yii中登陸後跳轉回登陸前請求的頁面

當咱們請求一個通過權限控制的請求不經過時，會跳轉到一個地方請求權限，請求結束後須要跳轉回以前的頁面。好比咱們請求一個須要登陸的action，會被跳轉到login頁面，咱們但願登陸成功後跳轉到咱們以前但願去的action頁面。要實現這個，只須要在login以後，執行如下這句便可：

Yii:app()->getRequest()-redirect(Yii::app()->user->getReturnUrl());

爲何呢？由於在請求一個須要登陸的aciton的跳轉到登陸頁面以前，yii會把當前請求的url存到user對象的returnUrl屬性中，方便後面的跳轉。有代碼爲證（來自Yii源碼）：

//先遭到CAccessControllFilter攔截，執行它的accessDenied方法

/**
 * Denies the access of the user.
 * This method is invoked when access check fails.
 * @param IWebUser $user the current user
 * @param string $message the error message to be displayed
 */
protected function accessDenied($user,$message)
{
	if($user->getIsGuest())
		$user->loginRequired();
	else
		throw new CHttpException(403,$message);
}

//而後執行CWebUser中的loginRequired方法

/**
 * Redirects the user browser to the login page.
 * Before the redirection, the current URL (if it's not an AJAX url) will be
 * kept in {@link returnUrl} so that the user browser may be redirected back
 * to the current page after successful login. Make sure you set {@link loginUrl}
 * so that the user browser can be redirected to the specified login URL after
 * calling this method.
 * After calling this method, the current request processing will be terminated.
 */
public function loginRequired()
{
	$app=Yii::app();
	$request=$app->getRequest();

	if(!$request->getIsAjaxRequest())
		$this->setReturnUrl($request->getUrl());
	elseif(isset($this->loginRequiredAjaxResponse))
	{
		echo $this->loginRequiredAjaxResponse;
		Yii::app()->end();
	}

	if(($url=$this->loginUrl)!==null)
	{
		if(is_array($url))
		{
			$route=isset($url[0]) ? $url[0] : $app->defaultController;
			$url=$app->createUrl($route,array_splice($url,1));
		}
		$request->redirect($url);
	}
	else
		throw new CHttpException(403,Yii::t('yii','Login Required'));
}

文章來源： http://sudodev.cn/articles/147.html