How to select from tableA sum of grouped numbers from tableB above their sums average in Oracle?

I have data like this:

tableA.ID
---------
1
2
3

tableB.ID tableB.NUM
--------------------
1         10
1         15
2         18
3         12
2         12
2         15
3         13
1         12

I need to select tableA IDs where the sum of their NUMs in tableB is above the average of all tableA IDs sums. In other words:

SUM ID=1 -> 10+15+12 = 37开发者_如何学C
SUM ID=2 -> 18+12+15 = 45
SUM ID=3 -> 12+13    = 25

AVG ALL IDs -> (37+45+25)/3 = 35

The SELECT must only show ID 1 and 2 because 37 > 35, 45 > 35 but 25 < 35.

This is my current query which is working fine:

SELECT tableA.ID
FROM tableA, tableB
WHERE tableA.ID = tableB.ID
HAVING SUM(tableB.NUM) > (
    SELECT AVG(MY_SUM)
    FROM (
        SELECT SUM(tableB.NUM) MY_SUM
        FROM tableA, tableB
        WHERE tableA.ID = tableB.ID
        GROUP BY tableA.ID
    )
)
GROUP BY tableA.ID

But I have a feeling there might be a better way without all those nested SELECTs. Perhaps 2, but 3 feels like too much. I'm probably wrong though.

For instance, why can't I do something simple like this:

SELECT tableA.ID
FROM tableA, tableB
WHERE tableA.ID = tableB.ID
HAVING SUM(tableB.NUM) > AVG(SUM(tableB.NUM))
GROUP BY tableA.ID

Or this:

SELECT tableA.ID, SUM(tableB.NUM) MY_SUM
FROM tableA, tableB
WHERE tableA.ID = tableB.ID
HAVING MY_SUM > AVG(MY_SUM)
GROUP BY tableA.ID

---

In SQL Server you can do this. Don't know if it works in Oracle. I'm sure I'll find out soon if it doesn't!

WITH cte As (
SELECT 
      tableA.ID, 
      SUM(tableB.NUM) AS MY_SUM, 
      AVG(SUM(tableB.NUM)) over() As Average
FROM tableA, tableB
WHERE tableA.ID = tableB.ID
GROUP BY tableA.ID
)
SELECT ID 
FROM cte 
WHERE MY_SUM > Average

---

This should reduce it by one level

SELECT tableA.ID
FROM tableA, tableB
WHERE tableA.ID = tableB.ID
GROUP BY tableA.ID
HAVING SUM(tableB.NUM) > (
    SELECT SUM(tableB.NUM)/COUNT(DISTINCT tableA.ID)
    FROM tableA, tableB
    WHERE tableA.ID = tableB.ID
)