SQL - 从同一列中减去值

Question

我有一张如下表

fab_id  x  y  z  m  
12      14 10 3  5
12      10 10 3  4

这里我在 id 上使用 group by 子句。现在我想减去那些具有相似 id 的列值。 例如，按 id (12) 分组。现在减去 (14-10)X, (10-10)Y, (3-3)z, (5-4)m

我知道有一个聚合函数 sum 用于加法，但是有什么函数可以用来减去这个值。

或者有没有其他方法可以达到效果。

注意--ve 中的值可能会有所变化。那么有什么函数可以处理这个吗？

再举一个例子 - （按correction_date desc 排序，结果将首先显示最近的更正）

fab_id  x  y  z  m  correction_date
14      20 12 4  4   2014-05-05 09:03
14      24 12 4  3   2014-05-05 08:05
14      26 12 4  6   2014-05-05 07:12

所以结果是在 id (14) 上实现分组。现在减去 (26-20)X, (12-12)Y, (4-4)z, (6-4)m

Answer 1

现在，您已经提供了有关如何处理更多记录的更多信息，并且您透露涉及时间列，这是一个可能的解决方案。查询选择每个 fab_id 的第一条和最后一条记录并减去这些值：

select 
  fab_info.fab_id,
  earliest_fab.x - latest_fab.x,
  earliest_fab.y - latest_fab.y,
  earliest_fab.z - latest_fab.z,
  earliest_fab.m - latest_fab.m
from 
(
  select 
    fab_id, 
    min(correction_date) as min_correction_date,
    max(correction_date) as max_correction_date
  from fab
  group by fab_id
) as fab_info
inner join fab as earliest_fab on 
  earliest_fab.fab_id = fab_info.fab_id and 
  earliest_fab.min_correction_date = fab_info.min_correction_date
inner join fab as latest_fab on 
  latest_fab.fab_id = fab_info.fab_id and 
  latest_fab.min_correction_date = fab_info.max_correction_date;

Answer 2

假设您总是想从最大值中减去最小值：

select 
  fab_id, 
  max(x) - min(x), 
  max(y) - min(y), 
  max(z) - min(z), 
  max(m) - min(m)
from fab
group by fab_id;

Answer 3

正如你所说，总会有两行，你可以简单地做一个“自连接”并相互减去值：

SELECT t1.fab_id, t1.x - t2.x as diffx, t1.y - t2.y as diffy, <remainder columns here>
from <table> t1
inner join <table> t2 on t1.fab_id = t2.fab_id and t1.correctiondate > t2.correctiondate

如果您有超过两行，那么您需要进行子查询或使用窗口排名函数来计算每个 fab_id 的最大和最小更正日期，然后您可以通过加入这两个子查询来执行与上述相同的操作在一起而不是

Answer 4

不幸的是，SQL Server 2012 有方便的FIRST_VALUE()/LAST_VALUE() OLAP 函数，所以在超过 2 行的情况下，我们必须做一些不同的事情：

SELECT fab_id, SUM(CASE WHEN latest = 1 THEN -x ELSE x END) AS x,
               SUM(CASE WHEN latest = 1 THEN -y ELSE y END) AS y,
               SUM(CASE WHEN latest = 1 THEN -z ELSE z END) AS z,
               SUM(CASE WHEN latest = 1 THEN -m ELSE m END) AS m

FROM (SELECT fab_id, x, y, z, m,
             ROW_NUMBER() OVER(PARTITION BY fab_id 
                               ORDER BY correction_date ASC) AS earliest,
             ROW_NUMBER() OVER(PARTITION BY fab_id 
                               ORDER BY correction_date DESC) AS latest
      FROM myTable) fab
WHERE earliest = 1
      OR latest = 1
GROUP BY fab_id
HAVING COUNT(*) >= 2

（和working fiddle。感谢@AK47 的初始设置。）

产生预期的结果：

FAB_ID   X   Y   Z   M
12       4   0   0   1
14       6   0   0   2

请注意，HAVING COUNT(*) >= 2 仅考虑发生更改的行（否则您会得到一些空结果列）。

Answer 5

;with Ordered as 
(
select 
    fab_id,x,y,z,m,date, 
    row_Number() over (partition by fab_id order by date desc) as Latest,
    row_Number() over (partition by fab_id order by date) as Oldest
from fab
)
select 
   O1.fab_id, 
   O1.x-O2.x,
   O1.y-O2.y, 
   O1.z-O2.z, 
   O1.m-O2.m
from Ordered O1 
join Ordered O2 on 
O1.fab_id = O2.fab_id
where O1.latest = 1 and O2.oldest = 1  

Answer 6

我认为如果你有一致的集合或两行，那么下面的代码应该适合你。

select fab_id ,max(x) - min(x) as x
        ,max(y) - min(y) as y
            ,max(z) - min(z) as z
            ,max(m) - main(m) as m
            from Mytable
group by fab_id

即使您在一个组中获得超过 2 行，它也会起作用，但减法将从最小值的最大值开始。希望对你有帮助。

编辑：SQL 小提琴DEMO

Answer 7

CTE 可以提供帮助：

WITH cte AS (
    SELECT
        -- Get the row numbers per fab_id ordered by the correction date
        ROW_NUMBER() OVER (PARTITION BY fab_id ORDER BY correction_date ASC) AS rid
        , fab_id, x, y, z, m
    FROM
        YourTable
)
SELECT
    fab_id
    -- If the row number is 1 then, this is our base value
    -- If the row number is not 1 then, we want to subtract it (or add the negative value)
    , SUM(CASE WHEN rid = 1 THEN x ELSE x * -1 END) AS x
    , SUM(CASE WHEN rid = 1 THEN y ELSE y * -1 END) AS y
    , SUM(CASE WHEN rid = 1 THEN z ELSE z * -1 END) AS z
    , SUM(CASE WHEN rid = 1 THEN m ELSE m * -1 END) AS m
FROM
    cte
GROUP BY
    fab_id

记住，40-10-20 等于 40 + (-10) + (-20)