如何在SQL中使用查询过分区来获得当前值、平均值和最大值？

我有一张表，其中显示了设备在某个区域和特定位置执行的一个点

working_date    device   points   area   location
19-06-2020        a        1       x       xa   
19-06-2020        a        2       x       xa 
19-06-2020        a        3       x       xa 
19-06-2020        a        4       x       xa
20-06-2020        a        5       x       xa
20-06-2020        a        6       x       xa
20-06-2020        a        7       x       xa
20-06-2020        a        8       x       xa
20-06-2020        a        9       x       xa

我想得到当前，平均值和最大值，按区域和位置分组。如果我选择任何一天，当前数量将显示最近工作日期的数量。同时，“平均数量”将显示设备工作的总体平均值。最后，最大数量将显示设备完成的总体最大点

根据上表，如果我选择21-06-2020，则预期结果：

working_date  area  location   device   current_qty  avg_qty   max_qty
21-06-2020     x       xa        a         5           4,5        5

平均数量来自日期的总数量/总数量，而最大数量来自所有日期的最大数量

到目前为止，我构建的查询是：

select t1.working_date, t1.device, t1.area, t1.location, t1.points_qty, t1.total_date,
sum(t1.pile_qty) over(partition by t1.working_date) / sum(t1.total_date) over(partition by t1.working_date) as avg_qty,
max(t1.pile_qty) over(partition by t1.working_date) as max_qty
from (
select working_date, device, points, area, location, count(points) as points_qty, count(distinct working_date) as total_date 
from table1 group by device, area, location
group by working_date, device, points, area, location) t1
group by working_date, device, points, area, location, pile_qty, total_date

通过上述查询，我得到：

working_date  area  location   device   current_qty  avg_qty   max_qty
21-06-2020     x       xa        a         5           5          5

我应该如何编写查询以获得所需的结果

提前谢谢

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然而，在这种情况下，还不清楚2020-06-20组的五项记录中应该取哪一项。您必须应用您的订购标准，才能将预期的订单订购到顶部。

我想，我有解决方案可供您选择。然而，我不确定答案是否能在不同的场景中提供正确的结果。下面是我的代码=> 请检查链接=>

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注意：-在这里，您可以看到，对于当前数量，我使用了您的输入日期21-06-2020，如从CTE选择GRP计数，其中工作日期您所需的输出显示21，这在示例数据中不可用。因此，很难理解计算出的数据……这意味着，如果我选择任何一天，所需的结果将始终为我提供最新的工作日期数量值从何而来？通过计算点，我不知道PostgreSQL，但在MS SQL Server中，聚合函数返回的数据类型与聚合列相同；我建议使用sum…*1.0/总和…答案有帮助吗？那请别忘了给他们投票！如果一个答案完全解决了您的问题，请不要忘记接受它来结束问题。

SELECT
    *,
    AVG(current_qty) OVER () as avg_qty,             -- 2
    MAX(current_qty) OVER () as max_qty
FROM (
    SELECT 
        working_date,
        area,
        location,
        device,
        COUNT(*) as current_qty                      -- 1
    FROM mytable
    GROUP BY working_date, device, area, location    -- 1
) s
WHERE working_date <= '2020-06-21'                   -- 3
ORDER BY working_date DESC
LIMIT 1

SELECT
    *,
    AVG(current_qty) OVER () as avg_qty,
    MAX(current_qty) OVER () as max_qty
FROM (
    SELECT 
        working_date,
        area,
        location,
        device,
        COUNT(*) OVER (PARTITION BY working_date) as current_qty
    FROM mytable
) s
WHERE working_date <= '2020-06-21'
ORDER BY working_date DESC
LIMIT 1

WITH CTE AS
    (
      SELECT working_date,area,location,device, 
             COUNT(working_date) GrpCount
      FROM MYTable 
      GROUP BY working_date,area,location,device
    
    ),y AS
    (SELECT area,location,device,GrpCount,
           (SELECT GrpCount FROM CTE WHERE working_date<TO_DATE('21-06-2020','DD-MM-YYYY') ORDER BY working_date DESC LIMIT 1)  current_qty  
    FROM CTE
    )
    SELECT TO_DATE('21-06-2020','DD-MM-YYYY'),area,location,device, 
           MAX(current_qty) current_qty,
           string_agg(GrpCount::text, ',') avg_qty,
           Max(GrpCount) max_qty
    FROM Y
    GROUP BY area,location,device