php - 如何合并2选择而不重复

迭代问题
是否可以在使用codeigniter Mysql的单个查询中使用codeigniter获得此输出？

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

订货表

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

房间桌子

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

我试着用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();

但我知道。。。

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...

非常感谢你们的帮助，谢谢！

最佳答案

你的想法是对的。但正如其他人所说，GROUP BY是你在这里最好的朋友。另外，使用DISTINCT来消除对同一个服务生的两次计数。这就是你的代码应该是什么样子

// 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

下面是原始SQL语句

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

关于php - 如何合并2选择而不重复，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/58780390/