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PHP Zend_Acl_Resource_Interface::getType方法代码示例

本文整理汇总了PHP中Zend_Acl_Resource_Interface::getType方法的典型用法代码示例。如果您正苦于以下问题:PHP Zend_Acl_Resource_Interface::getType方法的具体用法?PHP Zend_Acl_Resource_Interface::getType怎么用?PHP Zend_Acl_Resource_Interface::getType使用的例子?那么恭喜您, 这里精选的方法代码示例或许可以为您提供帮助。您也可以进一步了解该方法所在Zend_Acl_Resource_Interface的用法示例。


在下文中一共展示了Zend_Acl_Resource_Interface::getType方法的1个代码示例,这些例子默认根据受欢迎程度排序。您可以为喜欢或者感觉有用的代码点赞,您的评价将有助于系统推荐出更棒的PHP代码示例。

示例1: assert

 /**
  * Returns true if and only if the assertion conditions are met
  *
  * This method is passed the ACL, Role, Resource, and privilege to which
  * the authorization query applies. If the $role, $resource, or $privilege
  * parameters are null, it means that the query applies to all Roles,
  * Resources, or privileges, respectively.
  *
  * @param  Zend_Acl                    $acl
  * @param  Zend_Acl_Role_Interface     $role
  * @param  Zend_Acl_Resource_Interface $resource
  * @param  string                      $privilege
  * @return boolean
  */
 public function assert(Zend_Acl $acl, Zend_Acl_Role_Interface $role = null, Zend_Acl_Resource_Interface $resource = null, $privilege = null)
 {
     if (!$role instanceof UserModel) {
         throw new Exception('Role must be an instance of UserModel');
     }
     if (!$resource instanceof OrgModelAbstract) {
         throw new Exception('Resource must be an instance of OrgModelAbstract');
     }
     $type = $resource->getType();
     if (!isset($this->_types[$type])) {
         throw new Exception('Resource has an undefined organization type');
     }
     $org = $role->getOrganization();
     $orgType = $org->getType();
     return $this->_diff($this->_types[$orgType], $this->_types[$type]);
 }
开发者ID:SandeepUmredkar,项目名称:PortalSMIP,代码行数:30,代码来源:ChildOrganizationType.php


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