计算两行之间的时差

时间:2022-06-11 16:52:49

I have a table that contains the following:

我有一张表,其中包括:

DataDate                 Value
2010-03-01 08:31:32.000  100
2010-03-01 08:31:40.000  110
2010-03-01 08:31:42.000  95
2010-03-01 08:31:45.000  101
.                        .
.                        .
.                        .

I need to multiply the value column by the difference in time between the current and previous rows and sum that for the entire day.

我需要将值列乘以当前行和前行之间的时间差,然后对这一项进行一整天的求和。

I currently have the data set up to come in every 10 seconds which makes for a simple conversion in the query:

我目前设置的数据每10秒就会出现一次,这使得查询的转换变得非常简单:

SELECT Sum((Value/6) FROM History WHERE DataDate BETWEEN @startDate and @endDate

Where @startDate and @endDate are today's date at 00:00:00 and 11:59:59.

@startDate和@endDate是今天在00:00:00和11:59:59的日期。

Before I set the data to be collected every 10 seconds it was collected whenever the Value changed. There aren't any duplicate entries in terms of time, the minimum time difference is 1 second.

在我将数据设置为每10秒收集一次之前,每当值发生变化时,它都会被收集。时间上没有重复的条目,最小的时间差是1秒。

How can I set up a query to get the elapsed time between rows for the case when I don't know the time interval between readings?

当我不知道读取之间的时间间隔时,如何设置查询来获取行之间的运行时间?

I am using SQL Server 2005.

我正在使用SQL Server 2005。

2 个解决方案

#1


120  

WITH    rows AS
        (
        SELECT  *, ROW_NUMBER() OVER (ORDER BY DataDate) AS rn
        FROM    mytable
        )
SELECT  DATEDIFF(second, mc.DataDate, mp.DataDate)
FROM    rows mc
JOIN    rows mp
ON      mc.rn = mp.rn - 1

In SQL Server 2012+:

在SQL Server 2012 +:

SELECT  DATEDIFF(second, pDataDate, dataDate)
FROM    (
        SELECT  *,
                LAG(dataDate) OVER (ORDER BY dataDate) pDataDate
        FROM    rows
        ) q
WHERE   pDataDate IS NOT NULL

#2


2  

A little tweak on Quassnoi's query if you prefer not to use a Subselect would be:

如果您不喜欢使用子选择,那么对Quassnoi的查询做一个小小的调整就是:

SELECT
      DATEDIFF(second, LAG(dataDate) OVER (ORDER BY dataDate), dataDate)
FROM  rows
WHERE LAG(dataDate) OVER (ORDER BY dataDate) IS NOT NULL

#1


120  

WITH    rows AS
        (
        SELECT  *, ROW_NUMBER() OVER (ORDER BY DataDate) AS rn
        FROM    mytable
        )
SELECT  DATEDIFF(second, mc.DataDate, mp.DataDate)
FROM    rows mc
JOIN    rows mp
ON      mc.rn = mp.rn - 1

In SQL Server 2012+:

在SQL Server 2012 +:

SELECT  DATEDIFF(second, pDataDate, dataDate)
FROM    (
        SELECT  *,
                LAG(dataDate) OVER (ORDER BY dataDate) pDataDate
        FROM    rows
        ) q
WHERE   pDataDate IS NOT NULL

#2


2  

A little tweak on Quassnoi's query if you prefer not to use a Subselect would be:

如果您不喜欢使用子选择,那么对Quassnoi的查询做一个小小的调整就是:

SELECT
      DATEDIFF(second, LAG(dataDate) OVER (ORDER BY dataDate), dataDate)
FROM  rows
WHERE LAG(dataDate) OVER (ORDER BY dataDate) IS NOT NULL