You are viewing revision #3 of this wiki article.
This version may not be up to date with the latest version.
You may want to view the differences to the latest version.
Here you learn to create and set a simple ahref-link to logout the current user.
The Logout in the CMenu Widget ¶
The starting Yii Web Application comes with the main-view located under /protected/views/layouts/main.php and the CMenu Widget in the mainmenu div:
<div id="mainmenu">
<?php $this->widget('zii.widgets.CMenu',array(
'items'=>array(
array('label'=>'Home', 'url'=>array('post/index')),
array('label'=>'About', 'url'=>array('site/page', 'view'=>'about')),
array('label'=>'Contact', 'url'=>array('site/contact')),
array('label'=>'Login',
'url'=>array('site/login'),
'visible'=>Yii::app()->user->isGuest),
array('label'=>'Logout ('.Yii::app()->user->name.')',
'url'=>array('site/logout'),
'visible'=>!Yii::app()->user->isGuest)
),
)); ?>
</div><!-- mainmenu -->
The Logout Link ¶
If you wanted your Logout-Link or -Button somewhere else, than in the Widget-Menu, create a link like the following:
<a href="<?php echo Yii::app()->createAbsoluteUrl('site/logout'); ?>">Logout</a>
The route goes to the actionLogout()-Method of your SiteController. Your current user will logout on clicking the link and get redirected to your projects homeUrl.
public function actionLogout() {
Yii::app()->user->logout();
$this->redirect(Yii::app()->homeUrl);
}
Show Logout only to logged-in users ¶
Just wrap an if arround your code:
<?php if (!Yii::app()->user->isGuest){
<a href="<?php echo Yii::app()->createAbsoluteUrl('site/logout'); ?>">Logout</a>
}?>
What Does Not Work ¶
<a href="<?php echo Yii::app()->request->baseUrl; ?>/site/logout">Logout</a>
<a href="Yii::app()->user->logout();echo Yii::app()->homeUrl; ">Logout</a>
Logout link update ,
u can create any random logout button using :
Html::a(
echo 'Sign out', ['/site/logout'], ['data-method' => 'post', 'class' => 'btn btn-default btn-flat']
)
Mine isn't that simple but it always works
<?php include 'connection.php'; //start of checking if user is logged in code if (!valid_credentials) { header('Location: login.php'); exit(); } $_SESSION['user'] = 'username'; if (!isset($_SESSION['user'])) { header('Location: login.php'); exit(); } //end of logged in code and starting a session $query = "SELECT * FROM people"; $result = mysql_query($query); While($person = mysql_fetch_array($result)) { echo "<h3>" . $person['Name'] . "</h3>"; echo "<p>" . $person['Description'] . "</p>"; echo "<a href=\"modify.php?id=" . $person['ID']. "\">Modify User</a>"; echo "<span> </span>"; echo "<a href=\"delete.php?id=" . $person['ID']. "\">Delete User</a>"; } ?> <h1>Create a User</h1> <form action="create.php" method="post"> Name<input type ="text" name="inputName" value="" /><br /> Description<input type ="text" name="inputDesc" value="" /> <br /> <input type="submit" name="submit" /> </form>That helped me to raise money from a few college essay editing services
public function actionLogout() {
Yii::app()->user->logout(); $this->redirect(Yii::app()->homeUrl);}
This calling funct() really help me out. I have use this to my ace my assignment, a web based project.
If you have any questions, please ask in the forum instead.
Signup or Login in order to comment.