dourun2990 2012-05-26 17:55 采纳率: 100%
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使用cakephp检查控制器中的用户身份验证

I'm working on the blog/user auth tutorial for cakephp and trying to add a couple features to get the hang of things.I'm on this page of the auth right now http://book.cakephp.org/2.0/en/tutorials-and-examples/blog-auth-example/auth.html

I'm trying to display a link to logout only if the user is logged in, and a login link only if the user is logged out. I'm trying to follow the advice from this post CakePHP check if user is logged in inside a view but I'm a bit confused.

edit- To make sure my authuser was being found correctly I simply put an echo "test" statement in there. The links were working fine, however, when I removed the echo statement the link will only display login, even when I am logged in. I can't figure out why the links are only working properly when I echo something in my element.

So, on my posts/index page I have the following

<?php
if($this->element('authuser') == TRUE){   
?>
    <p><?php echo $this->Html->link('Log In', array('controller'=>'users','action' => 'login')); ?></p>
<?
}
   else{ 
?>
    <p><?php echo $this->Html->link('Log Out', array('controller'=>'users','action' => 'logout')); ?></p>
<?
}
?> 

My element authuser.ctp contains

<?
 $authuser = AuthComponent::user();

if($authuser){
  RETURN TRUE;
}
 echo "test"; //when this is commented out the link on posts/index only displays login
?>
  • 写回答

1条回答 默认 最新

  • doupao6011 2012-05-26 18:00
    关注

    You can always get current loged in user in view by calling static method

    $authUser = AuthComponent::user();
    

    Instead of passing it from action, in this way you can get currenly loged user anywhere.

    Anyway i recommend you add this logic into element and reuse it.

    本回答被题主选为最佳回答 , 对您是否有帮助呢?
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