sql - sql计算一天与前一天的列值之间的增量

Question

我有一个这样的查询：

;WITH A AS (SELECT * FROM T1 where T1.targetDate=@inputdate), 
      B AS (SELECT A.*, T2.SId, T2.Type, T2.Value
            FROM A 
            INNER JOIN T2 ON A.SId = T2.SId )
SELECT A.*, B.Type, B.Value 
FROM B

我的问题是，而不是获得Value如何@inputdate,获得与前一天Value之间的增量？@inputdate(DATEADD(day, -1, @inputdate ))

编辑：

抱歉不清楚，“值”是 int 类型。例如，如果@inputdate = '20130708'' Value20130708' 的值是 30，而前一天 '20130707' 的 'Value' 是 20，那么它应该返回 (30 - 20)，即 10。

Answer 1

像这样，并假设 Value 是 DATE 格式

;WITH A AS (SELECT * FROM T1 where T1.targetDate=@inputdate), 
      B AS (SELECT A.*, T2.SId, T2.Type, T2.Value
            FROM A 
            INNER JOIN T2 on A.SId = T2.SId )
SELECT A.*, T2.Type, T2.Value, DATEDIFF(DAY, b.Value, DATEDADD(DAY, -1,@InputDate)) AS Delta
FROM B

Answer 2

假设您有一个股票价格表：其中包含代码、日期和收盘价等，您可以使用以下内容：

select symb, ret_dt, close, (close-(lead(close,1) over (partition by symb order by ret_dt desc,close)))as difference, (lead(close,1) over 
(partition by symb order by ret_dt desc,close)) as lead
from stocks.nyse2010;

注意：这里ret_dt是日期，收盘价是收盘价，我已经添加了一个额外的引导列用于表示目的。