Codeigniter 模型多连接语句返回最后一条语句

Question

我有一个连接语句，它从具有 TeamID 的团队表中连接 HomeID 和 AwayID。当我连接这两个表时，它只返回最后一个连接语句的值。这是我的模型：-

function get_fixtures(){
    $where=array(
    'gameweek'=>1,
    );

    $this->db->select();
    $this->db->from('matches AS M');
    $this->db->join('team AS T2', 'T2.teamID=M.awayClubID' );
    $this->db->join('team AS T1', 'T1.teamID=M.homeClubID');
    $this->db->where($where);
    $query = $this->db->get();
    return $query->result_array();
}

当我打印出结果时，它只返回 T1 结果。如果有人可以提供帮助将不胜感激:-)

Answer 1

不能直接返回结果数组

 $this->db->select();
            $this->db->from('matches M');
            $this->db->join('team T2', 'T2.teamID=M.awayClubID' );
            $this->db->join('team T1', 'T1.teamID=M.homeClubID');
            $this->db->where($where);
            $query = $this->db->get();

            foreach($query->result_array() as $que)
            {
                  $n[] = $que;
            }
            retunr $n;

Answer 2

谢谢大家，我已经通过区分选择语句中的列自己弄清楚了。这是代码：-

function get_fixtures(){
                $where=array(
                'gameweek'=>1,
                );
                $this->db->select('m.*, T1.clubName T1name,T2.clubName T2name');
                $this->db->from('matches as m');
                $this->db->join('team AS T1', 'T1.teamID=M.homeClubID', 'left');
               $this->db->join('team AS T2', 'T2.teamID=M.awayClubID', 'left');


                $this->db->where($where);
                $query = $this->db->get();
                return $query->result_array();

     }

Answer 3

您正在为同一张表使用 2 aliases。因此不要为team 表使用别名。

$this->db->select('*');
$this->db->from('matches AS M');
$this->db->join('team', 'team.teamID=M.awayClubID', 'inner');
$this->db->join('team', 'team.teamID=M.homeClubID', 'inner');
$this->db->where($where);
$query = $this->db->get();
return $query->result_array();

或仅第二次使用（临时解决方案）。

$this->db->select('*');
$this->db->from('matches AS M');
$this->db->join('team', 'team.teamID=M.awayClubID', 'inner');
$this->db->join('team as t', 't.teamID=M.homeClubID', 'inner');
$this->db->where($where);
$query = $this->db->get();
return $query->result_array();

编辑：

$this->db->select('*');
$this->db->from('matches AS M');
$this->db->join('team as T', 'T.teamID=M.awayClubID OR T.teamID=M.homeClubID', 'inner');
$this->db->where($where);
$query = $this->db->get();
return $query->result_array();