fleaphp rolesNameField bug解决方法
摘要：复制代码代码如下:functionfetchRoles($user){if($this->existsLink($this->rolesFi...
复制代码 代码如下:
function fetchRoles($user)
{
if ($this->existsLink($this->rolesField)) {
$link =& $this->getLink($this->rolesField);
$rolenameField = $link->assocTDG->rolesNameField;
} else {
$rolenameField = 'rolename';
}
if (!isset($user[$this->rolesField]) ||
!is_array($user[$this->rolesField])) {
return array();
}
$roles = array();
foreach ($user[$this->rolesField] as $role) {
if (!is_array($role)) {
return array($user[$this->rolesField][$rolenameField]);
}
$roles[] = $role[$rolenameField];
}
return $roles;
}
在页面中定义了rolesNameField 也无效,因此在下面这段后面加多一行
复制代码 代码如下:
$rolenameField = $link->assocTDG->rolesNameField;
复制代码 代码如下:
$rolenameField = $rolenameField ? $rolenameField : 'rolename';
【fleaphp rolesNameField bug解决方法】相关文章：