Conditional order by in MYSQL. Should affect part of row - mysql

Question

The left table is a result of my query. And I need to sort it as the right table.
I need to order by p_id, if level >= 2. The blue box of right table is a target of order by.
Is it possible? Of course it is an example. Actual data is hundreds and really need to be sorted.
I searched a lot, but coudln't find the same case.
edit : this table will be returned as java.util.ArrayList. If this kind of 'order by' is not possilbe, is it possible in java.util.ArrayList?

Answer 1

I'm sure it's not possible in one query in MySQL.
In your diagram on the right, the ordering has been done in two separate steps:
Sort by id
Sort each block by p_id if level >= 2
That's quite difficult to do in MySQL as you would need to identify the blocks and then iterate over them, sorting each block separately.
I've done something similar where ordering within blocks was required and then selecting from those ordered blocks. You can view that here but as I said, I think that that SQL code is horribly complicated involving 5 temporary tables. You would probably need fewer temp tables, but it would still be a very complicated procedure, quite slow and hard to maintain.
"Actual data is hundreds and really need to be sorted."
Is there any reason why you can't just sort it as you want in code?
$blockStart = FALSE;
$count = 0;
foreach($dataArray as $data){
if($blockStart === FALSE){
$blockStart = $count;
}
if($data['level'] < 2){ //Block has finished
sortBlock($dataArray, $blockStart, $count);
$blockStart = $count;
}
$count++;
}
sortBlock($dataArray, $blockStart, $count - 1);
function sortBlock($dataArray, $indexStart, $indexEnd){
//Sort the elements of $dataArray, between $indexStart and $indexEnd inclusive
//by the value of p_id
}
Trying to solve a general programming problem in MySQL when you could solve it in 1/10th of the programmer time (and probably have it perform faster as well) in Java is not a good path to follow.

Answer 2

It is possible to do this in SQL, but it would be a very, very complicated query in MySQL. Here is the approach.
(1) Create a subquery that has the original ids and an indicator of whether something is in level 2 or not. The ids in this table are going to define the final order.
(2) Next, create a separate counter for each group in this above table. In other databases, you would use row_number(). In my SQL, this requires a correlated subquery. This provides the mapping from id to the new ordering.
(3) Next, create a counter for each group, but this time with the needed order (by id for the non-level2 group, by your rules for ordering).
(4) Join the tables together to get the matching.
(5) Order by the original id.
Here is an attempt:
select altord.*
from (select t.*,
(select count(*) from t t2 where t2.id <= t.id and ((t2.level = 2 and t1.level = 2) or (t2.level <> 2 and t1.level <> 2))
) as seqnum
from t
) ord join
(select t.*,
(select count(*) from t t2 where (t2.id <= t.id and t2.level <> 2 and t.level <> 2) or (t2.level = 2 and t.level = 2 and (t2.pid < t.pid or t2.pid = t.pid and t2.id < t.id)))
) as seqnum
) altord
on ord.seqnum = altord.seqnum
order by ord.id
I'm not sure if this SQL is correct, but the idea can be implemented in a single query.