Question

如果该组中的任何值为0，我想扫描一个表并将组内的值重新分配给0。我似乎无法通过搜索{{1}的各种组合来找出解决方案。 }，Group By和Partition。

我开始的数据看起来像

Any

我希望结果显示为（包括换行符以便于解释）

CREATE TABLE #QP
(
    [Method] VARCHAR(1),
    [Station] VARCHAR(1),
    [Instrument] VARCHAR(20),
    [LastAnalysis] DATE,
    [DaysPassed] INT
)

INSERT INTO #QP
(Method, Station, Instrument, LastAnalysis, DaysPassed)
VALUES
('A', 1, 'Polaris', '2016-07-19', 21),
('B', 1, 'Polaris', '2016-08-04', 5),
('C', 1, 'Polaris', '2016-07-31', 9),
('A', 2, 'Polaris', '2016-07-31', 9),
('B', 2, 'Polaris', '2016-08-09', 0),
('C', 2, 'Polaris', '2016-07-23', 17),
('A', 3, 'Polaris', '2016-08-09', 0),
('B', 3, 'Polaris', '2016-07-27', 13),
('C', 3, 'Polaris', '2016-07-19', 21)

我到目前为止最接近的是使用

Method  Station Instrument  LastAnalysis    DaysPassed  Weight
A             1    Polaris    2016-07-19            21      21 
B             1    Polaris    2016-08-04            5        5
C             1    Polaris    2016-07-31            9        6

A             2    Polaris    2016-07-31            9        0
B             2    Polaris    2016-08-09            0        0
C             2    Polaris    2016-07-23            17       0

A             3    Polaris    2016-08-09            0        0
B             3    Polaris    2016-07-27            13       0
C             3    Polaris    2016-07-19            21       0

但是，当SELECT *, CASE WHEN 0 = ANY(SELECT DaysPassed FROM #QP) THEN 0 ELSE DaysPassed END AS [Weight] FROM #QP WHERE Instrument = 'Polaris' ORDER BY Station, Method组中的值保持原样时，这会将Weight列中的每个值设置为0。

如果这已有答案，我很想知道用于查找答案的正确搜索字词。

Answer 1

我认为使用min()窗口功能会起作用。试一试：

SELECT *,
       case when min(DaysPassed) over (partition by station) = 0 then 0 else DaysPassed end as [Weight]
FROM #QP 
WHERE Instrument = 'Polaris'
ORDER BY Station, Method

Answer 2

java.time.YearMonth

返回

SELECT *
      ,Weight= Sign(min(DaysPassed) over (Partition By Station)) * DaysPassed
FROM #QP 
WHERE Instrument = 'Polaris'
ORDER BY Station, Method

当组中的任何值为0时，将组内的所有值都设置为0

2 个答案: