Get current URL/URI without some of $_GET variables

Question

How, in Yii, to get the current page\'s URL. For example:

http://www.yoursite.com/your_yii_application/?lg=pl&id=15

but excluding the

Answer 1

Most of the answers are wrong.

The Question is to get url without some query param .

Here is the function that works. It does more things actually. You can remove the param that you don't want and you can add or modify an existing one.

/**
 * Function merges the query string values with the given array and returns the new URL
 * @param string $route
 * @param array $mergeQueryVars
 * @param array $removeQueryVars
 * @return string
 */
public static function getUpdatedUrl($route = '', $mergeQueryVars = [], $removeQueryVars = [])
{
    $currentParams = $request = Yii::$app->request->getQueryParams();

    foreach($mergeQueryVars as $key=> $value)
    {
        $currentParams[$key] = $value;
    }

    foreach($removeQueryVars as $queryVar)
    {
        unset($currentParams[$queryVar]);
    }

    $currentParams[0] = $route == '' ? Yii::$app->controller->getRoute() : $route;

    return Yii::$app->urlManager->createUrl($currentParams);

}

usage:

ClassName:: getUpdatedUrl('',[],['remove_this1','remove_this2'])

This will remove query params 'remove_this1' and 'remove_this2' from URL and return you the new URL

Answer 2

Yii::app()->createAbsoluteUrl(Yii::app()->request->url)

This will output something in the following format:

http://www.yoursite.com/your_yii_application/

Answer 3

Yii2

Url::current([], true);

or

Url::current();