区间值总和oracle

izkcnapc 于 2022-11-03 发布在 Oracle

关注(0)|答案(4)|浏览(139)

查询：
第一个
从上面的查询中，我们每1分钟得到一次结果，我们从给定的时间戳开始，每5分钟寻找一次数据总和。
预期结果：

+-----------+-----------+----------+
| Timestamp | Counts1   | Counts2  |
+-----------+-----------+----------+
| 01:05     | 1125      | 302      |
| 01:10     | 750       | 182      |
| 01:15     | 1030      | 187      |
| 01:20     | 1340      | 265      |
+-----------+-----------+----------+

有人能帮忙吗
已尝试以下操作：

select to_char(date + interval '5' minute, 'HH24:MI') as Timestamp,
count(case when type = 5 then 1 end) as Counts1,
count(case when type = 6 then 1 end) as Counts2,
from data
where date >= to_date('2022-10-27 01:00', 'YYYY-MM-DD HH24:MI')
and date <= to_date('2022-10-27 01:20', 'YYYY-MM-DD HH24:MI')
and type IN (5,6)
group by to_char(date + interval '5' minute, 'HH24:MI')
order by to_char(date + interval '5' minute, 'HH24:MI')

下面是我们得到的结果：

+-----------+-----------+----------+
| Timestamp | Counts1   | Counts2  |
+-----------+-----------+----------+
| 01:05     | 125       | 5        |
| 01:06     | 130       | 10       |
| 01:07     | 140       | 12       |
| 01:08     | 150       | 35       |
| 01:09     | 160       | 47       |
| 01:10     | 170       | 78       |
| 01:11     | 180       | 125      |
| 01:12     | 190       | 5        |
| 01:13     | 210       | 10       |
| 01:14     | 220       | 12       |
| 01:15     | 230       | 35       |
| 01:16     | 240       | 47       |
| 01:17     | 260       | 78       |
| 01:18     | 270       | 125      |
| 01:19     | 280       | 5        |
| 01:20     | 290       | 10       |
+-----------+-----------+----------+

我们正在寻找每5分钟间隔的总和，预期结果如下：

+-----------+-----------+----------+
| Timestamp | Counts1   | Counts2  |
+-----------+-----------+----------+
| 01:05     | 1125      | 302      |
| 01:10     | 750       | 182      |
| 01:15     | 1030      | 187      |
| 01:20     | 1340      | 265      |
+-----------+-----------+----------+

oracle

来源：https://stackoverflow.com/questions/74277054/sum-of-interval-values-oracle

4条答案

按热度按时间

t0ybt7op1#

您可以使用CONNECT BY来产生周期：

WITH dat AS
(
  SELECT  TO_DATE('01:00','HH24:MI') AS TIMESTAMP, 200 AS counts1, 12 AS counts2  FROM dual UNION ALL
  SELECT  TO_DATE('01:01','HH24:MI'), 250, 35  FROM dual UNION ALL
  SELECT  TO_DATE('01:02','HH24:MI'), 300, 47  FROM dual UNION ALL
  SELECT  TO_DATE('01:03','HH24:MI'), 150, 78  FROM dual UNION ALL
  SELECT  TO_DATE('01:04','HH24:MI'), 100, 125 FROM dual UNION ALL
  SELECT  TO_DATE('01:05','HH24:MI'), 125, 5   FROM dual UNION ALL
  SELECT  TO_DATE('01:06','HH24:MI'), 130, 10  FROM dual UNION ALL
  SELECT  TO_DATE('01:07','HH24:MI'), 140, 12  FROM dual UNION ALL
  SELECT  TO_DATE('01:08','HH24:MI'), 150, 35  FROM dual UNION ALL
  SELECT  TO_DATE('01:09','HH24:MI'), 160, 47  FROM dual UNION ALL
  SELECT  TO_DATE('01:10','HH24:MI'), 170, 78  FROM dual UNION ALL
  SELECT  TO_DATE('01:11','HH24:MI'), 180, 125 FROM dual UNION ALL
  SELECT  TO_DATE('01:12','HH24:MI'), 190, 5   FROM dual UNION ALL
  SELECT  TO_DATE('01:13','HH24:MI'), 210, 10  FROM dual UNION ALL
  SELECT  TO_DATE('01:14','HH24:MI'), 220, 12  FROM dual UNION ALL
  SELECT  TO_DATE('01:15','HH24:MI'), 230, 35  FROM dual UNION ALL
  SELECT  TO_DATE('01:16','HH24:MI'), 240, 47  FROM dual UNION ALL
  SELECT  TO_DATE('01:17','HH24:MI'), 260, 78  FROM dual UNION ALL
  SELECT  TO_DATE('01:18','HH24:MI'), 270, 125 FROM dual UNION ALL
  SELECT  TO_DATE('01:19','HH24:MI'), 280, 5   FROM dual UNION ALL
  SELECT  TO_DATE('01:20','HH24:MI'), 290, 10  FROM dual
)
SELECT periods.p_end, SUM(counts1), SUM(counts2)
  FROM dat times
     , (SELECT TO_DATE('01:00','HH24:MI') + (DECODE(LEVEL,1,1,2)-1)/(24*60) +  5*(LEVEL-1)/(24*60) AS p_start --start time + offset
             , TO_DATE('01:00','HH24:MI') + 5*(LEVEL)/(24*60) AS p_end 
          FROM dual
       CONNECT BY LEVEL <= 100 --adjust according to your data 
       ) periods
 WHERE times.TIMESTAMP >= periods.p_start 
   AND times.TIMESTAMP <= periods.p_end
 GROUP BY periods.p_end;

可能不是最佳解决方案，但它可以得到您想要的结果。

赞(0）回复(0）举报 2022-11-03

8zzbczxx2#

使用您的示例数据：

WITH
    tbl AS
        (
            Select  '01:00' "TIMESTAMP", 200 "COUNTS1", 12 "COUNTS2" From Dual Union All
            Select  '01:01', 250, 35  From Dual Union All
            Select  '01:02', 300, 47  From Dual Union All
            Select  '01:03', 150, 78  From Dual Union All
            Select  '01:04', 100, 125 From Dual Union All
            Select  '01:05', 125, 5   From Dual Union All
            Select  '01:06', 130, 10  From Dual Union All
            Select  '01:07', 140, 12  From Dual Union All
            Select  '01:08', 150, 35  From Dual Union All
            Select  '01:09', 160, 47  From Dual Union All
            Select  '01:10', 170, 78  From Dual Union All
            Select  '01:11', 180, 125 From Dual Union All
            Select  '01:12', 190, 5   From Dual Union All
            Select  '01:13', 210, 10  From Dual Union All
            Select  '01:14', 220, 12  From Dual Union All
            Select  '01:15', 230, 35  From Dual Union All
            Select  '01:16', 240, 47  From Dual Union All
            Select  '01:17', 260, 78  From Dual Union All
            Select  '01:18', 270, 125 From Dual Union All
            Select  '01:19', 280, 5   From Dual Union All
            Select  '01:20', 290, 10  From Dual
        ),

...您可以以5分钟为一个步骤创建CTE：

grid AS
        (
            Select
                t.TIMESTAMP "TIMESTAMP",
                Min(s.STEP) "STEP"
            From 
                tbl t
            Inner Join
                ( Select LEVEL * 5 "STEP" From Dual d Connect By LEVEL <= 12  ) s ON(1 = 1)
            WHERE
                To_Number(SubStr(t.TIMESTAMP, 4, 2)) <= s.STEP
            GROUP BY
                t.TIMESTAMP
            Order By 
                t.TIMESTAMP
        )

......并将数据与CTE连接，按STEP分组并求和......

Select 
    SubStr(t.TIMESTAMP, 1, 3) || LPAD(STEP, 2, '0') "TIMESTAMP",
    Sum(t.COUNTS1) "SUM_1",
    Sum(t.COUNTS2) "SUM_2"
From 
    tbl t
INNER JOIN
    grid g ON(g.TIMESTAMP = t.TIMESTAMP)
GROUP BY
    SubStr(t.TIMESTAMP, 1, 3) || LPAD(STEP, 2, '0')
ORDER BY
    SubStr(t.TIMESTAMP, 1, 3) || LPAD(STEP, 2, '0')

结果应该是：
| 时间戳|总和_1|总和_2|
| - -|- -|- -|
| 01点05分|小行星1125| 302个|
| 01点10分|七百五十人|一百八十二|
| 01点15分|小行星103|一百八十七|
| 01点20分|小行星1340|二百六十五|
如果你想每10分钟做一次，那么只需将CTE的内部连接部分更改为如下所示：

...
( Select LEVEL * 10 "STEP" From Dual d Connect By LEVEL <= 6 )
...

...在这种情况下，结果将是：
| 时间戳|总和_1|总和_2|
| - -|- -|- -|
| 01点10分|小行星1875|四百八十四人|
| 01点20分|小行星2370|四百五十二|
此致。

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t1qtbnec3#

使用width_bucket函数分成5分钟的间隔。然后，将时段数乘以时段大小加到开始间隔。需要从时段数中减去1，因为第一个时段将是时段数1。

select   to_char(to_date('2022-10-27 01:00', 'YYYY-MM-DD HH24:MI')  -- interval start
                + (width_bucket(date
                               ,to_date('2022-10-27 01:00', 'YYYY-MM-DD HH24:MI') -- interval start
                               ,to_date('2022-10-27 01:20', 'YYYY-MM-DD HH24:MI') -- interval end
                               ,4 -- number of buckets 20/5 = 4 minutes
                               ) 
                   -1 -- for zero offset on first interval
                  )  
                  * 300/86400 -- to add 5 minute intervals to date
                ,'HH24:MI') timestamp, 
         count(case when type = 5 then 1 end) as Counts1,
         count(case when type = 6 then 1 end) as Counts2
from     data
where    date >= to_date('2022-10-27 01:00', 'YYYY-MM-DD HH24:MI')
and      date <= to_date('2022-10-27 01:20', 'YYYY-MM-DD HH24:MI')
and      type IN (5,6)
group by to_char(to_date('2022-10-27 01:00', 'YYYY-MM-DD HH24:MI')  -- interval start
                + (width_bucket(date
                               ,to_date('2022-10-27 01:00', 'YYYY-MM-DD HH24:MI') -- interval start
                               ,to_date('2022-10-27 01:20', 'YYYY-MM-DD HH24:MI') -- interval end
                               ,4 -- number of buckets 20/5 = 4 minutes
                               ) 
                   -1 -- for zero offset on first interval
                  )  
                  * 300/86400 -- to add 5 minute intervals to date
                ,'HH24:MI')
order by 1

也可以达到同样的结果没有函数。不需要计算桶数与此方法。

select to_char(to_date('2022-10-27 01:00', 'YYYY-MM-DD HH24:MI') -- interval start
               + floor(
                        (date - to_date('2022-10-27 01:00', 'YYYY-MM-DD HH24:MI')) -- date minus interval start
                        /(300/86400) -- divide into 5 minute intervals
                      ) -- gives interval number or bucket number
                      * 300/86400 -- multiply bucket number by bucket width
              ,'HH24:MI') timestamp, 
count(case when type = 5 then 1 end) as Counts1,
count(case when type = 6 then 1 end) as Counts2
from data
where date >= to_date('2022-10-27 01:00', 'YYYY-MM-DD HH24:MI')
and date <= to_date('2022-10-27 01:20', 'YYYY-MM-DD HH24:MI')
and type IN (5,6)
group by to_char(to_date('2022-10-27 01:00', 'YYYY-MM-DD HH24:MI') -- interval start
               + floor(
                        (date - to_date('2022-10-27 01:00', 'YYYY-MM-DD HH24:MI')) -- date minus interval start
                        /(300/86400) -- divide into 5 minute intervals
                      ) -- gives interval number or bucket number
                      * 300/86400 -- multiply bucket number by bucket width
              ,'HH24:MI')
order by 1

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jw5wzhpr4#

也许这能帮上忙。

ALTER SESSION SET NLS_DATE_FORMAT = 'DD-MON-YYYY HH24:MI:SS';

CREATE TABLE time_data (     
 seq_num NUMBER GENERATED BY DEFAULT AS IDENTITY (START WITH 1) NOT NULL,
   dt    DATE 
);

Generate a row for each minute in a day. This is my test data. I suspect you have plenty of data.

INSERT INTO time_data(dt)
select timestamp '2022-11-01 00:00:00' + numtodsinterval(rownum-1*1,'MINUTE')
    from dual
    CONNECT BY LEVEL <= (24*60);

Group into 5 minute intervals. Your data should show different counts.

select  trunc(dt,'hh24') + numtodsinterval(floor(to_char(dt,'mi') / 5) * 5,'minute') dt_5_min,
        count(*) cnt
  from  time_data
  group by trunc(dt,'hh24') + numtodsinterval(floor(to_char(dt,'mi') / 5) * 5,'minute')
  order by trunc(dt,'hh24') + numtodsinterval(floor(to_char(dt,'mi') / 5) * 5,'minute')
/

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区间值总和oracle

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