Question

我下面的ms-sql代码在sql server 2017上运行：

DECLARE @report_int TABLE
(
  atimestamp DATETIME,
  adate DATE,
  ahour INT,
  intvalue INT,
  attribute CHAR(254),
  total INT
)

INSERT INTO @report_int
SELECT
  el.servertimestamp,
  cast(el.servertimestamp as date) adate,
  (DATEPART(HH,el.servertimestamp)) ahour,
  el.intvalue,
  el.attributes,
  0
FROM
  eventlog el
where
  el.servertimestamp  BETWEEN '6/29/2017'
                  AND DATEADD(dd, +1, '6/29/2019')
  and (attributes LIKE '%N<=>OlympieioIn%' OR attributes LIKE '%N<=>OlympieioOut%')
order by
  el.servertimestamp

DECLARE @report TABLE
(
  adate datetime,
  ahour int,
  came_in bigint,
  gone_out bigint

)

--fill report with dates and hours
INSERT INTO @report (adate, ahour) SELECT DISTINCT adate, ahour FROM @report_int

--reset
UPDATE @report set came_in = 0, gone_out = 0

--update in totals, min and max times
UPDATE @report SET came_in = i.total from
@report r,
(
select adate, ahour, COUNT(*) total
FROM @report_int
WHERE attribute LIKE 'N<=>OlympieioIn'
GROUP BY adate, ahour
) i
WHERE
i.adate = r.adate
and i.ahour = r.ahour

--update out totals
UPDATE @report SET gone_out = i.total from
@report r,
(
SELECT adate, ahour, COUNT(*) total
FROM @report_int
WHERE attribute LIKE 'N<=>OlympieioOut'
GROUP BY adate, ahour
) i
where
i.adate = r.adate
and i.ahour = r.ahour

select
  adate,
  ahour,
  RIGHT('00'+cast(ahour as varchar),2)+':00 - '+RIGHT('00'+cast(ahour+1 as varchar),2)+':00' timeframe,
  came_in,
  gone_out
from
  @report

此代码显示每天有多少人进出建筑物。如您所见，进出的人数按小时分组显示。下面是执行代码的屏幕截图。

Here is the link where you can see the screenshot

我想做的是将这些数据分为2小时的分组，例如，上面执行的代码的第3和第4小时的分组将是：

日期（日期）：2018-05-12 小时（小时）： 15 时间范围：：15：00-17：00 came_in： 0 gone_out： 2

我们将非常感谢您的帮助

谢谢。

Answer 1

我只需手动创建一个时间表并将其加入。该时间表将有24行（每小时一排），可能如下所示：小时小时时间框架twohourtimeframe 13 13:00-14:00 13:00-15:00

Answer 2

将日期时间设置为最后2小时的一种方法是使用：

DATEADD(HOUR, DATEDIFF(HOUR, 0, YourDateColumn) - (DATEDIFF(HOUR, 0, YourDateColumn) % 2), 0)

如果我们随后获取一些（随机）样本数据，则会得到以下信息：

WITH VTE AS (
    SELECT CONVERT(datetime2(0),V.DT) AS DT
    FROM (VALUES('2018-07-04T00:11:32'),('2018-07-04T01:17:12'),('2018-07-04T02:00:01'),('2018-07-04T07:49:59'),('2018-07-04T07:59:58'),('2018-07-04T08:00:00')) V(DT))
SELECT DT, DATEADD(HOUR, DATEDIFF(HOUR, 0, DT) - (DATEDIFF(HOUR, 0, DT) % 2), 0) AS DT2h
FROM VTE;

哪个返回：

DT                          DT2h
--------------------------- -----------------------
2018-07-04 00:11:32         2018-07-04 00:00:00.000
2018-07-04 01:17:12         2018-07-04 00:00:00.000
2018-07-04 02:00:01         2018-07-04 02:00:00.000
2018-07-04 07:49:59         2018-07-04 06:00:00.000
2018-07-04 07:59:58         2018-07-04 06:00:00.000
2018-07-04 08:00:00         2018-07-04 08:00:00.000

如果您希望将COUNT放2小时：

WITH VTE AS (
    SELECT CONVERT(datetime2(0),V.DT) AS DT
    FROM (VALUES('2018-07-04T00:11:32'),('2018-07-04T01:17:12'),('2018-07-04T02:00:01'),('2018-07-04T07:49:59'),('2018-07-04T07:59:58'),('2018-07-04T08:00:00')) V(DT))
SELECT DATEADD(HOUR, DATEDIFF(HOUR, 0, DT) - (DATEDIFF(HOUR, 0, DT) % 2), 0) AS DT2h, COUNT(DT) AS C
FROM VTE
GROUP BY DATEADD(HOUR, DATEDIFF(HOUR, 0, DT) - (DATEDIFF(HOUR, 0, DT) % 2), 0);

哪个返回：

DT2h                    C
----------------------- -----------
2018-07-04 00:00:00.000 2
2018-07-04 02:00:00.000 1
2018-07-04 06:00:00.000 2
2018-07-04 08:00:00.000 1

ms sql组2小时

2 个答案: