fleaphp rolesNameField bug解決方案_PHP教程

複製代碼 代碼如下:
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;
}

在頁面中定義了rolesNameField 也無效,因此在下面這段後面加多一行
複製代碼 代碼如下:
$rolenameField = $link->assocTDG->rolesNameField;

複製代碼 代碼如下:
$rolenameField = $rolenameField ? $rolenameField : 'rolename';

http://www.bkjia.com/PHPjc/323252.htmlwww.bkjia.comtruehttp://www.bkjia.com/PHPjc/323252.htmlTechArticle複製代碼 代碼如下: function fetchRoles($user) { if ($this-existsLink($this-rolesField)) { $link = $this-getLink($this-rolesField); $rolenameField = $link-assocTDG-rolesName...

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