php - 如何在codeigniter中使用where子句连接两个表？

Question

我正在开发好友请求和通知系统!当用户发送好友请求时，第二个用户收到通知!当第二个用户接受它时，通知表的 did_read 字段中更新为 1。我已经尝试过，但它显示了所有通知，包括已读(did_read = 1)。请告诉我我的代码中的问题出在哪里。谢谢

我的函数如下

public function manage_friend_req($loginuser){


    $this->db->select('from_user,to_user,did_read,avatar');
    $this->db->from('notifications');
    $this->db->join('users','users.username = notifications.from_user');
    $this->db->where(array('notifications.to_user'=>$loginuser));
    $this->db->where('notifications.did_read',0);
    $query = $this->db->get();        
    return $query->result();

 }  // 'did_read' field in database is an enum.

Answer 1

1. $this->db->or_where('notifications.did_read',0);//使用或
2. $this->db->where('notifications.did_read',0);//使用和