SQL order by same values only insteading of alphabetic order - mysql

Question

Suppose I have a table like this:
Order_ID | Box_ID | Item_ID
------------------------------------
B | 1 | b
B | 2 | b
A | 2 | a
A | 3 | a
A | 4 | a
Now when I order the table by order ID,
SELECT * FROM table ORDER BY Order_ID, Box_ID;
the result will look like this:
Order_ID | Box_ID | Item_ID
------------------------------------
A | 2 | a
A | 3 | a
A | 4 | a
B | 1 | b
B | 2 | b
But now, I want the sql statement to first consider putting same value of Box_ID together,
such that the result will look like this:
Order_ID | Box_ID | Item_ID
------------------------------------
A | 2 | a
B | 2 | b
// despite order_ID being B, but box_ID is equal to previous value, so group together
A | 3 | a
A | 4 | a
B | 1 | b
But the following code,
SELECT * FROM table ORDER BY Box_ID, Order_ID
Will produce the following result since it lists Box_ID in alphabetical order first.
Order_ID | Box_ID | Item_ID
------------------------------------
B | 1 | b
A | 2 | a
B | 2 | b
A | 3 | a
A | 4 | a
How can I output my desired result?

Answer 1

I think you want:
order by min(order_id) over(partition by box_id), box_id
In earlier versions, you can use a subquery:
select *
from mytable t
order by
(select min(order_id) from mytable t1 where t1.box_id = t.box_id),
box_id

Answer 2

In order versions of MySQL, you can use a correlated subquery:
select t.*
from table t
order by (select min(t2.order_id) from table t2 where t2.box_id = t.box_id),
box_id;

Answer 3

If you want the rows with most occurrences of Box_ID at the top, then you need this query:
SELECT Box_ID, COUNT(*) counter
FROM tablename
GROUP BY Box_ID
joined to the table like this:
SELECT t.*
FROM tablename t
INNER JOIN (
SELECT Box_ID, COUNT(*) counter
FROM tablename
GROUP BY Box_ID
) g ON g.Box_ID = t.Box_ID
ORDER BY g.counter DESC, Order_ID, Box_ID
See the demo.
Results:
> Order_ID | Box_ID | Item_ID
> :------- | -----: | :------
> A | 2 | a
> B | 2 | b
> A | 3 | a
> A | 4 | a
> B | 1 | b