ids[]=30;
    $query = room::find();
    $dataProvider = new ActiveDataProvider([
        'query' => $query,
    ]);
    $this->load($params);
    if (!$this->validate()) {

     }
 $query->andFilterWhere([
        'price' => $this->price,
        'category_id' => $this->category_id,
        'id' => $this->id,
        'ci_id'=>$this->ci_id;

    ]);
    $query->andFilterWhere(['not in','id',$ids]);
    return $dataProvider;
}

这里的 Controller 操作是 Controller 操作，我不认为有什么奇怪的，只是发布以防万一出现问题。

public function actionIndex()
{
    $searchModel = new roomSearch();
    //$user = Yii::$app->user->identity;

    if ($searchModel->load(Yii::$app->request->post()) ) {
    $dataProvider = $searchModel->search(Yii::$app->request->queryParams);

    return $this->render('index', [
        'searchModel' => $searchModel,
        'dataProvider' => $dataProvider,
    ]);
}
else{

  //

    $dataProvider = $searchModel->search(Yii::$app->request->queryParams);


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

显示房间的查看代码如下:

  <?php \yii\widgets\Pjax::begin(['timeout' => 30000, 'clientOptions' => ['container' => 'pjax-container']]); 
?>

    <?php $form = ActiveForm::begin([ 'action' => ['index'],
'method' => 'get','options' => ['data-pjax' => true ],]); ?>
 <?= $form->field($searchModel, 'category_id')->dropdownList(MtCat::getHierarchy(false, [], false, false),['onchange'=>'this.form.submit()','prompt' => 'Όλες οι κατηγορίες']); ?>
 <?= $form->field($searchModel, 'ci_id')->dropdownList(City::getList(),['onchange'=>'this.form.submit()','prompt' => 'Όλες οι περιοχές']); ?>

   <?php ActiveForm::end(); ?>
</br>


<?=

ListView::widget([

    'layout' => "{pager}\n{summary}\n<div class='all-items'>{items}</div>\n{pager}",

    //'layout' => "{pager}\n<div class='all-items'>{items}</div>\n{pager}",

    'summary' => sprintf('<div class="summary"><b>%s</b> αποτελέσματα δωματίων</div>', $dataProvider->totalCount),

    'dataProvider' => $dataProvider,

     'showOnEmpty' => false,


    'itemOptions' => ['class' => 'item'],

    'itemView' => '_img', /* function ($model, $key, $index, $widget) { return Html::a(Html::encode($model->id), ['view', 'id' => $model->id]); }, */

    'pager' => [

        'maxButtonCount' => ceil($dataProvider->totalCount / $dataProvider->pagination->pageSize),

    ],

]);

?>
 <?php \yii\widgets\Pjax::end(); ?>

Answer 1

原因是您只选择了

room_id

这就是为什么它看不到

id

您可以像这样更改代码的前几行。

$ids = [];
$ids[] = 30;
$results = Yii::$app->db->createCommand('select id,room_id from y2gall_reservations')->queryAll();
$query = room::find();