Question

我在使用codeigniter加入这两个数据库时出现问题，我使用mysql尝试了它并且有效。

有人可以将其转换为codeigniter

这是我的代码

SELECT DATE_FORMAT(FROM_UNIXTIME(ex.extime), '%Y-%m-%d') as `export_report`, u.username as `Username`, ex.qdesc as `Search_Performed`,
                 ex.to as `Number_of_records_exported`, ex.format as `Format`
                 FROM cone.users u
                 INNER JOIN igdata.export_status as ex
                 ON u.id = ex.uid
                 WHERE DATE_FORMAT(FROM_UNIXTIME(ex.extime), '%Y-%m-%d') = '{$dateToday}'
                 ORDER BY u.username ASC

我也把它放在我的构造函数

中

private $cone_read;
private $ixdata;

function __construct()
{
    // Call the Model constructor
    parent::__construct();

    $this->cone_read = $this->load->database('cone_read', TRUE);
    $this->ixdata = $this->load->database('cone_us', TRUE);

}

Answer 1

试试这个：

        $this->db->select("DATE_FORMAT(FROM_UNIXTIME(ex.extime), '%Y-%m-%d') as export_report, u.username as Username, ex.qdesc as Search_Performed, ex.to as Number_of_records_exported, ex.format as Format");    
        $this->db->from('cone.users u');
        $this->db->join('igdata.export_status as ex', 'u.id = ex.uid');
        $this->db->where("DATE_FORMAT(FROM_UNIXTIME(ex.extime), '%Y-%m-%d') = '$dateToday'");
        $this->db->order_by("u.username", "asc");
        $query = $this->db->get();

如何在codeigniter中的两个数据库之间进行连接

1 个答案: