Generate rows with time intervals between 2 dates in Oracle - sql

I have table in which Sunday to Saturdy "Doctor Start" and "End Time" is given.
I want to create time slots of 15 minutes.
On the basis of that, the patient clicks on calendar datetime interval which shows slots that have already been booked.

The following example shows how to split time into slices of 15 minutes. It uses hierarchical query. A little bit of explanation:
line 2: trunc function, applied to a date value, returns "beginning" of that day (at midnight). Adding 15 / (24*60) adds 15 minutes (as there are 24 hours in a day and 60 minutes in an hour). Multiplying 15 by level works as a "loop", i.e. adds 15-by-15-by-15 ... minutes to previous value.
line 4: similar to line 2, but it makes sure that a day (24 hours * 60 minutes) is divided to 15-minutes parts
line 6: start time is trivial
line 7: end time just adds 15 minutes to start_time
line 9: return only time between 10 and 16 hours (you don't have patients at 02:15 AM, right?)
SQL> with fifteen as
2 (select trunc(sysdate) + (level * 15)/(24*60) c_time
3 from dual
4 connect by level <= (24*60) / 15
5 )
6 select to_char(c_time, 'hh24:mi') start_time,
7 to_char(c_time + 15 / (24 * 60), 'hh24:mi') end_time
8 from fifteen
9 where extract(hour from cast (c_time as timestamp)) between 10 and 15;
START_TIME END_TIME
---------- ----------
10:00 10:15
10:15 10:30
10:30 10:45
10:45 11:00
11:00 11:15
11:15 11:30
11:30 11:45
11:45 12:00
12:00 12:15
12:15 12:30
12:30 12:45
12:45 13:00
13:00 13:15
13:15 13:30
13:30 13:45
13:45 14:00
14:00 14:15
14:15 14:30
14:30 14:45
14:45 15:00
15:00 15:15
15:15 15:30
15:30 15:45
15:45 16:00
24 rows selected.
SQL>

Related

Split a time duration into time segments

I have an incident start time and end time e.g Sr=tart time of 15/01/2018 11:30 and end time of 16/01/2018 02:40 in an excel table.
How can I split this time range into different time segments. The segments are:-
06:00 - 11:59, 12:00 - 14:59, 15:00 - 17:59, 18:00 - 22:59, 23:00 - 05:59
For 06:00 - 11:59 I would expect 0.50 as this is 30 mins.
For 12:00 - 14:59 I would expect 3.0 for 3 hours.
For 15:00 - 17:59 I would expect 3.0 for 3 hours again.
For 18:00 - 22:59 this should be 5hrs
and for 23:00 - 02:40 should be 3.67 hrs.
What formula would I need to achieve this?
For the example you give with the layout as shown below this formula should work:
=MIN($D3-F$1,MIN(F$2-F$1,IF($C3<F$2,F$2-$C3+SUM($E3:E3),)))
The formula in C3 is:
=24*(MOD(A3,1))
and in D3:
=24*(MOD(B3,1)+INT(B3>INT(A3)))

oracle sql Time to Resolve Calculation

I have a question and hopefully someone can help, because i have been stuck on this for a long time.
I have a column with remaining minutes for a task to expire and i want to calculate when this task will expire within the business days timeframe starting from the current sysdate day lets say weekdays from 09:00 to 17:00.
| Task No | Minutes Remaining | Expiration date |
| Task1 | 1800 | 27-10-16 9:45 AM |
| Task2 | 3400 | 28-10-16 9:45 AM |
| Task3 | 400 | 29-10-16 9:45 AM |
| Task4 | 180 | 30-10-16 9:45 AM |
| Task5 | 8400 | 31-10-16 9:45 AM |
| Task6 | 5000 | 1-11-16 9:45 AM |
OK, this was a fun problem. To summarize: You are given a date (which in Oracle always includes the time-of-day) from which you start measurement, and an initial duration in minutes. You need to find the expiration date (meaning date and time-of-day as always), which is calculated by adding the duration in minutes to the "clock-starting" date, but the clock should only run during business hours - 9 to 17, Monday to Friday only (not on weekends).
I assume if the "minutes remaining" is 0, then the expiration should be the same as the "clock-starting" date if it falls within work hours, or 9 am on the next work day otherwise.
To understand the solution, let's break it down in two parts. First let's consider a very special case: the "clock starts" on a Monday at 9 am. Then break down minutes remaining into an integer multiple of 2400 (5*8*60 = 2400 minutes in a full work week), plus an integer multiple of 480 from what's left (480 minutes to a work day), plus whatever is left, if anything. Then: the expiration date is the "clock-starting" date, plus however many weeks, plus however many whole days (between 0 and 4), plus the remaining minutes. One exceptional case here: if the "minutes remaining" is an exact multiple of 480 minutes, then expiration is at 5 pm on a certain work day, and not 9 am on the next work day. This requires special handling in the formula. All this is done in the outer query (at the bottom of the solution below).
Then we need to reduce the general case to this special case. This is done in the subquery prep in the solution. I simply increase the "minutes remaining" by the work minutes elapsed from 9 am on Monday at the beginning of the week. This is a relatively simple computation. Note that if the "clock starting" date is after 5 pm on a Friday (or any time on Saturday or Sunday), I must add exactly 2400 minutes, a full work week.
In the solution, I show a variety of "clock starting" dates, dt, and minutes remaining, rm. I tested a variety of situations, and I think the solution is correct, but you may want to test on more data (other situations I didn't include in the tests).
with
inputs ( task, min_rem, dt ) as (
select 'Task1', 1800, to_date('27-10-16 9:45 AM', 'dd-mm-yy hh:mi AM') from dual union all
select 'Task2', 3400, to_date('28-10-16 9:45 AM', 'dd-mm-yy hh:mi AM') from dual union all
select 'Task3', 400, to_date('29-10-16 3:45 AM', 'dd-mm-yy hh:mi AM') from dual union all
select 'Task4', 180, to_date('30-10-16 9:45 AM', 'dd-mm-yy hh:mi AM') from dual union all
select 'Task5', 8400, to_date('31-10-16 9:45 PM', 'dd-mm-yy hh:mi AM') from dual union all
select 'Task6', 5000, to_date('01-11-16 5:00 PM', 'dd-mm-yy hh:mi AM') from dual union all
select 'Task7', 0, to_date('01-12-16 5:00 PM', 'dd-mm-yy hh:mi PM') from dual
),
prep ( task, min_rem, dt, adj_min, adj_dt ) as (
select task, min_rem, dt,
min_rem + case when dt > trunc(dt, 'iw') + 5 + 17/24 then 2400
else (trunc(dt) - trunc(dt, 'iw')) * 480 +
least(480, greatest(0, 1440 * (dt - trunc(dt) - 9/24)))
end,
trunc(dt, 'iw') + 9/24
from inputs
)
select task, min_rem, dt,
adj_dt + 7 * trunc(adj_min / 2400)
+ case when adj_min/480 = trunc(adj_min/480)
then mod(adj_min, 2400) / 480 - 1 + 8/24
else trunc(mod(adj_min, 2400) / 480) + mod(adj_min, 480) / 1440
end as expiration
from prep
order by task
;
Output:
TASK MIN_REM DT EXPIRATION
----- ---------- ----------------- -----------------
Task1 1800 27-10-16 09:45 AM 01-11-16 03:45 PM
Task2 3400 28-10-16 09:45 AM 08-11-16 10:25 AM
Task3 400 29-10-16 03:45 AM 31-10-16 03:40 PM
Task4 180 30-10-16 09:45 AM 31-10-16 12:00 PM
Task5 8400 31-10-16 09:45 PM 24-11-16 01:00 PM
Task6 5000 01-11-16 05:00 PM 16-11-16 12:20 PM
Task7 0 01-12-16 05:00 PM 01-12-16 05:00 PM
7 rows selected

How to count the records per half hour from a period (datetimefrom and datetimeto) field?

I have a table which looks like you can see below:
Id Date ScheduledTimeFrom ScheduledTimeTo ActualTimeFrom ActualTimeTo
1 2013-01-01 1899-12-30 07:00:00 1899-12-30 18:00:00 1899-12-30 07:23:00 1899-12-30 17:15:00
I need to calculate per half hour how many records exists, the output should be like:
Time Actual Count:
7:00 4
7:30 4
8:00 4
8:30 4
9:00 4
9:30 5
10:00 5
10:30 6
11:00 7
11:30 8
12:00 8
12:30 8
13:00 8
13:30 8
14:00 8
14:30 8
15:00 7
15:30 7
16:00 7
16:30 6
17:00 5
17:30 4
18:00 4
I already tried to make a helper table which should hold the times per halfhour. I have joined this helpertable with the table that contains the data and after that I tried to use a group by function but it was not working.
My query was like:
Create table period (timefrom datetime, timeto datetime)
insert into period
select '1899-12-30 07:00:00.000', '1899-12-30 07:30:00.000'
Union all
select '1899-12-30 07:30:00.000', '1899-12-30 08:00:00.000'
select *
from period p left join table1 t on t.ActualTimeFrom < p.timeto and t.ActualTimeTo >=p.timefrom
Grouping this give me no desired result....
Anyone an idea how to come to the result?
P.s. I am using sql server 2005.
After snooping around and testing it on my side, looks like this date function could be the answer:
DATEADD(mi,DATEDIFF(mi,0,YOUR_DATE_COLUMN)/30*30,0)

SQL code for Comparing date fields in different rows and combining the results

I need help for proper Oracle SQL code to combine rows for a crystal reports command object. This is a part of the bigger query I'm working on and got stuck for the past couple of days.
for eg. if the columns are like below
PatId In_time Out_time
151 01/01/2012 07:00:00 am 01/01/2012 10:00:00 am
151 01/01/2012 11:00:00 am 01/02/2012 08:00:00 am
151 01/02/2012 11:00:00 am 01/02/2012 01:00:00 pm
151 01/03/2012 08:00:00 am 01/03/2012 03:00:00 pm
151 01/06/2012 03:30:00 pm 01/09/2012 07:00:00 am
167 01/03/2012 01:30:00 pm 01/09/2012 07:00:00 am
167 01/13/2012 03:30:00 pm 01/14/2012 07:00:00 am
167 01/14/2012 11:30:00 am 01/15/2012 11:30:00 am
167 01/18/2012 12:00:00 pm 01/19/2012 03:00:00 am
Within a PatId, the code should compare the Out_time of one row to the In_time of the next row, and check whether the time gap is greater than 48 hours. If not, then it is considered part of the same visit. I want one result row per PatID & visit, with min(In_time) and max(Out_time). The time span of the visit (result row) itself may be greater than 48 hours.
For this example, for PatId 151 the time difference between the out_time of 1st row and In_time of 2nd row is less than 48 hours. The difference between Out_time of second row and In_time of 3rd row, as well as between the 3rd and 4th rows, is also less than 48 hours. After this the gap between Out_time of the 4th row and In_time of 5th row is greater than 48 hours. The result for PatId 151 should be as below and same for EmpId 167, the chaining should continue until a gap greater than 48 hours is found.
So the result for the above table should be displayed as,
PatId In_time Out_time
151 01/01/2012 07:00:00 am 01/03/2012 03:00:00 pm
151 01/06/2012 03:30:00 pm 01/09/2012 07:00:00 am
167 01/03/2012 01:30:00 pm 01/09/2012 07:00:00 am
167 01/13/2012 03:30:00 pm 01/15/2012 11:30:00 am
167 01/18/2012 12:00:00 pm 01/19/2012 03:00:00 am
I could not get the logic on how to compare and merge rows.
Thanks in Advance, Abhi
General example of subtracting time - copy/paste to see the output. This example will give you differences in hours, minutes, seconds between two dates. The basic formula is (end_date - start_date) * 86400 (number of seconds in 24 hrs)...:
SELECT trunc(mydate / 3600) hr
, trunc(mod(mydate, 3600) / 60) mnt
, trunc(mod(mydate, 3600) / 60 /60) sec
FROM
(
SELECT (to_date('01/03/2012 10:00:00', 'mm/dd/yyyy hh24:mi:ss') -
to_date('01/01/2012 07:00:00', 'mm/dd/yyyy hh24:mi:ss')) * 86400 mydate
FROM dual
)
/
HR | MNT | SEC
---------------
51 | 0 | 0
You need to check your example and logic. I could not understand what needs to be comnpared with what...

SQL How Many Employees Are Working, Group By Hour

I have a table, a timetable, with check-in and check-out times of the employees:
ID Date Check-in Check out
1 1-1-2011 11:00 18:00
2 1-1-2011 11:00 19:00
3 1-1-2011 16:00 18:30
4 1-1-2011 17:00 20:00
Now I want to know how many employees are working, every (half) hour.
The result I want to see:
Hour Count
11 2
12 2
13 2
14 2
15 2
16 3
17 3
18 2,5
19 1
Every 'Hour' you must read as 'till the next full hour', ex. 11 -> 11:00 - 12:00
Any ideas?
Build an additional table, called Hours, containing the following data:
h
00:00
00:30
01:00
...
23:30
then, run
Select h as 'hour' ,count(ID) as 'count' from timetable,hours where [Check_in]<=h and h<=[Check_out] group by h