對日期時間列表的累積頻率的SQL查詢

Question

我有一個數據庫列（表示訪問網站）的時間列表。

我需要他們組的間隔，然後獲取這些日期的「累積頻率」表。

比如我可能有：

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

，我想將其轉換成

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

我怎麼能這樣做？我可以在SQL中輕鬆實現這一點嗎？我可以很容易地做到這一點在C＃

Answer 1

創建一個表periods描述你想成劃分一天的時間段。

SELECT periods.name, count(time) 
    FROM periods, times 
WHERE period.start <= times.time 
    AND     times.time < period.end 
GROUP BY periods.name 

Answer 2

我應該指出的是，根據問題的規定「意圖」，做分析訪客流量 - 我寫了這個說法來概括制服團體計數。

要否則做（如在「實例」基團）在10分鐘的間隔5分鐘的時間間隔爲計數期間將被比較的計數 - 這是沒有意義的。

你神交到的用戶需求，而不是它的字面「讀書」，「意圖」。 :-)

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   

Answer 3

它使用了不少招數SQL（SQL Server 2005中）：

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 

注意，它把所有的時間在同一天，並假定他們是在一個日期時間列。它唯一不能做的就是從時間表示中去掉前導零。

Answer 4

創建一個包含你想在隨後的兩個表聯合起來獲得總計什麼間隔表。

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 

如果不是想要你的範圍內通緝總數累積總數，然後添加一個TIME_START列到TIME_INTERVAL表和查詢更改爲

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; 

Answer 5

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 

結果於：

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 

Answer 6

哦，太複雜了，所有這些東西。

正常化秒，通過你的水桶區間劃分，截短和remultiply：

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) 

使用羅恩野人的數據，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 | 
+----------+----------+ 

您不妨使用小區（）或圓形（ ）而不是floor（）。

更新：對於

create table d (
    d datetime 
);