获取记录创建

之间的平均时间，所以我有数据是这样的：获取记录创建

UserID CreateDate 
1  10/20/2013 4:05 
1  10/20/2013 4:10 
1  10/21/2013 5:10 
2  10/20/2012 4:03

我需要按每个用户得到CreateDates之间的平均时间。我想要的结果是这样的：

UserID AvgTime(minutes) 
1  753.5 
2  0

如何找到为用户分组返回的所有记录的CreateDates之间的区别？

编辑：

使用SQL Server 2012

来源

2013-10-21 RJP

结果应该不是'（5 + 1500）/ 2 = 752,5'？ – Lamak

@Lamak你是对的;我以为我用了4:10而不是5:10，谢谢。但它仍然（5 + 1500）/ 3，但该用户有三个结果平均返回501.67分钟三。 – RJP

@Tonnie - 不，它应该是'753.5' - 有三个_records_但只有两个_differences_。 –

试试这个：

SELECT A.UserID, 
     AVG(CAST(DATEDIFF(MINUTE,B.CreateDate,A.CreateDate) AS FLOAT)) AvgTime 
FROM #YourTable A 
OUTER APPLY (SELECT TOP 1 * 
      FROM #YourTable 
      WHERE UserID = A.UserID 
      AND CreateDate < A.CreateDate 
      ORDER BY CreateDate DESC) B 
GROUP BY A.UserID

来源

2013-10-21 18:31:09 Lamak

这给用户2提供了NULL。这不是OP想要的。 – geomagas

这个作品，谢谢！我可以将AVG语句包装在一个ISNULL中以将NULL变为0 – RJP

让我们希望不可能创建两个具有相同日期和用户ID的记录 –

事情是这样的：

;WITH CTE AS 
(
    SELECT 
     ROW_NUMBER() OVER (PARTITION BY UserID ORDER BY CreateDate) RN, 
     UserID, 
     CreateDate 
    FROM Tbl 
) 
SELECT 
    T1.UserID, 
    AVG(DATEDIFF(mi, ISNULL(T2.CreateDate, T1.CreateDate), T1.CreateDate)) AvgTime 
FROM CTE T1 
    LEFT JOIN CTE T2 
    ON T1.UserID = T2.UserID 
     AND T1.RN = T2.RN - 1 
GROUP BY T1.UserID

来源

2013-10-21 18:24:54

[为什么'mi'？为什么不拼写'MINUTE'？]（http://sqlblog.com/blogs/aaron_bertrand/archive/2011/09/20/bad-habits-to-kick-using-shorthand-with-date-time-operations。 aspx） –

@AaronBertrand区别在哪里？ –

@AaronBertrand我认为你必须根据[DATEDIFF（Transact-SQL）]（http://msdn.microsoft.com/en-us/library/ms189794.aspx）编辑上面提到的博客帖子。 –

这种做法应该ASLO工作。

Fiddle demo here：

;WITH CTE AS (
    Select userId, createDate, 
      row_number() over (partition by userid order by createdate) rn 
    from Table1 
) 
select t1.userid, 
     isnull(avg(datediff(second, t1.createdate, t2.createdate)*1.0/60),0) AvgTime 
from CTE t1 left join CTE t2 on t1.UserID = t2.UserID and t1.rn +1 = t2.rn 
group by t1.UserID;

更新时间：感谢@Lemark您指出number of diff = recordCount - 1

来源

2013-10-21 18:31:17 Kaf

您对'UserID ='的第一个值使用值为'0' 1'。这不应该在平均值 – Lamak

@Lamak上考虑; [检查这个]（http://sqlfiddle.com/#!6/84ba5/4），第二个是你的查询，它不给这些值。有点混淆的价值观。 – Kaf

因为op在他的计算上是错误的，我的观点是正确的：http://stackoverflow.com/questions/19501789/get-average-time-between-record-creation?noredirect=1#comment28928501_19501789 – Lamak

随着SQL 2012可以使用ROW_NUMBER功能和自联接查找 “前”每组中的排：

WITH Base AS 
(
SELECT 
    ROW_NUMBER() OVER (PARTITION BY UserID ORDER BY CreateDate) RowNum, 
    UserId, 
    CreateDate 
FROM Users 
) 
SELECT 
    B1.UserID, 
    ISNULL(
     AVG(
      DATEDIFF(mi,B2.CreateDate,B1.CreateDate) * 1.0 
     ) 
    ,0) [Average] 
FROM Base B1 
LEFT JOIN Base B2 
    ON B1.UserID = B2.UserID 
    AND B1.RowNum = B2.RowNum + 1 
GROUP BY B1.UserId

虽然我获得用户ID 1不同的答案 - 我得到的，因为你正在使用的2012（5 + 1500）/ 2 = 752

来源

2013-10-21 18:31:54

[为什么'mi'？为什么不拼写'MINUTE'？]（http://sqlblog.com/blogs/aaron_bertrand/archive/2011/09/20/bad-habits-to-kick-using-shorthand-with-date-time-operations。 ASPX） –

平均可使用铅（）来做到这一点

with cte as 
(select 

     userid, 
     (datediff(second, createdate, 
        lead(CreateDate) over (Partition by userid order by createdate) 
        )/60) datdiff 


From table1 
) 
select 
    userid, 
    avg(datdiff) 
from cte 
group by userid

Demo

来源

2013-10-21 18:40:42

在2012年这仅适用于您可以使用LEAD解析函数：

CREATE TABLE dates (
    id integer, 
    created datetime not null 
); 

INSERT INTO dates (id, created) 
SELECT 1 AS id, '10/20/2013 4:05' AS created 
UNION ALL SELECT 1, '10/20/2013 4:10' 
UNION ALL SELECT 1, '10/21/2013 5:10' 
UNION ALL SELECT 2, '10/20/2012 4:03'; 

SELECT id, isnull(avg(diff), 0) 
FROM (
    SELECT id, 
    datediff(MINUTE, 
      created, 
      LEAD(created, 1, NULL) OVER(partition BY id ORDER BY created) 
      ) AS diff 
    FROM dates 
) as diffs 
GROUP BY id;

http://sqlfiddle.com/#!6/4ce89/22

来源

2013-10-21 18:48:56 axblount

获取记录创建

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