詢問：
select to_char(date, '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, 'HH24:MI')
order by to_char(date, 'HH24:MI')
----------- ----------- ----------
| Timestamp | Counts1 | Counts2 |
----------- ----------- ----------
| 01:00 | 200 | 12 |
| 01:01 | 250 | 35 |
| 01:02 | 300 | 47 |
| 01:03 | 150 | 78 |
| 01:04 | 100 | 125 |
| 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 |
----------- ----------- ----------
從上面的查詢中，我們每 1 分鐘得到一次結果，我們正在尋找給定時間戳的每 5 分鐘間隔的資料總和。
預期結果：
----------- ----------- ----------
| Timestamp | Counts1 | Counts2 |
----------- ----------- ----------
| 01:05 | 1125 | 302 |
| 01:10 | 750 | 182 |
| 01:15 | 1030 | 187 |
| 01:20 | 1340 | 265 |
----------- ----------- ----------
有人可以幫忙嗎
在下面嘗試：
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')
下面是我們得到的結果：
----------- ----------- ----------
| 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 |
----------- ----------- ----------
我們正在尋找每 5 分鐘間隔的總和，預期結果如下：
----------- ----------- ----------
| Timestamp | Counts1 | Counts2 |
----------- ----------- ----------
| 01:05 | 1125 | 302 |
| 01:10 | 750 | 182 |
| 01:15 | 1030 | 187 |
| 01:20 | 1340 | 265 |
----------- ----------- ----------
uj5u.com熱心網友回復：
您可以CONNECT BY用來生成期間：
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;
可能不是最佳解決方案，但它會得到你想要的結果。
uj5u.com熱心網友回復：
使用您的樣本資料：
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
),
...您可以用 5 分鐘的步驟創建 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
)
...并將您的資料與 CTE 連接起來，按 STEP 對它們進行分組并求和...
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')
結果應該是：
| 時間戳 | SUM_1 | SUM_2 |
|---|---|---|
| 01:05 | 1125 | 302 |
| 01:10 | 750 | 182 |
| 01:15 | 1030 | 187 |
| 01:20 | 1340 | 265 |
如果您想每 10 分鐘執行一次，只需將 CTE 的 Inner Join 部分更改為如下所示：
...
( Select LEVEL * 10 "STEP" From Dual d Connect By LEVEL <= 6 )
...
...在這種情況下，結果將是：
| 時間戳 | SUM_1 | SUM_2 |
|---|---|---|
| 01:10 | 1875年 | 484 |
| 01:20 | 2370 | 452 |
問候...
uj5u.com熱心網友回復：
使用width_bucket函式以 5 分鐘為間隔。然后，將桶數乘以桶大小添加到開始間隔。需要從存盤桶編號中減去一個，因為第一個存盤桶將是存盤桶編號 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
不使用該功能也可以達到同樣的效果。不需要用這種方法計算桶的數量。
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
uj5u.com熱心網友回復：
也許這可能會有所幫助。
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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