Как объединение 2 выбрать без дублирования

Question

Повторный вопрос

Возможно ли получить эти выходные данные с использованием codeigniter в одном запросе с использованием Codeigniter Mysql?

oid |  count(waiters) as total_waiters
----+-------------------------------
1   |      1 <-- john will be count as 1 even if assigned to 2 room
2   |      1
3   |      2 <-- count is 2 because different waiters are assigned with different room
4   |      0

---

Таблица заказов

oid |  name 
----+-------
1   |   aa   
2   |   bb   
3   |   cc   
4   |   dd     

Стол в комнате

Rid |  oid  |  waiter_assigned
----+-------+-----------------
1   |   1   |     john
2   |   1   |     john
3   |   2   |     john
4   |   3   |     mike
5   |   3   |     dude

---

Я пытался использовать union

$this->db->select('o.oid, "" AS tot_book_thera');
$this->db->from('order o');
$query1 = $this->db->get_compiled_select();

$this->db->select('r.oid, count(r.waiter_assigned) AS total_waiters');
$this->db->from('room r');
$this->db->group_by('r.waiter_assigned');
$query2 = $this->db->get_compiled_select();

Но я получаю это ...

oid |  count(waiters) as total_waiters
----+-------------------------------
1   |      1 
2   |      1
3   |      2
1   |      '' <-- not sure how to combine it with the 1st or how to exclude this or remove this...

Любая помощь очень ценится, спасибо, ребята!

Answer 1

У вас была правильная идея. Но, как утверждают другие, GROUP BY ваш лучший друг здесь. Кроме того, используйте DISTINCT, чтобы избавиться от счета официанта дважды для одного и того же заказа. Вот как должен выглядеть ваш код

// An inner select query whose purpose is to count all waiter per room
// The key here is to group them by `oid` since we are interested in the order
// Also, in the count(), use DISTINCT to avoid counting duplicates
$this->db->select('room.oid, count(DISTINCT room.waiter_assigned) AS total_waiters');
$this->db->from('room');
$this->db->group_by('room.oid');
$query1 = $this->db->get_compiled_select();

// If you run $this->db->query($query1)->result(); you should see
oid |  total_waiters
----+-------------------------------
1   |      1
2   |      1
3   |      2

// This is how you would use this table query in a join.
// LEFT JOIN also considers those rooms without waiters
// IFNULL() ensures that you get a 0 instead of null for rooms that have no waiters
$this->db->select('order.oid, order.name, IFNULL(joinTable.total_waiters, 0) AS total_waiters');
$this->db->from('order');
$this->db->join('('.$query1.') joinTable', 'joinTable.oid = order.oid', 'left');
$this->db->get()->result();

// you should see
oid |  name     |  total_waiters
----+-----------+-------------------------
1   |  aa       |      1
2   |  bb       |      1
3   |  cc       |      2
4   |  dd       |      0

Вот необработанный оператор SQL

SELECT order.oid, order.name, IFNULL(joinTable.total_waiters, 0) AS total_waiters
FROM order
LEFT JOIN (
    SELECT room.oid, count(DISTINCT room.waiter_assigned) AS total_waiters  
    FROM room
    GROUP BY room.oid
) joinTable ON joinTable.oid = order.oid