SQL 使用 group by 逐列聚合

Question

场景：

下面是我的table。

+-------+-----------+--------+--------+
|col2   |col3       |col4    |col5    |
+-------+-----------+--------+--------+
|    1.0|          2|       a|      a1|
|    1.0|          1|       a|      a2|
|    1.0|          2|       b|      a3|
|    2.0|          1|       a|      a1|
|    2.0|          1|       a|      a2|
+-------+-----------+--------+--------+

我需要得到如下的聚合结果。

+-------+-----------+-----------+-----------+-----+
|col2   |col3       | field_name|field_value|count|
+-------+-----------+-----------+-----------+-----+
|    2.0|          1|   col3    |          1|    2|
|    1.0|          1|   col3    |          1|    1|
|    1.0|          2|   col3    |          2|    2|
|    2.0|          1|   col4    |          a|    2|
|    1.0|          1|   col4    |          a|    1|
|    1.0|          2|   col4    |          a|    1|
|    1.0|          2|   col4    |          b|    1|
|    1.0|          1|   col5    |         a2|    1|
|    1.0|          2|   col5    |         a1|    1|
|    1.0|          2|   col5    |         a3|    1|
|    2.0|          1|   col5    |         a1|    1|
|    2.0|          1|   col5    |         a2|    1|
+-------+-----------+-----------+-----------+-----+

实施的解决方案：

我通过创建三个不同的 tables[T1、T2、T3] 来实施解决方案。然后对于每个 table，我以编程方式创建了 [rowset1、rowset2、rowset3] 以将所有组合成一个 table

select col2, col3, col3, count(*) from calc group by col2, col3;
T1 :
|col2   |col3       |col3       |count|
+-------+-----------+-----------+-----+
|    2.0|          1|          1|    2|
|    1.0|          1|          1|    1|
|    1.0|          2|          2|    2|
+-------+-----------+-----------+-----+

rowset1 = [[2.0,1,col3,1,2,18000], [1.0,1,col3,1,1,18000], [1.0,2,col3,2,2,18000]]

select  col2, col3,  col4, count(*) from calc group by col2, col3, col4;
T2:
+-------+-----------+--------+-----+
|col2   |col3       |col4    |count|
+-------+-----------+--------+-----+
|    2.0|          1|       a|    2|
|    1.0|          1|       a|    1|
|    1.0|          2|       a|    1|
|    1.0|          2|       b|    1|
+-------+-----------+--------+-----+
rowset2 = [[2.0,1,col4,a,2,18000], [1.0,1,col4,a,1,18000], [1.0,2,col4,a,1,18000],[1.0,2,col4,b,1,18000]]

select col2, col3, col5 , count(*) from calc group by col2, col3, col5;
T3:
+-------+-----------+--------+-----+
|col2   |col3       |col5    |count|
+-------+-----------+--------+-----+
|    1.0|          1|      a2|    1|
|    1.0|          2|      a1|    1|
|    1.0|          2|      a3|    1|
|    2.0|          1|      a1|    1|
|    2.0|          1|      a2|    1|
+-------+-----------+--------+-----+

rowset3 = [1.0,2,col5,b,1,18000], [1.0,1,col5,a2,1,18000], [1.0,2,col5,a1,1,18000], [1.0,2,col5,a3,1,18000], [2.0,1,col5,a1,1,18000], [2.0,1,col5,a2,1,18000]]

问题:

我怎样才能在 SQL 中实现相同的目标，而无需创建行集 1、2、3 并将其组合成一个 table？

Answer 1

如果我没理解错的话，你想将三个聚合合二为一。一种方法是使用cross join带入信息来区分每个聚合：

select col2, col3, field_name,
       (case when field_name = 'col3' then col3
             when field_name = 'col4' then col4
             when field_name = 'col5' then col5
        end) as field_value,
       count(*) as cnt
from t cross join
     (select 'col3' as field_name union all
      select 'col4' as field_name union all
      select 'col5' as field_name
     ) f
group by col2, col3, field_name, field_value
order by field_name, col2 desc, col3;

Here 是一个 db<>fiddle.

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