Calculate the sum of the duration stored in format HH24
Dear professionals,
I use Oracle Database 11 g Enterprise Edition Release 11.2.0.4.0 - 64 bit Production. I stored length of certain events in the table TIME_DURATION as as follows (format h24:mi):
CREATE TABLE "TIME_DURATION" ("ID" NUMBER(11,0), "HOURSMINUTES" VARCHAR2(5)); Insert into TIME_DURATION (ID,HOURSMINUTES) values ('5','00:55'); Insert into TIME_DURATION (ID,HOURSMINUTES) values ('7','00:18'); Insert into TIME_DURATION (ID,HOURSMINUTES) values ('9','06:34'); Insert into TIME_DURATION (ID,HOURSMINUTES) values ('15','00:12'); Insert into TIME_DURATION (ID,HOURSMINUTES) values ('17','09:50'); INSERT INTO TIME_DURATION (ID,HOURSMINUTES) VALUES ('41','12:39');
select * from time_duration; ID HOURS ---------- ----- 5 00:55 7 00:18 9 06:34 15 00:12 17 09:50 41 12:39 6 rows selected.
Now, I want to calculate the total time for all events (sum of all specific times). In this case, it should be 30 hours and 38 minutes.
Any help would be much appreciated.
Thanks in advance.
Hello
So, you want to add a number of lines varibale. This sounds like a job for the SUM function. AMOUNT of work on numbers, no channels such as time, so use TO_NUMBER to convert strings to numbers, so you can add them. If you want to display the result as a string, you can use TO_CHAR to convert the sum into a string.
WITH got_total_minutes AS
(
SELECT SUM ((TO_NUMBER (SUBSTR (heure, 1, 2) * 60)))
+ TO_NUMBER (SUBSTR (hour 4))
) AS total_minutes
OF time_duration
)
SELECT TRUNC (total_minutes / 60). ':'
|| To_char (MOD (total_minutes, 60))
, "FM00.
) AS total_hours_minutes
OF got_total_minutes
;
The output is not quite what you asked:
TOTAL_HOURS_MINUTES
--------------------
30: 28
If you really want 30:38', explain how to get it.
This solution assumes that time always is always 5 characters (2 digits, a separator and another of 2 digits) as it is in your sample data.
If your actual data aren't like your sample data, the same approach still works, but the SUBSTR expressions will be more complicated.
adnanBIH wrote:
Dear professionals,
I use Oracle Database 11 g Enterprise Edition Release 11.2.0.4.0 - 64 bit Production. I stored length of certain events in the table TIME_DURATION as as follows (format h24:mi):
- CREATE TABLE 'TIME_DURATION' ('ID' NUMBER (11.0), VARCHAR2 (5)) 'HOUR '.
- Insert into TIME_DURATION (ID, TIME) values ('5',' 00:55 ');
- Insert into TIME_DURATION (ID, TIME) values ('7',' 00:18 ');
- ...
Thanks for posting the CREATE TABLE and INSERT.
ID is a NUMBER, so do not use quotes around it:
Insert into TIME_DURATION (ID, TIME) values (5, ' 00:55 ');
Insert into TIME_DURATION (ID, TIME) values (7, ' 00:18 ');
...
Depending on how you plan to use the hour, you may want to store a NUMBER, also, or maybe an INTERVAL DAY TO SECOND.
Tags: Database
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Hello
We are 11g.
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OlivierOf course; It's easy with the use of the function analytic lag:
WITH t AS (SELECT 'aaa' col1, to_date( '20100201 09:23:50', 'YYYYMMDD HH24:MI:SS') date1 FROM dual UNION ALL SELECT 'aaa' col1, to_date( '20100201 13:14:33', 'YYYYMMDD HH24:MI:SS') date1 FROM dual UNION ALL SELECT 'aaa' col1, to_date( '20100207 13:14:33', 'YYYYMMDD HH24:MI:SS') date1 FROM dual UNION ALL SELECT 'bbb' col1, to_date( '20100202 09:21:10', 'YYYYMMDD HH24:MI:SS') date1 FROM dual UNION ALL SELECT 'bbb' col1, to_date( '20100203 08:11:06', 'YYYYMMDD HH24:MI:SS') date1 FROM dual UNION ALL SELECT 'bbb' col1, to_date( '20100203 15:13:55', 'YYYYMMDD HH24:MI:SS') date1 FROM dual UNION ALL SELECT 'bbb' col1, to_date( '20100210 10:14:27', 'YYYYMMDD HH24:MI:SS') date1 FROM dual), t_diff as (SELECT col1, to_char( date1,'YYYYMMDD HH24:MI:SS') date1, (date1 - lag(date1, 1, date1) over (partition by col1 order by date1))*24*60*60 date_diff FROM t) select col1, date1, sum(date_diff) over (partition by col1 order by date1) duration from t_diff ORDER BY col1, date1;
Published by: Boneist on February 15, 2010 16:39
(I can't read; initially gave the cumulative total of the period. Oh!)ETA2: Take note of the additional parameters, I used in the lag() - the third parameterd manages what should be the value in the field if there is no previous rank, so there is no need to use nvl elsewhere in the application to handle this situation.
Published by: Boneist on February 15, 2010 16:42
-
data
From time to time
November 10, 2015 16:00 November 10, 2015 17:10
November 10, 2015 16:50 10 November 2015 18:00
November 10, 2015 17:15 November 10, 2015 23:00
November 10, 2015 17:30 November 10, 2015 22:20
November 10, 2015 19:35 November 10, 2015 19:50
from the first record, if I want to calculate the time duration that is, 70 min.
second disc begins to the first record. service time is now 18-16 120 min.
third record also begins in the second. duration is now 300 min 23-16.
fourth and fifth registration is the responsibility of the third record time.
so final duration is 18-16 = 120 min.
How can I achieve this. Please just write the sql query.
Thank you.
teefu.
Lahore. in Pakistan.
SQL> SQL> -- SQL> select * from dt_lines order by f_dntime, t_dntime; DT_ID F_DNTIME T_DNTIME DT_DUR ---------- -------------------- -------------------- ---------- 1 10-Nov-2015 16:00:00 10-Nov-2015 17:10:00 70 2 10-Nov-2015 16:50:00 10-Nov-2015 18:00:00 130 3 10-Nov-2015 17:15:00 10-Nov-2015 23:00:00 345 4 10-Nov-2015 17:30:00 10-Nov-2015 22:20:00 290 5 10-Nov-2015 19:35:00 10-Nov-2015 19:50:00 15 6 10-Nov-2015 23:15:00 10-Nov-2015 23:45:00 30 6 rows selected. SQL> SQL> -- SQL> with t as ( 2 select f_dntime, t_dntime, 3 case when min(f_dntime) over (order by f_dntime, t_dntime rows between unbounded preceding and 1 preceding) is null 4 then f_dntime 5 when f_dntime > max(t_dntime) over (order by f_dntime, t_dntime rows between unbounded preceding and 1 preceding) 6 then f_dntime 7 end as max_fdt_adjusted 8 from dt_lines 9 ), 10 t1 as ( 11 select f_dntime, t_dntime, max_fdt_adjusted, 12 last_value(max_fdt_adjusted ignore nulls) over (order by f_dntime) as min_from_ts 13 from t 14 ) 15 select min_from_ts, max(t_dntime) as max_to_ts, 16 round((max(t_dntime) - min_from_ts)*24*60) as duration_min 17 from t1 18 group by min_from_ts 19 order by min_from_ts 20 ; MIN_FROM_TS MAX_TO_TS DURATION_MIN -------------------- -------------------- ------------ 10-Nov-2015 16:00:00 10-Nov-2015 23:00:00 420 10-Nov-2015 23:15:00 10-Nov-2015 23:45:00 30 2 rows selected. SQL> SQL>
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Hello
I have a table as below:
create table select TEST_CASETBL (ID, CASE_NUM, CASE_STATUS, CASE_SUB_STATUS, LAST_UPD_DTTM)
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-------------------------------------------------------------------------------------------------------------------------------------------------------
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Should he not?...
with test_casetbl (id, case_num, case_status, case_sub_status, last_upd_dttm) as)
Select 112, 123-456', 'Open', 'Work', TO_DATE (11 March 2015 13:00 ',' dd/mm/yyyy hh24:mi:ss') of all the DOUBLE union
Select 113, 123-456', 'Open', 'pending on the admin', TO_DATE (10 January 2015 15:00 ',' dd/mm/yyyy hh24:mi:ss') of all the DOUBLE union
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Select 651, 123-456', 'Open', 'Work', TO_DATE (August 20, 2015 10:00 ',' the hh24: mi: ss' dd/mm/yyyy) double
),
t like)
Select case_num,
case_sub_status,
(last_upd_dttm, 1, sysdate) ahead of diff last_upd_dttm (partition by order of last_upd_dttm case_num).
of test_casetbl
)
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trunc (a) | "day (s). TO_CHAR (date ' 1-1-1' + a, "fmhh24" 'mi' minute (s) "ss" second (s) hour (s)"") "work."
trunc (b) | "day (s). TO_CHAR (day 1-1-1' + b, "fmhh24" 'mi' minute (s) "ss" second (s) hour (s)"") 'Waiting on admin',
trunc (c) | "day (s). TO_CHAR (day 1-1-1' + c, "fmhh24" 'mi' minute (s) "ss" second (s) hour (s)"") "waiting on customer."
t
pivot)
Sum (diff)
for case_sub_status in)
'Work' is.
B "waiting on admin',
C "customer expectation."
)
)
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Hi all
If there are several durations then how we can calculate the period of time that overlap between these times.
For example: for 3 time periods. 03/12/2015-16/08/2015, 05/01/2015 to 31/07/2015 and the 06/09/2015 to 30/11/2015, how the overlap period can be calculated?
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In your example:
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HelloI'll try to explain my problem.
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...
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end if
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You may need to replace:
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with
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