CODEWITHSUNDEEP
yii
Gihan
Gihan

Reputation: 4283

Yii default ajax search redirection

If only single result found from the default Yii ajax search, I want to redirect to it's view action. How can I trigger the redirection?. (This is similar to Google's "I'm feeling lucky" Option)

Upvotes: 0

Views: 597

Answers (2)

Gihan
Gihan

Reputation: 4283

Answering myself,

I endup with javascript hack coz I couldn't find a proper way to do it.

In my CGridView I used afterAjaxUpdate

$this->widget('zii.widgets.grid.CGridView', array(
    'dataProvider' => $dataProvider,
    'summaryCssClass'=>'summary fr',
    'id'=>'order-grid',
    'afterAjaxUpdate'=>'function(id, options){ if($("#order-grid tbody tr").length == 1 && $("#order-grid tbody tr:first td").length > 1) {window.location = $("#order-grid tbody tr td:first a").attr("href");}}',
    'columns' => array()
));

Hope this helps to someone.

Upvotes: 1

Pentium10
Pentium10

Reputation: 207830

public function actionAdmin() {
        $this->pageTitle = Yii::app()->name . ' - Customer Support';

        $model = new residence('customer');

        $model->unsetAttributes();

        if (isset($_GET['residence']))
            $model->attributes = $_GET['residence'];
        if (!Yii::app()->request->isAjaxRequest && isset($_GET['search_field'])) {
            $raw = $model->customer()->getData();
            if (count($raw) == 1) {
                $this->redirect(array("customer/details", "id" => $raw[0]->id));
            }
        }

        $this->render('admin', array(
            'model' => $model,
        ));
    }

where $model->customer() returns the DataProvider that the grid also uses.

Upvotes: 1

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