Sql 顺序链接数据的滚动求和

我正在处理一个大型的遗留数据集，其中包含顺序相关的数据，我没有文字来解释这些数据，因此我制作了一幅漂亮的绘画图像。这当然不是真正的数据集，但很接近。本例中有三个序列

每个记录都有一个ID和一个值。它还有一个指向下一个相关ID的指针。序列长度是随机的，当下一个相关ID达到0值时停止。所有记录在一个序列中只使用一次，这意味着它们不能合并或拆分。一个序列只能由一条记录组成

我需要完成的是使用SQL查询（SQL server 2014）获取序列中每个记录的滚动和。如果序列中有公共标识符，我知道如何执行此操作，但在本例中没有

我已经能够在Excel中通过查找上一个总和（如果存在）并添加当前值来完成它（不管它值多少）。但我无法将其转换为SQL。有人知道从哪里开始实现SQL中“滚动求和结果”列的最终目标吗

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您需要类似于递归查询的东西

你可以用CTE做这个。这是对您的数据的测试（您寻找的列是“cumul”，其他列用于帮助了解发生了什么）：

要按相反的顺序执行此操作。。。可能不止一种方法。从我的头顶上，我会得到每个链（由其lastid标识）的最小和最大累积值，然后我会应用阶梯算法——在这种情况下，递减滚动和是VALMIN+VALMAX-rolling

比如

WITH sequenza AS (
    SELECT       
        id, 
        value,
        nextid,
        id AS lastid,
        value as cumul
    FROM       
        items
    WHERE nextid = 0
    UNION ALL
    SELECT
        curr.id, 
        curr.value,
        curr.nextid,
        prev.lastid,
        prev.cumul + curr.value AS cumul
    FROM 
        items AS curr
        INNER JOIN sequenza AS prev
            ON prev.id = curr.nextid
),
sequenza2 AS (
    SELECT       
        id, 
        value,
        nextid,
        id AS lastid,
        value as cumul
    FROM       
        items
    WHERE nextid = 0
    UNION ALL
    SELECT
        curr.id, 
        curr.value,
        curr.nextid,
        prev.lastid,
        prev.cumul + curr.value AS cumul
    FROM 
        items AS curr
        INNER JOIN sequenza2 AS prev
            ON prev.id = curr.nextid
)
SELECT sequenza.*, m1+m2-cumul AS cumulasc FROM sequenza
JOIN (
  SELECT lastid, MIN(cumul) AS m1, MAX(cumul) AS m2
  FROM sequenza2
  GROUP BY lastid
) AS cirpo ON (sequenza.lastid = cirpo.lastid)
ORDER BY sequenza.lastid, cumul DESC

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是否有序列的合并，换句话说，两行可以使用同一个指针？@Charlieface这是一个重要的细节。它们不能合并。我已经把这篇文章放在了主要的帖子里。这看起来很有希望！但当我执行查询时，序列似乎发生了翻转（从序列中的最后一个计数到第一个计数），有没有办法扭转这种局面？@Monthy编辑的答案…谢谢！你的回答几乎就是我需要的。我注意到它将序列中的最后一项作为第一个和（例如：在id 31处，累积结果是82，这是序列中的最后一个值，下一个id将正确地添加29，这是id 31中的值）。为了使其与Excel示例完全相同，我从联接和公式中删除了m1，并添加了如下当前值：“选择sequenza.*，m2 cumul+值作为sequenza中的cumulasc”。这个现在很好用。再次感谢！

WITH sequenza AS (
    SELECT       
        id, 
        value,
        nextid,
        id AS lastid,
        value as cumul
    FROM       
        items
    WHERE nextid = 0
    UNION ALL
    SELECT
        curr.id, 
        curr.value,
        curr.nextid,
        prev.lastid,
        prev.cumul + curr.value AS cumul
    FROM 
        items AS curr
        INNER JOIN sequenza AS prev
            ON prev.id = curr.nextid
)
SELECT * FROM sequenza
WHERE id = 31;

WITH sequenza AS (
    SELECT       
        id, 
        value,
        nextid,
        id AS lastid,
        value as cumul
    FROM       
        items
    WHERE nextid = 0
    UNION ALL
    SELECT
        curr.id, 
        curr.value,
        curr.nextid,
        prev.lastid,
        prev.cumul + curr.value AS cumul
    FROM 
        items AS curr
        INNER JOIN sequenza AS prev
            ON prev.id = curr.nextid
),
sequenza2 AS (
    SELECT       
        id, 
        value,
        nextid,
        id AS lastid,
        value as cumul
    FROM       
        items
    WHERE nextid = 0
    UNION ALL
    SELECT
        curr.id, 
        curr.value,
        curr.nextid,
        prev.lastid,
        prev.cumul + curr.value AS cumul
    FROM 
        items AS curr
        INNER JOIN sequenza2 AS prev
            ON prev.id = curr.nextid
)
SELECT sequenza.*, m1+m2-cumul AS cumulasc FROM sequenza
JOIN (
  SELECT lastid, MIN(cumul) AS m1, MAX(cumul) AS m2
  FROM sequenza2
  GROUP BY lastid
) AS cirpo ON (sequenza.lastid = cirpo.lastid)
ORDER BY sequenza.lastid, cumul DESC