Question

我在列中有示例值，如下所示：

值上升然后下降到0并再次上升。我需要计算新列中行之间的差异以及所有值的这些差异之和（累积和）。值56和5是零的新差异总和是819。底部的例子＆gt; （23-2）+（43-23）+（109-43）+ .. +（654-643）+（5）+（23-5）+ ..

Answer 1

好的，这是我的尝试。但是，您需要添加一个标识字段（我称之为“AddSequence”），该字段以1开头表示第一个值（“2”），并为每个其他值加1。

SELECT SUM(C.Diff) FROM
(
 SELECT CASE WHEN (A.[Value] - (SELECT [Value] FROM [TestValue] AS B WHERE B.[AddSequence]= A.[AddSequence]-1)) > 0
    THEN (A.[Value] - (SELECT [Value] FROM [TestValue] AS D WHERE D.[AddSequence]= A.[AddSequence]-1))
    ELSE 0
    END AS Diff
 FROM [TestValue] AS A
) AS C

我忽略了第一个解决方案，即每当差异为负时我们必须重新开始。

Answer 2

我认为你正在寻找类似的东西：

SELECT SUM(a - b)) as sum_of_differences
FROM ...

Answer 3

已编辑 - 基于您的问题编辑（T-SQL）我不知道如何在不添加ID的情况下执行此操作。如果您添加了ID，则会显示您在编辑之前发布的确切输出。可能有更好的方法，但这很快又很脏 - 一次性拍摄。使用SELF JOIN。差异是最初新列的名称。

UPDATE A
SET differences = CASE WHEN A.[values] > B.[Values] THEN A.[values] -  B.[Values]
ELSE A.[values] END
FROM SO_TTABLE A
JOIN SO_TTABLE B ON A.ID = (B.ID - 1)

输出

Select [Values], differences FROM SO_TTABLE 


   [values]     differences   
    ------------------------
    89          24             
    65           9            
    56          56             
    78           4             
    74           1            
    73          28             
    45          22             
    23          18              
     5           5              
   654          11           
   643         100            
   543         198           
   345          90            
   255          22            
   233         124            
   109          66             
    43          20             
    23          21              
     2           0

Answer 4

我认为你想要这个差异，我已经在sqlite中测试过了

SELECT CASE WHEN (v.value - val) < 0 THEN 0 ELSE (v.value - val) END AS differences
FROM v,
(SELECT rowid, value AS val FROM v WHERE rowid > 1) as next_val
WHERE v.rowid = next_val.rowid - 1

总和

SELECT SUM(differences) FROM
(
    SELECT CASE WHEN (v.value - val) < 0 THEN 0 ELSE (v.value - val) END AS differences
    FROM v,
    (SELECT rowid, value AS val FROM v WHERE rowid > 1) AS next_val
    WHERE v.rowid = next_val.rowid - 1
)

行差异的总和

4 个答案: