如何使用 PostgreSQL 中另一列的小计订购 id

Question

I have a table returned by a select query. Example:

id |  day   | count | 
-- | ------ | ----- | 
1  |  71    |   3   | 
1  |  70    |   2   |
1  |Subtotal|   5   |
2  |  70    |   5   |
2  |  71    |   2   |
2  |  69    |   2   |
2  |Subtotal|   9   |
3  |  69    |   1   |
3  |  70    |   1   |
3  |Subtotal|   2   |

the day column contains text values (so varchar)

subtotal is the sum of the counts for an id (eg id 2 has subtotal of 5 + 2 + 2 = 9)

I now want to order this table so the id's with the lowest subtotal count come first, and then ordered by day with subtotal at the end (like before)

Expected output:

id |  day   | count | 
-- | ------ | ----- | 
3  |  69    |   1   | 
3  |  70    |   1   |
3  |Subtotal|   2   |
1  |  70    |   2   |
1  |  71    |   3   |
1  |Subtotal|   5   |
2  |  69    |   2   |
2  |  70    |   5   |
2  |  71    |   2   |
2  |Subtotal|   9   |

I can't figure out how to order based on subtotal only?

i've tried multiple order by (eg: ORDER BY day = 'Subtotal' & a mix of others) and using window functions but none are helping. Cheers !

Answer 1

Not sure if it's directly applicable to your source query (since you haven't included it) however the ordering you require on the sample data can be done with:

order by Max(count) over(partition by id), day

Note - ordering by day works with your sample data but as it's a string it will not honour numeric ordering, this should really be ordered by the source of the numerical value - again since we don't have your actual query I can't suggest anything more applicable but I'm sure you can substitute the correct column/expression.

Answer 2

I just crated table with 3 columns and tried to reproduce your expected result. I assume that there might be a problem ordering by day, subtotal would be always on top, but it seems as working solution.

create table test
(
    id int,
    day varchar(15),
    count int
)

insert into test
values 
(1,'71',3),
(1,'70',2),
(2,'70',5),
(2,'71',2),
(2,'69',2),
(3,'69',1),
(3,'70',1)

select id, day, count
from
(
    select id, day, sum(count) as count
    from test
    group by id, rollup(day)
) as t
order by Max(count) over(partition by id), day