Oracle CASE next business day - sql

I want to calculate next business day in Oracle query. So on Monday - Thursday it should be sysdate+1, on Friday it should be sysdate+3, on Saturday it should be sysdate+2 and on Sunday it should be sysdate+1.
I want to do it dynamically, rather than have lots of where statements covering the different days.
where order_date = CASE WHEN (1 + TRUNC (SYSDATE) - TRUNC (SYSDATE, 'IW')) < 5
THEN TRUNC (SYSDATE) + 1
ELSE TRUNC (SYSDATE + 4), 'IW')
END
I found this from the answer Next business day (Monday - Friday) in Oracle? but it seems error in query and I can't work it out.


You can use to_char(sysdate, 'fmday') which returns the string name of the day:
where order_date =
case
when to_char(sysdate, 'fmday', 'nls_date_language=AMERICAN') = 'friday' then trunc(sysdate) + 3
when to_char(sysdate, 'fmday', 'nls_date_language=AMERICAN') = 'saturday' then trunc(sysdate) + 2
else trunc(sysdate) + 1
end
and better:
where order_date =
trunc(sysdate) + case
when to_char(sysdate, 'fmday', 'nls_date_language=AMERICAN') = 'friday' then 3
when to_char(sysdate, 'fmday', 'nls_date_language=AMERICAN') = 'saturday' then 2
else 1
end


to have the results from the statement, use below statement.
where order_date =
(Select
CASE WHEN (1 + TRUNC (SYSDATE) - TRUNC (SYSDATE, 'IW')) < 5
THEN TRUNC (SYSDATE) + 1
ELSE TRUNC ((SYSDATE + 4), 'IW')
END from dual);

Related

I'm unable to find tabular format result in oracle sql developer

I'm unable to get output as tabular format.
DECLARE
prev VARCHAR2(10) := 'sysdate+1';
tdy VARCHAR2(10) := 'sysdate';
v_dt VARCHAR2(10);
result resultset;
BEGIN
v_dat :=
CASE
WHEN trunc(sysdate) = trunc(currnet_date) THEN
tdy
ELSE prev
END;
WITH
yesterday AS (
SELECT
src_dim_id,
rec_count
FROM
tab_total_last_updated
WHERE
to_char(dw_insert_datetime, 'mm-dd-yy') = CASE
WHEN TRIM(to_char(v_dt, 'DAY', 'NLS_DATE_LANGUAGE=ENGLISH')) =
'MONDAY' THEN
to_char(v_dt - 3, 'mm-dd-yy')
ELSE
to_char(v_dt - 1, 'mm-dd-yy')
END
AND table_type = 'EXT'
)
SELECT
*
INTO result
FROM
yesterday;
RETURN result;
END;
I'm trying to pass value sysdate into v_dt as per the condition but I'm unable to proceed with my code please help on this one.

You have many typos in your code v_dat should be v_dt, current_date is misspelt.
You cannot RETURN a result as you are using an anonymous PL/SQL block and not a function.
Also, you probably do not want to use PL/SQL and just want an SQL query (and do not need the sub-query factoring clause):
SELECT src_dim_id,
rec_count
FROM tab_total_last_updated
WHERE dw_insert_datetime >= CASE TRUNC(SYSDATE) - TRUNC(SYSDATE, 'IW')
WHEN 0 THEN TRUNC(SYSDATE) - 3 -- Monday
WHEN 6 THEN TRUNC(SYSDATE) - 2 -- Sunday
ELSE TRUNC(SYSDATE) - 1
END
AND dw_insert_datetime < CASE TRUNC(SYSDATE) - TRUNC(SYSDATE, 'IW')
WHEN 0 THEN TRUNC(SYSDATE) - 2 -- Monday
WHEN 6 THEN TRUNC(SYSDATE) - 1 -- Sunday
ELSE TRUNC(SYSDATE)
END
AND table_type = 'EXT';
It is unclear what you are trying to achieve using:
v_dt := CASE
WHEN trunc(sysdate) = trunc(current_date)
THEN SYSDATE
ELSE SYSDATE + 1
END;
But you appear to be trying to correct for differences between the dates on the server and the client. In which case either:
Use CURRENT_DATE throughout the above code instead of SYSDATE; or
Use a specific time zone and replace SYSDATE with, for example, SYSTIMESTAMP AT TIME ZONE 'UTC'.

Case when time is beween and on weekends

I am trying to do a case when on one of my columns, but the time has to be between two times during the week and different on the weekends. So if sched_time is between 8:30:00 and 14:45:00 during the week (M-F) then 'ABC' else 'DEF'. The sched_time column looks like this 2020-03-27 09:29:00. Here is what I have so far:
SELECT Client_Last_Name,
Client_First_Name,
Sched_Time,
CASE WHEN Sched_time IS BETWEEN '08:30:00' AND '14:45:00'
THEN 'ABC'
ELSE 'DEF'
END
Field_Name,
Recoded_Response,
Dlsequence
FROM DAILY_LOG_CUSTOM_DATA
WHERE SERVICE_NAME = 'Medical'
AND FIELD_CATEGORY = 'Background Information'
AND Field_Name = 'Restraint?'
AND Recoded_Response = 1
And Sched_Time >= TRUNC(SYSDATE + 1, 'IW') - 8 AND
Sched_Time <= TRUNC(SYSDATE + 1, 'IW') - 1

Assuming that SCHED_TIME is of type DATE then you can take advantage of INTERVAL arithmetic:
SELECT Client_Last_Name,
Client_First_Name,
Sched_Time,
CASE
WHEN Sched_time BETWEEN TRUNC(SCHED_TIME) + INTERVAL '8' HOUR + INTERVAL '30' MINUTE
AND TRUNC(SCHED_TIME) + INTERVAL '14' HOUR + INTERVAL '45' MINUTE AND
TO_CHAR(SCHED_TIME, 'DY') IN ('MON', 'TUE', 'WED', 'THU', 'FRI')
THEN 'ABC'
ELSE 'DEF'
END
Field_Name,
Recoded_Response,
Dlsequence
FROM DAILY_LOG_CUSTOM_DATA
WHERE SERVICE_NAME = 'Medical'
AND FIELD_CATEGORY = 'Background Information'
AND Field_Name = 'Restraint?'
AND Recoded_Response = 1
AND Sched_Time >= TRUNC(SYSDATE + 1, 'IW') - 8 AND
Sched_Time <= TRUNC(SYSDATE + 1, 'IW') - 1

You can use to_char(). Assuming English language settings:
CASE WHEN TO_CHAR(Sched_time, 'HH24:MI:SS') BETWEEN '08:30:00' AND '14:45:00' AND
TO_CHAR(Sched_time, 'DY') NOT IN ('SAT', 'SUN')
THEN 'ABC'
ELSE 'DEF'
END
You can adjust this to include international settings, so the value is in English -- or your native language -- if that is preferable.

Business days between two dates with calendar table [duplicate]

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,

how to find the difference between two dates using Oracle sql by excluding weekends [duplicate]

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,

Find Quarter from first day of the week

I am trying to find the Quarter from the firstday of the week.
For example, The first day of the week is 12/31/2012
I want the quarter to say Q1 2013. But if I find the Qtr from the specific date it says Q4 2012.

I'm not sure this will work for all circumstances in your case, so please test out with different dates:
DECLARE #date DATETIME
SET #date = '2012-12-31'
SELECT
qtr = DATEPART(quarter, DATEADD(DAY, 7 - DATEPART(weekday, #Date), #date)),
YearNum = DATEPART(YEAR, DATEADD(DAY, 7 - DATEPART(weekday, #Date), #date))

This is for Oracle - annual week table... May give you some ideas:
-- ISO_WK# --
SELECT mydate
, TRUNC(mydate, 'w') wk_starts
, TRUNC(mydate, 'w') + 7 - 1/86400 wk_ends
, TO_NUMBER (TO_CHAR (mydate, 'IW')) ISO_wk#
FROM
(
SELECT TRUNC(SYSDATE, 'YEAR')-1 + LEVEL AS mydate -- 1st day of yr-1 day --
FROM dual
CONNECT BY LEVEL <=
(-- First day of curr year – first day of past year --
SELECT TRUNC(SYSDATE, 'YEAR')-TRUNC(ADD_MONTHS (SYSDATE, -12), 'YEAR') "Num of Days"
FROM dual
)
)
/