Question

我有这样的查询：

;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，不是Value @inputdate, Value如何在@inputdate和前一天(DATEADD(day, -1, @inputdate ))之间获得@inputdate = '20130708'的增量？

被修改：

很抱歉不清楚，'Value'是int类型。例如，如果Value，'20130708'的{{1}}为30，前一天'20130707'的'值'为20，则返回（30 - 20）为10。

Answer 1

像这样的东西，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，您可以使用以下代码：

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：：此处ret_dt是日期，收盘价是收盘价，为了说明性目的，我添加了一个额外的潜在客户栏。

sql计算一天和前一天的列值之间的差值

2 个答案: