Home >php教程 >php手册 >fleaphp rolesNameField bug解决方法

fleaphp rolesNameField bug解决方法

WBOY
WBOYOriginal
2016-06-13 12:10:071392browse

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';

Statement:
The content of this article is voluntarily contributed by netizens, and the copyright belongs to the original author. This site does not assume corresponding legal responsibility. If you find any content suspected of plagiarism or infringement, please contact admin@php.cn