Oracle sql：使用GROUP BY ROLLUP进行排序

Question

I'm looking everywhere for an answer but nothing seems to compare with my problem. So, using rollup with query:

select year, month, count (sale_id) from sales 
group by rollup (year, month);

Will give the result like:

  YEAR      MONTH     TOTAL
  2015          1     200
  2015          2     415
  2015       null     615
  2016          1     444
  2016          2     423
  2016       null     867
  null       null    1482

And I would like to sort by total desc, but I would like year with biggest total to be on top (important: with all records that compares to that year), and then other records for other years. So I would like it to look like:

  YEAR      MONTH     TOTAL
  null       null    1482
  2016       null     867
  2016          1     444
  2016          2     423
  2015       null     615
  2015          2     415
  2015          1     200

Or something like that. Main purpose is to not "split" records comparing to one year while sorting it with total. Can somebody help me with that?

Answer 1

Hmmm. I think this does what you want:

select year, month, count(sale_id) as cnt
from sales 
group by rollup (year, month)
order by sum(count(sale_id)) over (partition by year) desc, year;

Actually, I've never use window functions in an order by with a rollup query. I wouldn't be surprised if a subquery were necessary.

Answer 2

Try using window function max to get max of total for each year in the order by clause:

select year, month, count(sale_id) total
from sales
group by rollup(year, month)
order by max(total) over (partition by year) desc, total desc;

Answer 3

I think you need to used GROUPING SETS and GROUP_ID's. These will help you determine a NULL caused by a subtotal. Take a look at the doc: https://docs.oracle.com/cd/B19306_01/server.102/b14223/aggreg.htm