Codeigniter，調用多個數據庫的

Question

我對codeigniter很新，但處理有關多個數據庫的問題。這些數據庫位於同一主機上。

我已經改變了database.php中的文件，從「默認」新組預留有一個額外的組被稱爲「社會」

$active_group = 'default'; 
$active_group = 'social'; 
$active_record = TRUE; 

$db['default']['hostname'] = 'xxxx'; 
$db['default']['username'] = 'xxx'; 
$db['default']['password'] = 'xxxx'; 
$db['default']['database'] = 'mysql'; 
$db['default']['dbdriver'] = 'mysqli'; 
$db['default']['dbprefix'] = ''; 
$db['default']['pconnect'] = TRUE; 
$db['default']['db_debug'] = FALSE; 
$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'] = FALSE; 

$db['social']['hostname'] = 'xxx'; 
$db['social']['username'] = 'xxx'; 
$db['social']['password'] = 'xxx'; 
$db['social']['database'] = 'social'; 
$db['social']['dbdriver'] = 'mysqli'; 
$db['social']['dbprefix'] = ''; 
$db['social']['pconnect'] = TRUE; 
$db['social']['db_debug'] = FALSE; 
$db['social']['cache_on'] = FALSE; 
$db['social']['cachedir'] = ''; 
$db['social']['char_set'] = 'utf8'; 
$db['social']['dbcollat'] = 'utf8_general_ci'; 
$db['social']['swap_pre'] = ''; 
$db['social']['autoinit'] = TRUE; 
$db['social']['stricton'] = FALSE; 

現在，在我的代碼調用DBS

public function __construct() 
{ 
    $dbgroup = !file_exists(".git") ? "default" : "development"; 
    $this->_time = microtime(true); 
    $this->_db =& DB($dbgroup); 

    $dbgroup = "social"; 
    $this->_time = microtime(true); 
    $this->_dbs =& DB($dbgroup); 

我做錯了什麼，爲什麼我不能使用_dbs從第二個數據庫獲取數據？請記住......在這裏完全編碼新手！這不是我的日常工作:)

更新！ 按照stackoverflow.com/a/8269596中所述嘗試調用此函數中的db，但仍然無法使其工作。

public function getUsers($userid = NULL) 
{ 
    if (empty($userid)) { 
       $rs = $this->load->database('social', TRUE); 
     ->select("id, username") 
     ->from("user") 
     ->get(); 
     $userid = $rs->result(); 
    return $userid; 
    } 
    elseif (isset($userid)) { 
       $rs = $this->load->database('social', TRUE); 
     ->select("id, username") 
     ->from("user") 
     ->where('id', $userid) 
     ->get(); 
     $userid = $rs->result(); 
    return $userid; 
    } 
} 

Answer 1

我認爲這應該工作：

public function getUsers($userid = NULL) 
{ 
    $rs = $this->load->database('social', TRUE); 
    if (empty($userid)) { 
     $rs->select("id, username") 
     ->from("user") 
     ->get(); 
     $userid = $rs->result(); 
    return $userid; 
    } 
    elseif (isset($userid)) { 
     $rs->select("id, username") 
     ->from("user") 
     ->where('id', $userid) 
     ->get(); 
     $userid = $rs->result(); 
    return $userid; 
    } 
}