4

我在数据库列中有一个时间列表（代表对网站的访问）。

我需要按时间间隔对它们进行分组，然后得到这些日期的“累积频率”表。

例如我可能有：

9:01
9:04
9:11
9:13
9:22
9:24
9:28

我想把它转换成

9:05 - 2
9:15 - 4
9:25 - 6
9:30 - 7

我怎样才能做到这一点？我什至可以在 SQL 中轻松实现这一点吗？我可以很容易地在 C# 中做到这一点

4

6 回答 6

8 
create table accu_times (time_val datetime not null, constraint pk_accu_times primary key (time_val));
go

insert into accu_times values ('9:01');
insert into accu_times values ('9:05');
insert into accu_times values ('9:11');
insert into accu_times values ('9:13');
insert into accu_times values ('9:22');
insert into accu_times values ('9:24');
insert into accu_times values ('9:28'); 
go

select rounded_time,
    (
    select count(*)
    from accu_times as at2
    where at2.time_val <= rt.rounded_time
    ) as accu_count
from (
select distinct
  dateadd(minute, round((datepart(minute, at.time_val) + 2)*2, -1)/2,
    dateadd(hour, datepart(hour, at.time_val), 0)
  ) as rounded_time
from accu_times as at
) as rt
go

drop table accu_times

结果是：

rounded_time            accu_count
----------------------- -----------
1900-01-01 09:05:00.000 2
1900-01-01 09:15:00.000 4
1900-01-01 09:25:00.000 6
1900-01-01 09:30:00.000 7
于 2008-10-03T03:56:28.403 回答
3

我应该指出，基于问题的陈述“意图”，对访问者流量进行分析 - 我写了这个声明来总结统一组的计数。

否则（如在“示例”组中）会将 5 分钟间隔内的计数与 10 分钟间隔内的计数进行比较 - 这是没有意义的。

您必须了解用户需求的“意图”，而不是字面意义上的“阅读”。:-)

    create table #myDates
       (
       myDate       datetime
       );
    go

    insert into #myDates values ('10/02/2008 09:01:23');
    insert into #myDates values ('10/02/2008 09:03:23');
    insert into #myDates values ('10/02/2008 09:05:23');
    insert into #myDates values ('10/02/2008 09:07:23');
    insert into #myDates values ('10/02/2008 09:11:23');
    insert into #myDates values ('10/02/2008 09:14:23');
    insert into #myDates values ('10/02/2008 09:19:23');
    insert into #myDates values ('10/02/2008 09:21:23');
    insert into #myDates values ('10/02/2008 09:21:23');
    insert into #myDates values ('10/02/2008 09:21:23');
    insert into #myDates values ('10/02/2008 09:21:23');
    insert into #myDates values ('10/02/2008 09:21:23');
    insert into #myDates values ('10/02/2008 09:26:23');
    insert into #myDates values ('10/02/2008 09:27:23');
    insert into #myDates values ('10/02/2008 09:29:23');
    go

    declare @interval int;
    set @interval = 10;

    select
       convert(varchar(5), dateadd(minute,@interval - datepart(minute, myDate) % @interval, myDate), 108) timeGroup,
       count(*)
    from
       #myDates
    group by
       convert(varchar(5), dateadd(minute,@interval - datepart(minute, myDate) % @interval, myDate), 108)

retuns:

timeGroup             
--------- ----------- 
09:10     4           
09:20     3           
09:30     8
于 2008-10-03T03:45:50.160 回答
2

哦，所有这些东西都太复杂了。

标准化为秒，除以您的存储桶间隔，截断并重新乘法：

select sec_to_time(floor(time_to_sec(d)/300)*300), count(*)
from d
group by sec_to_time(floor(time_to_sec(d)/300)*300)

使用 Ron Savage 的数据，我得到

+----------+----------+
| i        | count(*) |
+----------+----------+
| 09:00:00 |        1 |
| 09:05:00 |        3 |
| 09:10:00 |        1 |
| 09:15:00 |        1 |
| 09:20:00 |        6 |
| 09:25:00 |        2 |
| 09:30:00 |        1 |
+----------+----------+

您可能希望使用 ceil() 或 round() 而不是 floor()。

更新：对于使用创建的表

create table d (
    d datetime
);
于 2008-10-03T12:13:32.107 回答
1

创建一个表格periods，描述您希望将一天分成的时段。

SELECT periods.name, count(time)
  FROM periods, times
 WHERE period.start <= times.time
   AND                 times.time < period.end
 GROUP BY periods.name
于 2008-10-03T03:44:26.840 回答
1

创建一个表格，其中包含您想要获得总计的间隔，然后将两个表格连接在一起。

如：

time_entry.time_entry
-----------------------
2008-10-02 09:01:00.000
2008-10-02 09:04:00.000
2008-10-02 09:11:00.000
2008-10-02 09:13:00.000
2008-10-02 09:22:00.000
2008-10-02 09:24:00.000
2008-10-02 09:28:00.000

time_interval.time_end
-----------------------
2008-10-02 09:05:00.000
2008-10-02 09:15:00.000
2008-10-02 09:25:00.000
2008-10-02 09:30:00.000

SELECT 
    ti.time_end, 
    COUNT(*) AS 'interval_total' 
FROM time_interval ti
INNER JOIN time_entry te
    ON te.time_entry < ti.time_end
GROUP BY ti.time_end;


time_end                interval_total
----------------------- -------------
2008-10-02 09:05:00.000 2
2008-10-02 09:15:00.000 4
2008-10-02 09:25:00.000 6
2008-10-02 09:30:00.000 7

如果您不想要累积总计，而是想要某个范围内的总计，那么您将 time_start 列添加到 time_interval 表并将查询更改为

SELECT 
    ti.time_end, 
    COUNT(*) AS 'interval_total' 
FROM time_interval ti
INNER JOIN time_entry te
    ON te.time_entry >= ti.time_start
            AND te.time_entry < ti.time_end
GROUP BY ti.time_end;
于 2008-10-03T03:54:07.097 回答
0

这使用了很多 SQL 技巧（SQL Server 2005）：

CREATE TABLE [dbo].[stackoverflow_165571](
    [visit] [datetime] NOT NULL
) ON [PRIMARY]
GO

;WITH buckets AS (
    SELECT dateadd(mi, (1 + datediff(mi, 0, visit - 1 - dateadd(dd, 0, datediff(dd, 0, visit))) / 5) * 5, 0) AS visit_bucket
            ,COUNT(*) AS visit_count
    FROM stackoverflow_165571
    GROUP BY dateadd(mi, (1 + datediff(mi, 0, visit - 1 - dateadd(dd, 0, datediff(dd, 0, visit))) / 5) * 5, 0)
)
SELECT LEFT(CONVERT(varchar, l.visit_bucket, 8), 5) + ' - ' + CONVERT(varchar, SUM(r.visit_count))
FROM buckets l
LEFT JOIN buckets r
    ON r.visit_bucket <= l.visit_bucket
GROUP BY l.visit_bucket
ORDER BY l.visit_bucket

请注意，它将所有时间都放在同一天，并假设它们在日期时间列中。唯一不像您的示例那样做的是从时间表示中去除前导零。

于 2008-10-03T03:47:10.360 回答