Yii2: Method Not Allowed (#405) while logout user

10
votes

I am loging out user through following code. This is my view code behind logout button:

<li>
   <a href="<?= Url::to(['site/logout'])?>">
    <i class="fa fa-sign-out"></i> Log out
   </a>
 </li>

My controller code is:

public function actionLogout()
{
    Yii::$app->user->logout();

    $model = new LoginForm();
    $this->layout = 'index';
    return $this->render('login', ['model' => $model]);
}

In the logout it shows me:

Method Not Allowed. This url can only handle the following request methods: POST.

What is it?

7
yii2

7 Answers

39
votes

Seems like you have VerbFilter attached to logout action in your SiteController:

/**
 * @inheritdoc
 */
public function behaviors()
{
    return [            
        'verbs' => [
            'class' => VerbFilter::className(),
            'actions' => [
                'logout' => ['post'],
            ],
        ],
    ];
}

That means this action can requested only with POST method, and you are requesting with GET, that's why exception #405 is thrown.

Either remove this from VerbFilter or add data-method attribute to request with POST:

<a href="<?= Url::to(['site/logout'])?>" data-method="post">...</a>

Update: Another reason of this problem can be missing dependency for yii\web\YiiAsset. Make sure it's included in AppAsset:

public $depends = [
    'yii\web\YiiAsset',
    ...
];

YiiAsset provides data-method attribute feature which gives possibility to link act as a form with action post by writing less code. Without asset obviously link will act as regular link and standard GET request will be sent.

2
votes

You can also use a custom template 

                     'items' => [
                        [
                            'label' => 'Logout',
                            'url' => ['/user/security/logout'],
                            'template' => '<a href="{url}" data-method="post">{label}</a>',
                        ],
                    ]
1
votes

u can change the view code and echo instead of 

<li>
   <a href="<?= Url::to(['site/logout'])?>">
    <i class="fa fa-sign-out"></i> Log out
   </a>
 </li>

this one: 

    <?= Html::a('<i class="fa fa-sign-out"></i>',
           ['/site/logout'],
           ['class'=>'btn btn-default btn-flat']), //optional* -if you need to add style
           ['data' => ['method' => 'post',]])
        ?>
1
votes

You must only replace 'logout' => ['post'], with 'logout' => ['get']. In this way your error will be solved.

This way works only with Yii Framework version 2.

0
votes

If you are using Nav::widget to generate menus, the logout item should have linkOptions specified:

[
  'label' => '<i class="fa fa-sign-out"></i>Logout',
  'url' => ['/logout'],
  'linkOptions' => ['data-method' => 'post'],
],
0
votes

Following works too assuming you might extra class and data-method attribute. 

<?= 
    Html::a(
        'Logout (' . Yii::$app->user->identity->username . ')', 
        ['/site/logout'], 
        ['class' => 'ui inverted button', 'data-method' => 'post']                
    );
?>
0
votes

this code is working for AdminLTE template.

['label' => 'Sign out (' . Yii::$app->user->identity->username . ')','url' => ['/site/logout'],'template' => '<a href="{url}" data-method="post">{label}</a>',],