php - Codeigniter，调用多个数据库

Question

我是 codeigniter 的新手，但正在处理有关多个数据库的问题。数据库位于同一台主机上。

我已经更改了 database.php 文件以包含一个额外的组，除了“默认”新组称为“社会”

$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;

现在在我的代码中调用 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);

我做错了什么，为什么我不能使用 _dbs 从第二个数据库中获取数据？请记住……这里完全是编码新手!这不是我的日常工作:)

更新!尝试按照 stackoverflow.com/a/8269596 中的描述在此函数中调用数据库，但我仍然无法让它工作。

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

我认为这应该可行:

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;
    }
}