Yii2：如何在Yii2中显示没有gridview的搜索结果

Question

Trying to implement search box for my page in which user has to enter job_code to get its order status. 尝试为我的页面实现搜索框，其中用户必须输入job_code才能获取其订单状态。 I'm getting result but this result in grid view I don't want this grid view. 我正在获取结果，但是此结果显示在网格视图中，我不需要此网格视图。 So, that I can place various attributes on page at different positions. 因此，我可以在页面上的不同位置放置各种属性。

Search form In which user has to enter job_code 搜索表单，用户必须在其中输入job_code

<?php $form = ActiveForm::begin(['action' => ['tracking'],'method' => 'get','class'=>'lockscreen-credentials']); ?>
<?= $form->field($searchModel, 'job_code')->textInput(array('placeholder' => 'Job Code..'))->label(false); ?>
<?= Html::submitButton('Search', ['class' => 'btn btn-primary btn-block btn-flat']) ?>
<?= Html::resetButton('Reset', ['class' => 'btn btn-primary btn-block btn-flat']) ?> 
<?php ActiveForm::end(); ?>

在此处输入图片说明

Result Page on which Search Result will display 将在其上显示搜索结果的结果页面

<body class="lockscreen">
<?= GridView::widget([
        'dataProvider' => $dataProvider,
        'filterModel' => $searchModel,
        'columns' => [
            'job_code',
            'client_code',
            'company_name',
            'job_description:ntext',
            'status',
            'emp_email:email',
            'emp_mobile',
            'emp_first_name',
            'emp_last_name',  
        ],
    ]); ?>      
</body>

在此处输入图片说明

Controllers 控制器

public function actionIndex()
    {
        $model = new Status();
        $searchModel = new StatusSearch();
        $dataProvider = $searchModel->search(Yii::$app->request->queryParams);

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

    public function actionTracking()
    {
         $model = new Status();
         $searchModel = new StatusSearch();
         $dataProvider = $searchModel->search(Yii::$app->request->queryParams);

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

How to achieve this? 如何实现呢？

Answer 1

You can access the data provider models data in this way 您可以通过这种方式访问数据提供者模型数据

for a dump result you don't need to format your code with html and limit the view to the first (or the only) line of data available. 对于转储结果，您不需要使用html格式化代码并将视图限制为可用数据的第一行（或唯一）。 Then for sample i use use only echo 然后我使用示例仅使用echo

<body class="lockscreen">
   <?php 

     echo 'job_code = ' . $dataProvider->models[0]->job_code .' - ' ;
     echo 'client_code = ' . $dataProvider->models[0]->client_code .' - ';
     echo 'company_name = ' .  $dataProvider->models[0]->company_name .' - ';
     echo 'job_description = ' . $dataProvider->models[0]->job_description .' - ';
     echo 'status = ' . $dataProvider->models[0]->status .' - ';
     // and so on  



</body>

I you need to show more then a row of data, you must obviously loop for all the models you prefer 我需要显示更多数据，然后再显示一行数据，显然必须为您喜欢的所有模型循环

Yii2：如何在Yii2中显示没有gridview的搜索结果

问题描述

1 个解决方案

解决方案1
1 已采纳 2015-07-18 15:53:07

Yii2：如何在Yii2中显示没有gridview的搜索结果

问题描述

1 个解决方案

解决方案1 1 已采纳 2015-07-18 15:53:07

解决方案1
1 已采纳 2015-07-18 15:53:07