通过SUM排序Oracle查询，而无需选择SUM [英] Order Oracle query by SUM without selecting the SUM

问题描述

我有一张类似于以下表格的表格:

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
 ...   |  ...     |  ...

我想对两行之间的批次和缺陷相等的数量求和，然后按其数量的总和排序(lot3 = 9 + 5 + 5 = 19，lot4 = 12 ，lot1 = 7 + 2 = 9，lot2 = 4)，然后是数量(在每手内部)，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.

因此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

您的问题似乎与排序结果有关.解决方案是在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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