Question

我有以下查询，其中的目的是显示每条记录的时间，直到下一条记录

数据：

gid         time                name
1010883478  29/03/2016 0:00:02  John
1010883527  29/03/2016 0:00:04  John
1010883578  29/03/2016 0:00:06  John

SQL：

  SELECT A.[gid]
       ,A.[time]
       ,A.[name]
       ,(B.[time] - A.[time]) as timeTilNext
  FROM [location] A CROSS JOIN [location] B 
  WHERE B.[gid] IN (
      SELECT MIN(C.[gid]) 
      FROM [location] C 
      WHERE C.[gid] > A.[gid] AND C.[name] = A.[name] )
  ORDER BY A.[gid]

当前输出：

gid         time                    name    timeTilNext
1010883478  2016-03-29 00:00:02.000 John    1900-01-01 00:00:02.000
1010883527  2016-03-29 00:00:04.000 John    1900-01-01 00:00:02.000

预期产出：

gid         time                    name    timeTilNext
1010883478  2016-03-29 00:00:02.000 John    1900-01-01 00:00:02.000
1010883527  2016-03-29 00:00:04.000 John    1900-01-01 00:00:02.000
1010883578  2016-03-29 00:00:06.000 John    -1 (or whatever)

但是，它没有显示给定[名称]的最高[gid]的记录（仅次于第二高）。

我希望最高[gid]为timeTilNext显示-1，表示没有更多事件。

有关如何修改查询的任何想法？

Answer 1

在SQL Server 2012中，您可以使用LEAD窗口函数来获取＆＃34; next＆＃34;的值。行。

DECLARE @location TABLE ([gid] int, [time] datetime, [name] varchar(50));

INSERT INTO @location ([gid], [time], [name]) VALUES
(1010883478, '2016-03-29 00:00:02', 'John'),
(1010883527, '2016-03-29 00:00:04', 'John'),
(1010883578, '2016-03-29 00:00:06', 'John');

SELECT 
    A.[gid]
    ,A.[time]
    ,A.[name]
    ,LEAD(A.[time]) OVER(PARTITION BY A.[name] ORDER BY A.[gid]) AS NextTime
    ,ISNULL(DATEDIFF(second, A.[time], 
        LEAD(A.[time]) OVER(PARTITION BY A.[name] ORDER BY A.[gid])), -1) AS SecondsTillNext
FROM @location A
ORDER BY A.[gid];

<强>结果

+------------+-------------------------+------+-------------------------+-----------------+
|    gid     |          time           | name |        NextTime         | SecondsTillNext |
+------------+-------------------------+------+-------------------------+-----------------+
| 1010883478 | 2016-03-29 00:00:02.000 | John | 2016-03-29 00:00:04.000 |               2 |
| 1010883527 | 2016-03-29 00:00:04.000 | John | 2016-03-29 00:00:06.000 |               2 |
| 1010883578 | 2016-03-29 00:00:06.000 | John | NULL                    |              -1 |
+------------+-------------------------+------+-------------------------+-----------------+

如果＆＃34;下一个＆＃34;行无法使用，LEAD将返回NULL。如果需要，可以使用ISNULL()将其替换为某些非空值。

Answer 2

select 
*,-1 as 'time until next ' from location t1
where time=(select max(time) from location t2 where t1.name=t2.name) b

Answer 3

SELECT A.gid，A.name，A.time，（（SELECT MIN（B.time）FROM [location] B WHERE B.time＆gt; A.time AND B.name = A.name） - A.time ）作为timeTilNext

FROM [location] A

SQL“Where In”表示空子查询

3 个答案: