SQL子查询按列值获取行

Question

我有这样的PostgreSQL表：

CREATE TABLE foo(man_id, subgroup, power, grp)
AS VALUES
    (1, 'Sub_A',  4, 'Group_A'),
    (2, 'Sub_B', -1, 'Group_A'),
    (3, 'Sub_A', -1, 'Group_B'),
    (4, 'Sub_B',  6, 'Group_B'),
    (5, 'Sub_A',  5, 'Group_A'),
    (6, 'Sub_B',  1, 'Group_A'),
    (7, 'Sub_A', -1, 'Group_B'),
    (8, 'Sub_B',  2, 'Group_B'),
    (9, 'Sub_C',  2, 'Group_B');

功率计算如下：

• grp Group_A中的Subgroup Sub_A的总功率为（4 + 5）= 9
• grp Group_A中子组Sub_B的总功率为（（-1）+ 1）= 0
• grp Group_B中的Subgroup Sub_A的总功率为（（-1）+（-1））= -2
• grp Group_B中子组Sub_B的总功率为（6 + 2）= 8

因此，Group_A中Sub_A的功率不等于Group_B中Sub_A的功率

因此，Group_A中Sub_B的幂不等于Group_B中Sub_B的幂

我想查询数据库并获取行，对于相同的subgroup名称，所有其他grp名称的总power不相等。

推荐的方法是什么？

我可以找到总功率的总和：

SELECT sum(power) AS total_power
FROM   foo
GROUP  BY grp

MySQL解决方案也将被接受。

Answer 1

单程：

SELECT f.*
FROM  (
   SELECT subgroup
   FROM  (
      SELECT subgroup, grp, sum(power) AS total_power
      FROM   foo
      GROUP  BY subgroup, grp
      ) sub
   GROUP  BY 1
   HAVING min(total_power) <> max(total_power)  -- can fail for NULL values;
   ) sg
JOIN foo f USING (subgroup);

在您的示例中，除最后一行带有“ Sub_C”外，所有行均符合条件。

与您之前的问题密切相关：

• 对于给定的子组，所有组的总功率是否相等？

类似的解释和注意事项。

db <> 在这里拨弄

Answer 2

我认为解决您的问题的一种方法是，您希望合计一组中子组的力量，然后查找具有相同名称的子组是否存在于另一组具有不同功效的组中。

第一步是合计所需的力量：

SELECT grp, subgroup, sum(power) as power
FROM foo
GROUP BY grp, subgroup

那应该给你这样的结果：

grp      subgroup  power
-------  --------  -----
Group_A  Sub_A      9
Group_A  Sub_B      0
Group_B  Sub_A     -2
Group_B  Sub_B      8
Group_B  Sub_C      2

有了这些信息后，您就可以使用CTE将结果与自身连接起来，进行比较以获得所需的结果。 您无需指定是否要显示Sub_C ，如果“不存在”被视为具有“不同的总能力”，那么您将需要使用左联接并检查别名b空值。 连接中的<使得它使得每个差异仅在低阶组为grp1时出现一次。

WITH totals AS (
    SELECT grp, subgroup, sum(power) as power
    FROM foo
    GROUP BY grp, subgroup
    ORDER BY grp, subgroup
)
SELECT a.subgroup,
       a.grp as grp1, a.power as Power1,
       b.grp as grp2, b.power as Power2
FROM totals a
    INNER JOIN totals b ON b.subgroup = a.subgroup
                           and a.grp < b.grp
WHERE b.power <> a.power
ORDER BY a.subgroup, a.grp, b.grp 

Answer 3

with totals as (
    select grp, subgroup, sum(power) as total_power
    from foo
    group by grp, subgroup
)
select * from totals t1
where t1.total_power <> all (
    select t2.total_power from totals t2
    where t2.subgroup = t.subgroup and t2.grp <> t1.grp
)

要么

with totals as (
    select grp, subgroup, sum(power) as total_power
    from foo
    group by grp, subgroup
), matches as (
    select grp, subgroup, count(*) over (partition by subgroup, total_power) as matches
)
select * from counts where matches = 1;

Answer 4

我将使用窗口函数：

select f.*
from (select f.*,
             min(sum_value)) over (partition by group) as min_sum_value,
             max(sum_value)) over (partition by group) as max_sum_value,
      from (select f.*,
                   sum(value) over (partition by subgroup, group) as sum_value
            from foo f
           ) f
      ) f
where min_sum_value <> max_sum_value;