I have a varchar5 column with times from
00:00
00:01
00:02
00:03
... all the way to
23:59
How would I count how many minutes are in an hour? To get the result
00 60
01 60
02 60
and so on...
SQL:
select 24_HOUR_CLOCK
From time table
Group by ...
Order by 24_HOUR_CLOCK ASC
Means to count records
I think you can use substr to extract first two characters from the time string and group on that.
select substr(col, 1, 2) hour, count(*) minutes
from your_table
group by substr(col, 1, 2)
order by hour
or find substr inside a subquery as #Mathguy suggested:
select hour,
count(*) minutes
from (
select substr(col, 1, 2) hour
from your_table
)
group by hour
order by hour
Related
I'm developing code calculation service availability based on events, so I need to split events into daily "sub-events" and calculate duration of then.
So as input I have set of events like (EVENT_ID, START_TIME, END_TIME):
'event1';2021-05-01 12:30;2021-05-01 13:00
'event2';2021-05-03 10:55;2021-05-05 12:01
As output I'd like to get (EVENT_ID, DAY, DURATION_MINUTES):
'event1'; 2021-05-01; 30
'event2'; 2021-05-03; 785
'event2'; 2021-05-04; 1440
'event2'; 2021-05-05; 721
I can get it using procedures and cursor but this is not effective (the events database is quite big), so is there a way to do it using oracle sql query ? Any idea?
You appear to want a recursive query:
WITH days ( event_id, day, start_time, end_time ) AS (
SELECT event_id,
TRUNC( start_time ),
start_time,
end_time
FROM table_name
UNION ALL
SELECT event_id,
day + INTERVAL '1' DAY,
start_time,
end_time
FROM days
WHERE day + INTERVAL '1' DAY < end_time
)
SELECT event_id,
day,
ROUND(
(
LEAST(end_time, day + INTERVAL '1' DAY)
- GREATEST(start_time, day)
) * 24 * 60
) AS duration_minutes
FROM days
Which, for the sample data:
CREATE TABLE table_name ( event_id, start_time, end_time ) AS
SELECT 'event1', DATE '2021-05-01' + INTERVAL '12:30' HOUR TO MINUTE, DATE '2021-05-01' + INTERVAL '13:00' HOUR TO MINUTE FROM DUAL UNION ALL
SELECT 'event2', DATE '2021-05-03' + INTERVAL '10:55' HOUR TO MINUTE, DATE '2021-05-05' + INTERVAL '12:01' HOUR TO MINUTE FROM DUAL;
Outputs:
EVENT_ID
DAY
DURATION_MINUTES
event1
2021-05-01
30
event2
2021-05-03
785
event2
2021-05-04
1440
event2
2021-05-05
721
db<>fiddle here
If your Oracle version is 12 or higher, you can use a lateral join (in any of several equivalent formulations/syntaxes) to make the query faster. For example (using the table set up in MT0's answer):
select event_id, day, round(1440 * duration_days) as duration_minutes
from table_name cross join lateral
( select trunc(start_time) + level - 1 as day,
case when level = 1 and connect_by_isleaf = 1
then end_time - start_time
when level = 1 then 1 - (start_time - trunc(start_time))
when connect_by_isleaf = 1 then end_time - trunc(end_time)
else 1 end as duration_days
from dual
connect by level <= 1 + trunc(end_time) - trunc(start_time)
)
where duration_days != 0
order by event_id, day
;
The where clause is used when the end_time is midnight (at the beginning of an otherwise "new" day); in that case, presumably, you don't want to include that "new day" in the output, with a duration of 0 minutes.
In the lateral view, level = 1 corresponds to the first date in the interval, while connect_by_isleaf = 1 is for the last date in the interval. A special calculation is made when the end_time and start_time are on the same date. The query computes the difference in days first, then converts to minutes. Note that date calculations aren't 100% precise; I used round so I don't get results like 33.9999999999938020 minutes. If the inputs are in hh24:mi, we know beforehand that the answer (in minutes) should be an integer, so round seems fine there.
I have a table called Orders, i want to get maximum number of orders for each day with respect to hours with following query
SELECT
trunc(created,'HH') as dated,
count(*) as Counts
FROM
orders
WHERE
created > trunc(SYSDATE -2)
group by trunc(created,'HH') ORDER BY counts DESC
this gets the result of all hours, I want only max hour of a day e.g.
Image
This result looks good but now i want only rows with max number of count for a day
e.g.
for 12/23/2019 max number of counts is 90 for "12/23/2019 4:00:00 PM",
for 12/22/2019 max number of counts is 25 for "12/22/2019 3:00:00 PM"
required dataset
1 12/23/2019 4:00:00 PM 90
2 12/24/2019 12:00:00 PM 76
3 12/22/2019 1:00:00 PM 25
This could be the solution and in my opinion is the most trivial.
Use the WITH clause to make a sub query then search for the greatest value in the data set on a specific date.
WITH ORD AS (
SELECT
trunc(created,'HH') as dated,
count(*) as Counts
FROM
orders
WHERE
created > trunc(SYSDATE-2)
group by trunc(created,'HH')
)
SELECT *
FROM ORD ord
WHERE NOT EXISTS (
SELECT 'X'
FROM ORD ord1
WHERE trunc(ord1.dated) = trunc(ord.dated) AND ord1.Counts > ord.Counts
)
Use ROW_NUMBER analytic function over your original query and filter the rows with number 1.
You need to partition on the day, i.e. TRUNC(dated) to get the correct result
with ord1 as (
SELECT
trunc(created,'HH') as dated,
count(*) as Counts
FROM
orders
WHERE
created > trunc(SYSDATE -2)
group by trunc(created,'HH')
),
ord2 as (
select dated, Counts,
row_number() over (partition by trunc(dated) order by Counts desc) as rn
from ord1)
select dated, Counts
from ord2
where rn = 1
The advantage of using the ROW_NUMBER is that it correct handels ties, i.e. cases where there are more hour in a day with the same maximal count. The query shows only one record and you can controll with the order by e.g. to show the first / last hour.
You can use the analytical function ROW_NUMBER as following to get the desired result:
SELECT DATED, COUNTS
FROM (
SELECT
TRUNC(CREATED, 'HH') AS DATED,
COUNT(*) AS COUNTS,
ROW_NUMBER() OVER(
PARTITION BY TRUNC(CREATED)
ORDER BY COUNT(*) DESC NULLS LAST
) AS RN
FROM ORDERS
WHERE CREATED > TRUNC(SYSDATE - 2)
GROUP BY TRUNC(CREATED, 'HH'), TRUNC(CREATED)
)
WHERE RN = 1
Cheers!!
I have a requirement to take every 3 seconds data within the specific time interval in SQL. I am new to SQL so can anyone help me on the scenario
This is my select query which returns all the values but i need data for every 3 seconds only
SELECT ton_nbr
FROM
icr_file_interface
WHERE
(
reading_dttm BETWEEN
TO_DATE(concat('2016-10-19',to_char(0930)),'yyyy-mm-dd HH24MISS')
AND TO_DATE(concat('2016-10-19',to_char('0945')),'yyyy-mm-dd HH24MISS')
)
AND
(
ton_nbr BETWEEN
(SELECT value FROM text_para WHERE para_cd='ICR_ST_RNG')
AND (SELECT value FROM text_para WHERE para_cd='ICR_ED_RNG')
)
If you only need to subtract 3 seconds from a date, you can use the following:
SQL> select to_char(sysdate, 'yyyy-mm-dd hh24:mi:ss'),
2 to_char(sysdate - 3*1/24/60/60, 'yyyy-mm-dd hh24:mi:ss')
3 from dual;
TO_CHAR(SYSDATE,'YY TO_CHAR(SYSDATE-3*1
------------------- -------------------
2016-10-19 09:38:17 2016-10-19 09:38:14
Given that sysdate -1 means "subtract one day to sysdate", you can derive the number of seconds you need with a bit af arithmetic
This selects you data between last 3 seconds. Hope you got the idea.
select ton_nbr
from icr_file_interface
where reading_dttm between dateadd(ss, -3, getdate()) and getdate() ;
I have table name Record which has the following columns, Empid in number column, dat in timestamp
Which has the following values
empid dat
====== ====
101 4/9/2012 9:48:54 AM
101 4/9/2012 9:36:28 AM
101 4/9/2012 6:16:28 PM
101 4/10/2012 9:33:48 AM
101 4/10/2012 12:36:28 PM
101 4/10/2012 8:36:12 PM
101 4/11/2012 9:36:28 AM
101 4/11/2012 4:36:22 PM
Here I need to display the following columns,
empid,min(dat) as start,max(dat) as end and difference(max(dat)-min(dat) for each day
Here 3 different days are exists so It should return 3 records with the above mentioned columns.
Please give some ways to get this.
Simply subtract them: max(dat) - min(dat)
SELECT empid,
min(dat) as strt,
max(dat) as end,
max(dat) - min(dat) as diff
FROM the_table
GROUP BY empid;
If you want to group by the day instead of the empid, use this one:
select trunc(dat),
min(dat) as strt,
max(dat) as end,
max(dat) - min(dat) as diff
from the_table
group by trunc(dat)
Date arithmetic is pretty straightforward in Oracle: the difference betwwen two dates is returned as the number of days. Values of less than a day are returned as *decimals". That is, 75 minutes is 1.25 hours not 1.15. If you want it as hours and minutes you need to work with an interval.
The inner query calculates the difference between the minimum and maximum data for each employee for each day, and converts it to a DAY interval. The outer query extracts the HOUR and MINUTES from that interval.
select empid
, tday
, sdt
, edt
, extract(hour from diff) diff_hours
, extract (minute from diff) diff_minutes
from (
select empid
, trunc(dat) tday
, min(dat) sdt
, max(dat) edt
, numtodsinterval(max(dat) - min(dat), 'DAY') diff
from t42
group by empid, trunc(dat)
)
The question I need to answer is this "What is the maximum number of page requests we have ever received in a 60 minute period?"
I have a table that looks similar to this:
date_page_requested date;
page varchar(80);
I'm looking for the MAX count of rows in any 60 minute timeslice.
I thought analytic functions might get me there but so far I'm drawing a blank.
I would love a pointer in the right direction.
You have some options in the answer that will work, here is one that uses Oracle's "Windowing Functions with Logical Offset" feature instead of joins or correlated subqueries.
First the test table:
Wrote file afiedt.buf
1 create table t pctfree 0 nologging as
2 select date '2011-09-15' + level / (24 * 4) as date_page_requested
3 from dual
4* connect by level <= (24 * 4)
SQL> /
Table created.
SQL> insert into t values (to_date('2011-09-15 11:11:11', 'YYYY-MM-DD HH24:Mi:SS'));
1 row created.
SQL> commit;
Commit complete.
T now contains a row every quarter hour for a day with one additional row at 11:11:11 AM. The query preceeds in three steps. Step 1 is to, for every row, get the number of rows that come within the next hour after the time of the row:
1 with x as (select date_page_requested
2 , count(*) over (order by date_page_requested
3 range between current row
4 and interval '1' hour following) as hour_count
5 from t)
Then assign the ordering by hour_count:
6 , y as (select date_page_requested
7 , hour_count
8 , row_number() over (order by hour_count desc, date_page_requested asc) as rn
9 from x)
And finally select the earliest row that has the greatest number of following rows.
10 select to_char(date_page_requested, 'YYYY-MM-DD HH24:Mi:SS')
11 , hour_count
12 from y
13* where rn = 1
If multiple 60 minute windows tie in hour count, the above will only give you the first window.
This should give you what you need, the first row returned should have
the hour with the highest number of pages.
select number_of_pages
,hour_requested
from (select to_char(date_page_requested,'dd/mm/yyyy hh') hour_requested
,count(*) number_of_pages
from pages
group by to_char(date_page_requested,'dd/mm/yyyy hh')) p
order by number_of_pages
How about something like this?
SELECT TOP 1
ranges.date_start,
COUNT(data.page) AS Tally
FROM (SELECT DISTINCT
date_page_requested AS date_start,
DATEADD(HOUR,1,date_page_requested) AS date_end
FROM #Table) ranges
JOIN #Table data
ON data.date_page_requested >= ranges.date_start
AND data.date_page_requested < ranges.date_end
GROUP BY ranges.date_start
ORDER BY Tally DESC
For PostgreSQL, I'd first probably write something like this for a "window" aligned on the minute. You don't need OLAP windowing functions for this.
select w.ts,
date_trunc('minute', w.ts) as hour_start,
date_trunc('minute', w.ts) + interval '1' hour as hour_end,
(select count(*)
from weblog
where ts between date_trunc('minute', w.ts) and
(date_trunc('minute', w.ts) + interval '1' hour) ) as num_pages
from weblog w
group by ts, hour_start, hour_end
order by num_pages desc
Oracle also has a trunc() function, but I'm not sure of the format. I'll either look it up in a minute, or leave to see a friend's burlesque show.
WITH ranges AS
( SELECT
date_page_requested AS StartDate,
date_page_requested + (1/24) AS EndDate,
ROWNUMBER() OVER(ORDER BY date_page_requested) AS RowNo
FROM
#Table
)
SELECT
a.StartDate AS StartDate,
MAX(b.RowNo) - a.RowNo + 1 AS Tally
FROM
ranges a
JOIN
ranges b
ON a.StartDate <= b.StartDate
AND b.StartDate < a.EndDate
GROUP BY a.StartDate
, a.RowNo
ORDER BY Tally DESC
or:
WITH ranges AS
( SELECT
date_page_requested AS StartDate,
date_page_requested + (1/24) AS EndDate,
ROWNUMBER() OVER(ORDER BY date_page_requested) AS RowNo
FROM
#Table
)
SELECT
a.StartDate AS StartDate,
( SELECT MIN(b.RowNo) - a.RowNo
FROM ranges b
WHERE b.StartDate > a.EndDate
) AS Tally
FROM
ranges a
ORDER BY Tally DESC