按 SUM 排序 Oracle 查询而不选择 SUM
我有一张类似于以下的表:
I have a table somewhat like the following one:
lot | defect | quantity
-------+----------+-----------
lot1 | c | 7
lot1 | c | 2
lot3 | e | 5
lot3 | b | 9
lot3 | a | 5
lot2 | d | 4
lot4 | c | 12
... | ... | ...
我想对行之间批次和缺陷数量相等的数量求和，然后按数量总和对批次进行排序(lot3=9+5+5=19，lot4=12, lot1=7+2=9, lot2=4)，然后是数量(每批内)，然后是缺陷.
I want to sum the quantities where the lot and defect are equal between rows and then order the lot by the sum of its quantity (lot3=9+5+5=19, lot4=12, lot1=7+2=9, lot2=4), then the quantity (inside each lot), and then the defect.
所以它应该导致以下结果:
So it should result in the following:
lot | defect | SUM(quantity)
-------+----------+----------------
lot3 | b | 9
lot3 | a | 5
lot3 | e | 5
lot4 | c | 12
lot1 | c | 9
lot2 | d | 4
... | ... | ...
我能想到的最接近的是以下查询:
The closest I can think of is the following query:
SELECT lot, defect, SUM(quantity)
FROM table
GROUP BY lot, defect
ORDER BY SUM(quantity), lot, defect
结果如下:
lot | defect | SUM(quantity)
-------+----------+----------------
lot4 | c | 12
lot1 | c | 9
lot3 | b | 9
lot3 | a | 5
lot3 | e | 5
lot2 | d | 4
... | ... | ...
推荐答案
您的问题似乎是关于对结果进行排序.解决办法是在ORDER BY中使用窗口函数:
Your question seems to be about ordering the results. The solution is to use window functions in ORDER BY:
SELECT lot, defect, SUM(quantity)
FROM table
GROUP BY lot, defect
ORDER BY SUM(SUM(quantity)) OVER (PARTITION BY lot) DESC,
lot, SUM(quantity) DESC, defect;
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