Question

我已从按钮中选择了ID。每次我选择要复制到另一个页面的ID。就像购物车一样，您可以在其中看到所选产品。

这就是我所做的

控制器中的

<ul id="payment_form_quickpaypayment_payment">
  <li>
    <table>
      <tbody>
        <tr>
          <td>
            <input type="radio" class="radio" name="qpayment_type" value="dankort" checked="checked">
          </td>
          <td>
            <img src="/dankort.png" title="Dankort">
          </td>
        </tr>
        <tr>
          <td>
            <input type="radio" class="radio" name="qpayment_type" value="mastercard">
          </td>
          <td>
            <img src="/mastercard.png" title="MasterCard">
          </td>
        </tr>
        <tr>
          <td>
            <input type="radio" class="radio" name="qpayment_type" value="mastercard-debet">
          </td>
          <td>
            <img src="/mastercarddebet.png" title="MasterCard Debet DK">
          </td>
        </tr>
      </tbody>
    </table>
  </li>
</ul>
<script>
  function payLoad() {
    var selectedImage = document.getElementById("payment_form_quickpaypayment_payment").getElementsByTagName('img');

    var i;
    for (i = 0; i < selectedImage.length; i++) {
      var newSpan = document.createElement("span");
      newSpan.innerText = selectedImage[i].title
      selectedImage[i].parentNode.appendChild(newSpan);
    }
  }
  window.onload = payLoad;
</script>

在视图中

public function actionClone($id)
{
    $model = $this->findModel($id);

    $newModel = new Masa();
    $newModel->attributes = $model->attributes;        
    $newModel->save(false);
    $newModel->save();

}

public function actionTabel()
{
    $searchModel = new MasaSearch();
    $dataProvider = $searchModel->search(Yii::$app->request->queryParams);

    foreach( $model->models as $id) {

        $newModel = new Masa();

        $newModel->attributes = $id->attributes; 
        $newModel->save();
     }

     return $this->render('tabel',[
         'models'=>$newModel,
     ]);
}

为简单起见，我有两个属性id和name。

Answer 1

尝试将方法更改为GET，现在数据将在url中显示，当您返回或返回时，数据不会丢失。

Html::a('<span class="glyphicon glyphicon-floppy-open">Clonare</span>',
    Yii::$app->urlManager->createUrl(['masa/clone', 'id' => $model->id]),
    [
        'title' => Yii::t('yii', 'Clonare'),
        'url' => Url::to(["/masa/clone", 'id' => $model->id]),
    ]
)

如果您不想显示身份证，可以加密

$decrypt = \Yii::$app->security->decryptByKey($_GET['id'], \Yii::$app->request->cookieValidationKey);
$encrypt=\Yii::$app->security->encryptByKey($id, \Yii::$app->request->cookieValidationKey);

如果要显示新视图，则必须在与另一个视图相同的路径中创建一个视图＆＃34; masa＆＃34;视图

新视图 masa / view.php

<div class="post-view">

    <div class="box box-primary">
        <div class="box-header with-border">
            <h3 class="box-title">Informaction of the Masa</h3>
        </div>
        <div class="box-body">
            <?= DetailView::widget([
                'model' => $model,
                'attributes' => [
                    [
                        'attribute' => 'MassaId',
                        'value' => $model->id,
                    ]
                   //....
                ]
            ])
            ?>
        </div>
    </div>

</div>

控制器

public function actionClone($id)
{
    $model = $this->findModel($id);

    $newModel = new Masa();
    $newModel->attributes = $model->attributes;        
    $newModel->save(false);
    $newModel->save();

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

我想从griview中复制数据并放在另一页Yii2上？

1 个答案: