php – 连接到第二个不在CodeIgniter中工作的数据库
发布时间:2020-12-13 16:20:35 所属栏目:PHP教程 来源:网络整理
导读:我试图用代码点火器集成phpbb3.我非常成功但我试图访问论坛数据库,它无法正常工作.这是我目前在数据库文件中的内容. /** FORUM DATABASE **/$active_group = 'forum';$active_record = TRUE;$db['forum']['hostname'] = 'localhost';$db['forum']['username'
我试图用代码点火器集成phpbb3.我非常成功但我试图访问论坛数据库,它无法正常工作.这是我目前在数据库文件中的内容.
/** FORUM DATABASE **/ $active_group = 'forum'; $active_record = TRUE; $db['forum']['hostname'] = 'localhost'; $db['forum']['username'] = 'root'; $db['forum']['password'] = 'root'; $db['forum']['database'] = 'phpbb'; $db['forum']['dbdriver'] = 'mysqli'; $db['forum']['dbprefix'] = 'phpbb'; $db['forum']['pconnect'] = FALSE; $db['forum']['db_debug'] = TRUE; $db['forum']['cache_on'] = FALSE; $db['forum']['cachedir'] = ''; $db['forum']['char_set'] = 'utf8'; $db['forum']['dbcollat'] = 'utf8_general_ci'; $db['forum']['swap_pre'] = ''; $db['forum']['autoinit'] = TRUE; $db['forum']['stricton'] = TRUE; /** CMS DATABASE **/ $active_group = 'default'; $active_record = TRUE; $db['default']['hostname'] = 'localhost'; $db['default']['username'] = 'root'; $db['default']['password'] = 'root'; $db['default']['database'] = 'cms'; $db['default']['dbdriver'] = 'mysql'; $db['default']['dbprefix'] = ''; $db['default']['pconnect'] = FALSE; $db['default']['db_debug'] = TRUE; $db['default']['cache_on'] = FALSE; $db['default']['cachedir'] = ''; $db['default']['char_set'] = 'utf8'; $db['default']['dbcollat'] = 'utf8_general_ci'; $db['default']['swap_pre'] = ''; $db['default']['autoinit'] = TRUE; $db['default']['stricton'] = TRUE; 这是一个尝试访问数据库的方法.它一直返回null. public function getUserGroupMembership() { $forum = $this->load->database('forum',TRUE); global $table_prefix; $userId = $this->_user->data['user_id']; $this->forum->select('g.group_name'); $this->forum->from($table_prefix . 'groups g'); $this->forum->from($table_prefix . 'user_group u'); $this->forum->where('u.user_id',$userId); $this->forum->where('u.group_id','g.group_id',FALSE); $query = $this->forum->get(); foreach ($query->result_array() as $group) { $groups[] = $group['group_name']; } return $groups; } 解决方法
数据库对象加载在$forum变量中,但$this->论坛变量用于与数据库交互.它不会起作用,因为$forum是一个局部变量而$this->论坛是一个类变量,它们不一样.要修复您的代码,您应该将$forum更改为$this->论坛或$this->论坛更改为$forum.你不能同时使用它们.
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