1
我有這個工作查詢:SQL查詢連接用笨的活動記錄
$q = $this->db->query('SELECT u.name FROM users u
JOIN user_group ug ON u.id = ug.user_id
JOIN groups g ON g.id = ug.group_id WHERE ug.group_id = "4" ');
而且我想aactive記錄來取代它。我想出了這樣的事情,但顯然它不工作:
$this->db->select('name');
$this->db->from('users');
$this->db->join('user_group', 'user_group.user_id = users.id);
$this->db->join('groups', 'groups.id = user_group.group_id');
$q = $this->db->get();
感謝 Leron
呀,這工作得很好。謝謝! – Leron 2012-03-29 20:50:05