如何使用codeigniter中的连接查询从表中选择行？

Question

我有两张桌子competition_registration 和competition_schedule。我有从competition_registration 中选择行的选择查询。这是我的代码，如下所示：

$this->db->select('competition_registration.permenent_registration_number');
    $this->db->from('competition_registration');
    $where = "permenent_registration_number is  NOT NULL";
    $this->db->where($where);
    $this->db->join('competition_schedule', 'competition_schedule.competition_schedule_id = competition_registration.competition_schedule_id');
    $this->db->join('competition_schedule', 'competition_schedule.period_id = 6');
    $this->db->join('competition_schedule', 'competition_schedule.competition_level_id = 3');
    $query = $this->db->get();  echo $this->db->last_query();exit;

但这显示错误。任何人都可以检查这个查询并为我纠正它吗？

我想从competition_schedule_id 和competition_level_id 等于3 和period_id 等于6 的competition_schedule 表中选择两个表中的列。

Answer 1

希望对您有所帮助：

将competition_schedule.period_id = 6 和这个competition_schedule.competition_level_id = 3 放在 where 子句中，如下所示：

$this->db->select('competition_registration.permenent_registration_number');
$this->db->from('competition_registration');
$this->db->join('competition_schedule', 'competition_schedule.competition_schedule_id = competition_registration.competition_schedule_id');
$where = "competition_registration.permenent_registration_number is  NOT NULL";

$this->db->where($where);
$this->db->where('competition_schedule.period_id' , '6');
$this->db->where('competition_schedule.competition_level_id','3');

//$this->db->join('competition_schedule', 'competition_schedule.period_id = 6');
//$this->db->join('competition_schedule', 'competition_schedule.competition_level_id = 3');
$query = $this->db->get();  
echo $this->db->last_query();
exit;

更多：https://www.codeigniter.com/user_guide/database/query_builder.html