I have made this query but when I execute it, I got error about "invalid number".
But in SQL Developer for Oracle, there is no error; I got the result that I want but in Toad I got 'Invalid Number' .
DECLARE v_rep number;
BEGIN
EXECUTE IMMEDIATE
'SELECT to_number(REPLACE(max(substr(to_char(r_timestamp_arr,''HH24:MI''),1,2) ||
ltrim(to_char(round(to_number(Substr(to_char(r_timestamp_arr, ''HH24:MI''),4,2)) /
60,2),''.00''))), ''.'', '','')) -
to_number(REPLACE(MIN(substr(to_char(r_timestamp_arr,''HH24:MI''),1,2) ||
ltrim(to_char(round(to_number(Substr(to_char(r_timestamp_arr, ''HH24:MI''),4,2)) /
60,2),''.00''))), ''.'', '',''))
FROM TV_MAX
WHERE TV_UID = ''7a87e8e4861a4d0aae65da1a7248b256'''
INTO v_rep;
END ;
You don't need EXECUTE IMMEDIATE and don't need to use strings:
Oracle Setup:
CREATE TABLE tv_max ( tv_uid, r_timestamp_arr ) AS
SELECT '7a87e8e4861a4d0aae65da1a7248b256', DATE '2019-12-27' + INTERVAL '00:00' HOUR TO MINUTE FROM DUAL UNION ALL
SELECT '7a87e8e4861a4d0aae65da1a7248b256', DATE '2019-12-27' + INTERVAL '01:30' HOUR TO MINUTE FROM DUAL;
Query 1:
If you want to ignore the date component of the date & time:
DECLARE
v_rep NUMBER;
BEGIN
SELECT ( MAX( r_timestamp_arr - TRUNC( r_timestamp_arr ) )
- MIN( r_timestamp_arr - TRUNC( r_timestamp_arr ) )
) * 24
INTO v_rep
FROM tv_max
WHERE TV_UID = '7a87e8e4861a4d0aae65da1a7248b256';
DBMS_OUTPUT.PUT_LINE( v_rep );
END;
/
Query 2:
If you want the min/max respecting the date component then the query can be even simpler:
DECLARE
v_rep NUMBER;
BEGIN
SELECT ( MAX( r_timestamp_arr ) - MIN( r_timestamp_arr ) ) * 24
INTO v_rep
FROM tv_max
WHERE TV_UID = '7a87e8e4861a4d0aae65da1a7248b256';
DBMS_OUTPUT.PUT_LINE( v_rep );
END;
/
Output:
For the test data, both output:
1.5
db<>fiddle here
Looks like You want to know the difference between max and min hour (including minutes, excluding seconds), date part truncated. So take truncated times, subtract as dates, you will get result in days, multiply by 24, result will be in hours. Query does not depend on NLS settings:
select 24 * (to_date(max(to_char(r_timestamp_arr, 'hh24:mi')), 'hh24:mi')
- to_date(min(to_char(r_timestamp_arr, 'hh24:mi')), 'hh24:mi')) as diff
from tv_max
where tv_uid = '7a87e8e4861a4d0aae65da1a7248b256'
dbfiddle
Business days are Monday through Friday.
Given I have a datetime field scheduled_for, how can I find the next business date and return that in a column alias?
I've tried something from another SO answer but it doesn't work as intended.
EXTRACT(ISODOW FROM v.scheduled_for)::integer) % 7 as next_business_day,
Error:
Query 1 ERROR: ERROR: syntax error at or near ")"
LINE 3: EXTRACT(ISODOW FROM v.scheduled_for)::integer % 7) as next...
^
Edit:
Thanks for the suggestions, I've attempted this:
SELECT
v.id AS visit_id,
(IF extract(''dow'' from v.scheduled_for) = 0 THEN
return v.scheduled_for + 1::integer;
ELSIF extract(''dow'' from v.scheduled_for) = 6 THEN
return v.scheduled_for - 1::integer;
ELSE
return v.scheduled_for;
) as next_business_day,
'' as invoice_ref_code,
The error I get is:
Query 1 ERROR: ERROR: syntax error at or near ")"
LINE 1: ) as next_business_day,
^
To generalize you need to create a function to calculate the next business day from a given date.
create or replace function utl_next_business_day(date_in date default current_date)
returns date
language sql immutable leakproof strict
as $$
with cd as (select extract(isodow from date_in)::integer d)
select case when d between 1 and 4
then date_in + 1
else date_in + 1 + (7-d)
end
from cd;
$$;
--- any single date
select current_date, utl_next_business_day();
-- over time span (short)
select gdate::date for_date, utl_next_business_day(gdate::date) next_business_day
from generate_series( current_date, current_date + 14, interval '1 day') gdate;
-- around year end over a time span
with test_date (dt) as
( values (date '2019-12-31')
, (date '2020-12-31'), (date '2021-12-31'),(date '2022-12-31')
, (date '2021-01-01'), (date '2022-01-01'),(date '2023-01-01')
)
select dt, utl_next_business_day(dt) from test_date
order by dt;
Alternatively with the calendar table suggestion from #Eric we get.
-- create and populate work table
create table bus_day_calendar ( bus_day date);
insert into bus_day_calendar (bus_day)
select utl_next_business_day(gdate::date)
from generate_series( date '2018-12-31', date '2023-01-01', interval '1 day') gdate
where extract(isodow from gdate)::integer not in (6,7) ;
--- Function to return next business day
create or replace function utl_next_cal_business_day(date_in date default current_date)
returns date
language sql stable leakproof strict
as $$
select min(bus_day)
from bus_day_calendar
where bus_day > date_in;
$$;
--- any single date
select current_date, utl_next_cal_business_day();
-- over time span (short)
select gdate::date for_date, utl_next_cal_business_day(gdate::date) next_business_day
from generate_series( current_date, current_date + 14, interval '1 day') gdate;
-- around year end over a time span
with test_date (dt) as
( values (date '2019-12-31')
, (date '2020-12-31'), (date '2021-12-31'),(date '2022-12-31')
, (date '2021-01-01'), (date '2022-01-01'),(date '2023-01-01')
)
select dt, utl_next_cal_business_day(dt) from test_date
order by dt;
Neither of these as they currently stand handle a non-business day that falls on Mon-Fri, but both can be modified to do so. Since the calendar table requires only deleting roes I think that becomes the superior method if this is necessary.
I am trying to calculate business days between two dates in Oracle select. I got to the point when my calculation gives most results correct for given dates (I compare it with NETWORKDAYS in excel) but sometimes it varies from 2 days to -2 days - and I don't know why...
Here's my code:
SELECT
((to_char(CompleteDate,'J') - to_char(InstallDate,'J'))+1) - (((to_char(CompleteDate,'WW')+ (52 * ((to_char(CompleteDate,'YYYY') - to_char(InstallDate,'YYYY'))))) - to_char(InstallDate,'WW'))*2) as BusinessDays
FROM TABLE
Thanks!
The solution, finally:
SELECT OrderNumber, InstallDate, CompleteDate,
(TRUNC(CompleteDate) - TRUNC(InstallDate) ) +1 -
((((TRUNC(CompleteDate,'D'))-(TRUNC(InstallDate,'D')))/7)*2) -
(CASE WHEN TO_CHAR(InstallDate,'DY','nls_date_language=english')='SUN' THEN 1 ELSE 0 END) -
(CASE WHEN TO_CHAR(CompleteDate,'DY','nls_date_language=english')='SAT' THEN 1 ELSE 0 END) as BusinessDays
FROM Orders
ORDER BY OrderNumber;
Thanks for all your responses !
I took into account all the different approaches discussed above and came up with a simple query that gives us the number of working days in each month of the year between two dates:
WITH test_data AS
(
SELECT TO_DATE('01-JAN-14') AS start_date,
TO_DATE('31-DEC-14') AS end_date
FROM dual
),
all_dates AS
(
SELECT td.start_date, td.end_date, td.start_date + LEVEL-1 as week_day
FROM test_data td
CONNECT BY td.start_date + LEVEL-1 <= td.end_date)
SELECT TO_CHAR(week_day, 'MON'), COUNT(*)
FROM all_dates
WHERE to_char(week_day, 'dy', 'nls_date_language=AMERICAN') NOT IN ('sun' , 'sat')
GROUP BY TO_CHAR(week_day, 'MON');
Please feel free to modify the query as needed.
Try this:
with holidays as
(
select d from (
select minDate + level -1 d
from (select min(submitDate) minDate, max (completeDate) maxDate
from t)
connect by level <= maxDate - mindate + 1)
where to_char(d, 'dy', 'nls_date_language=AMERICAN') not in ('sun' , 'sat')
)
select t.OrderNo, t.submitDate, t.completeDate, count(*) businessDays
from t join holidays h on h.d between t.submitDate and t.completeDate
group by t.OrderNo, t.submitDate, t.completeDate
order by orderno
Here is a sqlfiddle demo
I changed my example to more readable and to return count of bus. days between. I do not know why you need 'J'- Julian format. All it takes is start/Install and end/Complete dates. You will get correct number of days between 2 dates using this. Replace my dates with yours, add NLS if needed...:
SELECT Count(*) BusDaysBtwn
FROM
(
SELECT TO_DATE('2013-02-18', 'YYYY-MM-DD') + LEVEL-1 InstallDate -- MON or any other day
, TO_DATE('2013-02-25', 'YYYY-MM-DD') CompleteDate -- MON or any other day
, TO_CHAR(TO_DATE('2013-02-18', 'YYYY-MM-DD') + LEVEL-1, 'DY') InstallDay -- day of week
FROM dual
CONNECT BY LEVEL <= (TO_DATE('2013-02-25', 'YYYY-MM-DD') - TO_DATE('2013-02-18', 'YYYY-MM-DD')) -- end_date - start_date
)
WHERE InstallDay NOT IN ('SAT', 'SUN')
/
SQL> 5
I see that marked final solution is not correct always. Suppose, InstallDate is 1st of the month (if falls on Saturday) and CompleteDate is 16th of the month (if falls on Sunday)
In that case, actual Business Days is 10 but the marked query result will give the answer as 12. So, we have to treat this type of cases too, which I used
(CASE WHEN TO_CHAR(InstallDate,'DY','nls_date_language=english')='SAT' AND TO_CHAR(CompleteDate,'DY','nls_date_language=english')='SUN' THEN 2 ELSE 0 END
line to handle it.
SELECT OrderNumber, InstallDate, CompleteDate,
(TRUNC(CompleteDate) - TRUNC(InstallDate) ) +1 -
((((TRUNC(CompleteDate,'D'))-(TRUNC(InstallDate,'D')))/7)*2) -
(CASE WHEN TO_CHAR(InstallDate,'DY','nls_date_language=english')='SUN' THEN 1 ELSE 0 END) -
(CASE WHEN TO_CHAR(CompleteDate,'DY','nls_date_language=english')='SAT' THEN 1 ELSE 0 END) -
(CASE WHEN TO_CHAR(InstallDate,'DY','nls_date_language=english')='SAT' AND TO_CHAR(CompleteDate,'DY','nls_date_language=english')='SUN' THEN 2 ELSE 0 END)as BusinessDays
FROM Orders
ORDER BY OrderNumber;
The accepted solution is quite close but seems wrong in some cases (e.g., 2/1/2015 through 2-28/2015 or 5/1/2015 through 5/31/2015). Here's a refined version...
end_date-begin_date+1 /* total days */
- TRUNC(2*(end_date-begin_date+1)/7) /* weekend days in whole weeks */
- (CASE
WHEN TO_CHAR(begin_date,'D') = 1 AND REMAINDER(end_date-begin_date+1,7) > 0 THEN 1
WHEN TO_CHAR(begin_date,'D') = 8 - REMAINDER(end_date-begin_date+1,7) THEN 1
WHEN TO_CHAR(begin_date,'D') > 8 - REMAINDER(end_date-begin_date+1,7) THEN 2
ELSE 0
END) /* weekend days in partial week */
AS business_days
The part that handles the multiples of 7 (whole weeks) is good. But, when considering the partial week portion, it depends on both the day-of-week offset and the number of days in the partial portion, according to the following matrix...
654321
1N 111111
2M 100000
3T 210000
4W 221000
5R 222100
6F 222210
7S 222221
To just remove sundays and saturdays you can use this
SELECT Base_DateDiff
- (floor((Base_DateDiff + 0 + Start_WeekDay) / 7))
- (floor((Base_DateDiff + 1 + Start_WeekDay) / 7))
FROM (SELECT 1 + TRUNC(InstallDate) - TRUNC(InstallDate, 'IW') Start_WeekDay
, CompleteDate - InstallDate + 1 Base_DateDiff
FROM TABLE) a
Base_DateDiff counts the number of days between the two dates
(floor((Base_DateDiff + 0 + Start_WeekDay) / 7)) counts the number of sundays
(floor((Base_DateDiff + 1 + Start_WeekDay) / 7)) counts the number of saturdays
1 + TRUNC(InstallDate) - TRUNC(InstallDate, 'IW') get 1 for mondays to 7 for sunday
This query can be used to go backward N days from the given date (business days only)
For example, go backward 15 days from 2017-05-17:
select date_point, closest_saturday - (15 - offset + floor((15 - offset) / 6) * 2) from(
select date_point,
closest_saturday,
(case
when weekday_num > 1 then
weekday_num - 2
else
0
end) offset
from (
select to_date('2017-05-17', 'yyyy-mm-dd') date_point,
to_date('2017-05-17', 'yyyy-mm-dd') - to_char(to_date('2017-05-17', 'yyyy-mm-dd'), 'D') closest_saturday,
to_char(to_date('2017-05-17', 'yyyy-mm-dd'), 'D') weekday_num
from dual
))
Some brief explanation: suppose we want to go backward N days from a given date
- Find the closest Saturday that is less than or equal to the given date.
- From the closest Saturday, go back ward (N - offset) days. offset is the number of business days between the closest Saturday and the given date (excluding the given date).
*To go back M days from a Saturday (business days only), use this formula DateOfMonthOfTheSaturday - [M + Floor(M / 6) * 2]
Here is a function that is fast and flexible. You can count any weekday in a date range.
CREATE OR REPLACE FUNCTION wfportal.cx_count_specific_weekdays( p_week_days VARCHAR2 DEFAULT 'MON,TUE,WED,THU,FRI'
, p_start_date DATE
, p_end_date DATE)
RETURN NUMBER
IS
/***************************************************************************************************************
*
* FUNCTION DESCRIPTION:
*
* This function calculates the total required week days in a date range.
*
* PARAMETERS:
*
* p_week_days VARCHAR2 The week days that need to be counted, comma seperated e.g. MON,TUE,WED,THU,FRU,SAT,SUN
* p_start_date DATE The start date
* p_end_date DATE The end date
*
* CHANGE history
*
* No. Date Changed by Change Description
* ---- ----------- ------------- -------------------------------------------------------------------------
* 0 07-May-2013 yourname Created
*
***************************************************************************************************************/
v_date_end_first_date_range DATE;
v_date_start_last_date_range DATE;
v_total_days_in_the_weeks NUMBER;
v_total_days_first_date_range NUMBER;
v_total_days_last_date_range NUMBER;
v_output NUMBER;
v_error_text CX_ERROR_CODES.ERROR_MESSAGE%TYPE;
--Count the required days in a specific date ranges by using a list of all the weekdays in that range.
CURSOR c_total_days ( v_start_date DATE
, v_end_date DATE ) IS
SELECT COUNT(*) total_days
FROM ( SELECT ( v_start_date + level - 1) days
FROM dual
CONNECT BY LEVEL <= ( v_end_date - v_start_date ) + 1
)
WHERE INSTR( ',' || p_week_days || ',', ',' || TO_CHAR( days, 'DY', 'NLS_DATE_LANGUAGE=english') || ',', 1 ) > 0
;
--Calculate the first and last date range by retrieving the first Sunday after the start date and the last Monday before the end date.
--Calculate the total amount of weeks in between and multiply that with the total required days.
CURSOR c_calculate_new_dates ( v_start_date DATE
, v_end_date DATE ) IS
SELECT date_end_first_date_range
, date_start_last_date_range
, (
(
( date_start_last_date_range - ( date_end_first_date_range + 1 ) )
) / 7
) * total_required_days total_days_in_the_weeks --The total amount of required days
FROM ( SELECT v_start_date + DECODE( TO_CHAR( v_start_date, 'DY', 'NLS_DATE_LANGUAGE=english')
, 'MON', 6
, 'TUE', 5
, 'WED', 4
, 'THU', 3
, 'FRI', 2
, 'SAT', 1
, 'SUN', 0
, 0 ) date_end_first_date_range
, v_end_date - DECODE( TO_CHAR( v_end_date, 'DY', 'NLS_DATE_LANGUAGE=english')
, 'MON', 0
, 'TUE', 1
, 'WED', 2
, 'THU', 3
, 'FRI', 4
, 'SAT', 5
, 'SUN', 6
, 0 ) date_start_last_date_range
, REGEXP_COUNT( p_week_days, ',' ) + 1 total_required_days --Count the commas + 1 to get the total required weekdays
FROM dual
)
;
BEGIN
--Verify that the start date is before the end date
IF p_start_date < p_end_date THEN
--Get the new calculated days.
OPEN c_calculate_new_dates( p_start_date, p_end_date );
FETCH c_calculate_new_dates INTO v_date_end_first_date_range
, v_date_start_last_date_range
, v_total_days_in_the_weeks;
CLOSE c_calculate_new_dates;
--Calculate the days in the first date range
OPEN c_total_days( p_start_date, v_date_end_first_date_range );
FETCH c_total_days INTO v_total_days_first_date_range;
CLOSE c_total_days;
--Calculate the days in the last date range
OPEN c_total_days( v_date_start_last_date_range, p_end_date );
FETCH c_total_days INTO v_total_days_last_date_range;
CLOSE c_total_days;
--Sum the total required days
v_output := v_total_days_first_date_range + v_total_days_last_date_range + v_total_days_in_the_weeks;
ELSE
v_output := 0;
END IF;
RETURN v_output;
EXCEPTION
WHEN OTHERS
THEN
RETURN NULL;
END cx_count_specific_weekdays;
/
Here you go...
First check how many days you got in the holiday table, excluding weekend days.
Get business days (MON to FRI) between the 2 dates and after that subtract the holiday days.
create or replace
FUNCTION calculate_business_days (p_start_date IN DATE, p_end_date IN DATE)
RETURN NUMBER IS
v_holidays NUMBER;
v_start_date DATE := TRUNC (p_start_date);
v_end_date DATE := TRUNC (p_end_date);
BEGIN
IF v_end_date >= v_start_date
THEN
SELECT COUNT (*)
INTO v_holidays
FROM holidays
WHERE day BETWEEN v_start_date AND v_end_date
AND day NOT IN (
SELECT hol.day
FROM holidays hol
WHERE MOD(TO_CHAR(hol.day, 'J'), 7) + 1 IN (6, 7)
);
RETURN GREATEST (NEXT_DAY (v_start_date, 'MON') - v_start_date - 2, 0)
+ ( ( NEXT_DAY (v_end_date, 'MON')
- NEXT_DAY (v_start_date, 'MON')
)
/ 7
)
* 5
- GREATEST (NEXT_DAY (v_end_date, 'MON') - v_end_date - 3, 0)
- v_holidays;
ELSE
RETURN NULL;
END IF;
END calculate_business_days;
After that you can test it out, like:
select
calculate_business_days('21-AUG-2013','28-AUG-2013') as business_days
from dual;
There is another easier way, using connect by and dual...
with t as (select to_date('30-sep-2013') end_date, trunc(sysdate) start_date from dual)select count(1) from dual, t where to_char(t.start_date + level, 'D') not in (1,7) connect by t.start_date + level <= t.end_date;
with connect by you get all the dates from start_date till the end_date. Then you can exclude the dates you don't need and count only the needed.
This would return business days:
(CompleteDate-InstallDate)-2*FLOOR((CompleteDate-InstallDate)/7)-
DECODE(SIGN(TO_CHAR(CompleteDate,'D')-
TO_CHAR(InstallDate,'D')),-1,2,0)+DECODE(TO_CHAR(CompleteDate,'D'),7,1,0)-
DECODE(TO_CHAR(InstallDate,'D'),7,1,0) as BusinessDays,
I need to calculate the number of days between dates as detailed below using MSSQL
Each month should be considered as if it has 30 days (even if it doesn't)
The difference between 2 January, 2013 to 2 March, 2013 will be
(30-2) + 30 + 2 days
where (30-2) will be for January
30 will be for February
2 will be for March
create or replace function datediff( p_what in varchar2,
p_d1 in date,
p_d2 in date ) return number
as
l_result number;
begin
select (p_d2-p_d1) *
decode( upper(p_what),
'DAY', 1, 'SS', 24*60*60, 'MI', 24*60, 'HH', 24, NULL )
into l_result from dual;
return l_result;
end;
/
This is what I do in Oracle (Courtesy: ASKTOM).
I get either days, hours, minutes or seconds in difference.
In MS SQL, either
PRINT DATEDIFF(DAY, '1/1/2011', '3/1/2011')
This gives the number of times the midnight boundary is crossed between the two dates. You may decide to need to add one to this if you're including both dates in the count - or subtract one if you don't want to include either date.
OR
DECLARE #startdate datetime2 = '2007-05-05 12:10:09.3312722';
DECLARE #enddate datetime2 = '2009-05-04 12:10:09.3312722';
SELECT DATEDIFF(day, #startdate, #enddate);
Using this you can manipulate.
Looks like you want to get a result similar to Oracle's MONTHS_BETWEEN in SQL Server.
This is a SQL function i wrote in Teradata, you probably just have to change EXTRACT to YEAR/MONTH/DAY(date)
REPLACE FUNCTION MONTHS_BETWEEN(date1 DATE, date2 DATE)
RETURNS FLOAT
SPECIFIC months_between_DT
RETURNS NULL ON NULL INPUT
CONTAINS SQL
DETERMINISTIC
COLLATION INVOKER
INLINE TYPE 1
RETURN
(EXTRACT(YEAR FROM date1) * 12 + EXTRACT(MONTH FROM date1))
- (EXTRACT(YEAR FROM date2) * 12 + EXTRACT(MONTH FROM date2))
+ CASE
WHEN EXTRACT(MONTH FROM date2) <> EXTRACT(MONTH FROM date2+1) AND
EXTRACT(MONTH FROM date1) <> EXTRACT(MONTH FROM date1+1)
THEN 0
ELSE (CAST(1 AS FLOAT))/31 * (EXTRACT(DAY FROM date1) - EXTRACT(DAY FROM date2))
END
;
Then you simply multiply the result * 30 and cast it to an INT.