多个 GROUP BY 并根据第一个 GROUP BY 的结果选择值

Posted 2023-02-16

【中文标题】多个 GROUP BY 并根据第一个 GROUP BY 的结果选择值

【英文标题】：Multiple GROUP BY and select value dependent on result of the first GROUP BY

【发布时间】：2014-11-22 22:34:39

【问题描述】：

我得到以下数据库表：

combination_id  |   weight  |   group   |   std
-------------------------------------------------
    1           |   50      |   3       |   7   
    2           |   30      |   3       |   19
    3           |   30      |   3       |   19
    4           |   25      |   4       |   7

我按 group 和 std 列对条目进行分组，并总结 weight 列的值：

SELECT SUM(weight) as weight_sum, group, std FROM weight_table 
WHERE combination_id IN (1, 2, 3) 
GROUP BY group, std
ORDER BY weight_sum DESC;

结果如下：

    weight  |   group   |   std
-----------------------------------------------
    60      |   3       |   19  
    50      |   3       |   7
    25      |   4       |   7

现在我想做第二个 GROUP BY，但只在 group 列上，并在 weight 列上求和。结果中std列的值应该是上次查询中权重最高且同组的条目的std列的值。 所以对于 3 组，我希望为 std 选择 19，因为 60 是最高的 3 组的权重：

    weight  |   group   |   std
-----------------------------------------------
    110     |   3       |   19  
    25      |   4       |   7

我怎样才能做到这一点？ 我正在使用 sqlite 3。

【参考方案1】：

我想你想要这个：

SELECT SUM(weight) as weight_sum, group, max(std) as std FROM weight_table 
WHERE combination_id IN (1, 2, 3) 
GROUP BY group
ORDER BY weight_sum DESC;

换句话说，不要认为这是“多个分组”。将其视为单个聚合，您可以在其中获得权重总和以及 std 的最大值。

编辑：

我似乎误解了这个问题。这在 SQL lite 中有点痛苦。这是一种方法：

with w as (
      SELECT SUM(weight) as weight_sum, group, std
      FROM weight_table 
      WHERE combination_id IN (1, 2, 3) 
      GROUP BY group, std
     ),
     wmax as (
      SELECT group, MAX(weight_sum) as maxws
      FROM w
      GROUP BY gruop
     )
select w.group, sum(w.weight_sum) as weight_sum,
       max(case when w.weight_sum = wmax.weight_sum then std end) as std
from w left join
     wmax
     on w.group = wmax.group
group by w.group
order by weight_sum DESC;

【讨论】：

他想要std 基于最大weight_sum，这使它成为greatest-n-per-group，而不是简单的聚合。

@Clockwork-Muse 。 . .谢谢你。我确定了答案。