How sort this
a 1 15
a 2 3
a 3 34
b 1 55
b 2 44
b 3 8
to (by third column sum):
b 1 55
b 2 44
b 3 8
a 1 15
a 2 3
a 3 34
since (55+44+8) > (15+3+34)
If you are using SQL Server/Oracle/Postgresql you could use windowed SUM:
SELECT *
FROM tab
ORDER BY SUM(col3) OVER(PARTITION BY col) DESC, col2
LiveDemo
Output:
╔═════╦══════╦══════╗
║ col ║ col2 ║ col3 ║
╠═════╬══════╬══════╣
║ b ║ 1 ║ 55 ║
║ b ║ 2 ║ 44 ║
║ b ║ 3 ║ 8 ║
║ a ║ 1 ║ 15 ║
║ a ║ 2 ║ 3 ║
║ a ║ 3 ║ 34 ║
╚═════╩══════╩══════╝
You can do this using ANSI standard window functions. I prefer to use a subquery although this is not strictly necessary:
select col1, col2, col3
from (select t.*, sum(col3) over (partition by col1) as sumcol3
from t
) t
order by sumcol3 desc, col3 desc;
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