多行加入

Question

我有這個疑問/ PHP代碼：

public function get_days_appointments($day) { 
    $day_start = date('Y-m-d 00:00:00', $day); 
    $day_end = date('Y-m-d 23:59:59', $day); 
     return $this->db->select('e.id, 
      e.start_datetime, 
      e.end_datetime, 
      e.hash, 
      e.id_google_calendar, 
      u1.first_name AS customer_first_name, 
      u1.last_name AS customer_last_name, 
      u1.email AS customer_email, 
      u1.phone_number AS customer_phone_number, 
      u2.first_name AS provider_first_name, 
      u2.last_name AS provider_last_name, 
      u2.email AS provider_email, 
      u2.phone_number AS provider_phone_number, 
      s.name, 
      s.duration, 
      s.price, 
      s.description, 
      t.push_identifier') 
     ->from('ea_appointments e') 
     ->join('ea_users u1', 'e.id_users_customer = u1.id','left') 
     ->join('ea_users u2', 'e.id_users_provider = u2.id','left') 
     ->join('ea_services s', 'e.id_services = s.id','left') 
     ->join('ea_push_ids t', 'e.id_users_customer = t.user_id','left') 
     //->join('ea_cellcarrier cc', 'u1.id_cellcarrier = cc.id','left') 
     ->where('e.start_datetime >',$day_start) 
     ->where('e.start_datetime <',$day_end) 
    ->get()->result(); 
} 

我的問題是，有多個push_identifiers（ea_push_ids），但此查詢只檢索一個。我如何檢索屬於某個用戶的所有推送標識符而不會搞亂查詢的其餘部分？

Answer 1

您可以使用GROUP_CONCAT將所有ID放入以逗號分隔的字符串中。 （注意：我改變了別名push_identifiers與s）

public function get_days_appointments($day) { 
    $day_start = date('Y-m-d 00:00:00', $day); 
    $day_end = date('Y-m-d 23:59:59', $day); 
     return $this->db->select('e.id, 
      e.start_datetime, 
      e.end_datetime, 
      e.hash, 
      e.id_google_calendar, 
      u1.first_name AS customer_first_name, 
      u1.last_name AS customer_last_name, 
      u1.email AS customer_email, 
      u1.phone_number AS customer_phone_number, 
      u2.first_name AS provider_first_name, 
      u2.last_name AS provider_last_name, 
      u2.email AS provider_email, 
      u2.phone_number AS provider_phone_number, 
      s.name, 
      s.duration, 
      s.price, 
      s.description, 
      t.push_identifiers') 
     ->from('ea_appointments e') 
     ->join('ea_users u1', 'e.id_users_customer = u1.id','left') 
     ->join('ea_users u2', 'e.id_users_provider = u2.id','left') 
     ->join('ea_services s', 'e.id_services = s.id','left') 
     ->join('(SELECT GROUP_CONCAT(push_identifier) AS push_identifiers, user_id 
     FROM ea_push_ids 
     GROUP BY user_id) t', 'e.id_users_customer = t.user_id','left') 
     ->where('e.start_datetime >',$day_start) 
     ->where('e.start_datetime <',$day_end) 
    ->get()->result(); 
} 

或者這樣

public function get_days_appointments($day) { 
    $day_start = date('Y-m-d 00:00:00', $day); 
    $day_end = date('Y-m-d 23:59:59', $day); 
     return $this->db->select('e.id, 
      e.start_datetime, 
      e.end_datetime, 
      e.hash, 
      e.id_google_calendar, 
      u1.first_name AS customer_first_name, 
      u1.last_name AS customer_last_name, 
      u1.email AS customer_email, 
      u1.phone_number AS customer_phone_number, 
      u2.first_name AS provider_first_name, 
      u2.last_name AS provider_last_name, 
      u2.email AS provider_email, 
      u2.phone_number AS provider_phone_number, 
      s.name, 
      s.duration, 
      s.price, 
      s.description, 
      (SELECT GROUP_CONCAT(t.push_identifier) 
     FROM ea_push_ids t 
     WHERE t.user_id = e.id_users_customer) AS push_identifiers') 
     ->from('ea_appointments e') 
     ->join('ea_users u1', 'e.id_users_customer = u1.id','left') 
     ->join('ea_users u2', 'e.id_users_provider = u2.id','left') 
     ->join('ea_services s', 'e.id_services = s.id','left') 
     ->where('e.start_datetime >',$day_start) 
     ->where('e.start_datetime <',$day_end) 
    ->get()->result(); 
}