如何解决SQL Server查询以查找4个连续值的总和
| 有人可以帮我找到四个连续值的总和,即最后四个值的总和。 喜欢:VALUE SUM
1 NULL
2 NULL
3 NULL
4 10
5 14
6 18
7 22
8 26
9 30
10 34
11 38
12 42
13 46
14 50
15 54
16 58
17 62
18 66
19 70
20 74
21 78
22 82
23 86
24 90
25 94
26 98
27 102
28 106
29 110
30 114
31 118
32 122
33 126
34 130
35 134
36 138
37 142
38 146
谢谢,
解决方法
select sum(select top 4 Value from [table] order by Value Desc)
也许
select sum(value)
from [Table]
where Value >= (Max(Value) - 4)
我实际上还没有尝试过,而且目前也不能尝试,但是它们应该可以使您更加接近。
, 快速尝试,获取您在问题中发布的结果(第1行,第3行不是NULL)。假设VALUE字段是唯一的且按升序排列:
-- Create test TABLE with 38 values in
DECLARE @T TABLE (Value INTEGER)
DECLARE @Counter INTEGER
SET @Counter = 1
WHILE (@Counter <= 38)
BEGIN
INSERT @T VALUES(@Counter)
SET @Counter = @Counter + 1
END
-- This gives the results
SELECT t1.VALUE,x.Val
FROM @T t1
OUTER APPLY(SELECT SUM(VALUE) FROM (SELECT TOP 4 VALUE FROM @T t2 WHERE t2.VALUE <= t1.VALUE ORDER BY t2.VALUE DESC) x) AS x(Val)
ORDER BY VALUE
至少,您应该看到我要走的方向。
, 假设ID可以给您最后4行。
SELECT SUM([SUM])
FROM
(
SELECT TOP 4 [SUM] FROM myTable ORDER BY ID DESC
) foo
每次查询时,它将读取最后4行。
如果这是错误的(例如,您需要连续四行的总和),请提供示例输出
, 如果您的“ 5”列是连续的,则可以执行以下操作
;WITH q (Value) AS (
SELECT 1
UNION ALL
SELECT q.Value + 1
FROM q
WHERE q.Value < 38
)
SELECT q.Value,CASE WHEN q.Value >= 4 THEN q.Value * 4 - 6 ELSE NULL END
FROM q
否则你可能会使用这样的东西
;WITH q (Value) AS (
SELECT 1
UNION ALL
SELECT q.Value + 1
FROM q
WHERE q.Value < 38
),Sequential (ID,Value) AS (
SELECT ID = ROW_NUMBER() OVER (ORDER BY Value),Value
FROM q
)
SELECT s1.Value,[SUM] = s1.Value + s2.Value + s3.Value + s4.Value
FROM Sequential s1
LEFT OUTER JOIN Sequential s2 ON s2.ID = s1.ID - 1
LEFT OUTER JOIN Sequential s3 ON s3.ID = s2.ID - 1
LEFT OUTER JOIN Sequential s4 ON s4.ID = s3.ID - 1
请注意,示例中的表ѭ8是实际表的存根。实际的陈述变成
;WITH Sequential (ID,Value
FROM YourTable
)
SELECT s1.Value,[SUM] = s1.Value + s2.Value + s3.Value + s4.Value
FROM Sequential s1
LEFT OUTER JOIN Sequential s2 ON s2.ID = s1.ID - 1
LEFT OUTER JOIN Sequential s3 ON s3.ID = s2.ID - 1
LEFT OUTER JOIN Sequential s4 ON s4.ID = s3.ID - 1
版权声明:本文内容由互联网用户自发贡献,该文观点与技术仅代表作者本人。本站仅提供信息存储空间服务,不拥有所有权,不承担相关法律责任。如发现本站有涉嫌侵权/违法违规的内容, 请发送邮件至 dio@foxmail.com 举报,一经查实,本站将立刻删除。