Yii2 Redirect to previous page after update - redirect

Question

I have a application where after update user should be redirected to previous page from pagination.
let's say there is a gridview and user is at page 3. Then he update some record at that page. There should be a redirect to index page 3. What if, while user is updating record, before save, he opens another controller/action in new tab. Then ReturnUrl is now that new action and after save the record he is updating, he is redirected to that new url.
I've tried to set in every action "index" Url::remember(); and then in action "update" - return $this->goBack().
Also return $this->redirect(Yii::$app->request->referrer);, but it stays at same page.
There is a way to store every index URL in session, but in large project that means many sessions.

Answer 1

You could provide the returnUrl to the link, say:
Url::to(['update','id'=>$model->url,'returnUrl'=> Yii::$app->request->url]);
Then in your controller, use $this->request->queryParams['returnUrl'] to redirect to the previousUrl.
To take it one step further, to always provide the returnUrl, you could extend the Url Helper class:
namespace app\helpers;
use yii\helpers;
class Url extends yii\helpers\Url
public function toRouteAndReturn($route, array $params = [], $scheme = false) {
$params['returnUrl'] = Yii::$app->request->url;
return parent::toRoute($route,$params,$scheme);
}
You could provide in your main config:
'on afterAction' => function($event) {
if(!Yii::$app->getResponse()->isSent && !empty(Yii::$app->getRequest()->queryParams['returnUrl']) {
Yii::$app->getResponse()->redirect(Yii::$app->getRequest()->queryParams['returnUrl']);
}
}
Then you could use app\helpers\Url::toRouteAndReturn() instead of yii\helpers\Url::toRoute() to have it return to the previous url.

Answer 2

You can try below Solution.
First in your index page, get current page url and encode it.
$current_url=base64_encode(\Yii::$app->request->getUrl());
Append this url with your update link as below.
'urlCreator' => function ($action, $model, $key, $index) use ($current_url) {
if ($action === 'update') {
$url = Yii::$app->request->baseUrl . '/controllerName/update?id=' . $model->id.'&prev='.$current_url;
return $url;
}
// ......
}
In Controller, in Update method decode url as below and use for redirection.
public function actionUpdate($id)
{
$model = $this->findModel($id);
$prev=base64_decode($_REQUEST['prev']);
// ......
return $this->redirect($prev); // you will redirect from where update method is called
// ......
}

Answer 3

Isn't it quite easy to pass page param into your update url (<model/update>) like <model>/update?id=<id>&page=<page>?
in your index.php view, edit your ActionColumn as follow:
[
'class' => 'yii\grid\ActionColumn',
'urlCreator' => function ($action, $model, $key, $index) {
return \yii\helpers\Url::to([$action, 'id' => $model->id, 'page' => Yii::$app->request->getQueryParam('page', null)]);
},
],
As you can see, I'm getting page param from request url and pass it to models' action buttons
And when you click to update model, the page that we entered from is stored/placed in url.
Controller:
public function actionUpdate($id, $page = null)
{
$model = $this->findModel($id);
...
if($model->save()) {
return $this->redirect(['index', 'page' => $page]);
}
...
}
Finally, after we successfully update the model, the action redirects us to previous index page.