Can can someone explain me the query in the function - oracle

Question

CREATE OR REPLACE FUNCTION exlude_weekends (p_date_start DATE,
p_date_end DATE)
RETURN NUMBER
AS
l_no_of_days NUMBER := NULL;
BEGIN
SELECT COUNT ( * ) INTO l_no_of_days
FROM (SELECT date_extraction, TO_CHAR (date_extraction, 'DAY')
FROM (SELECT TO_DATE(p_date_start,'DD-MON-RRRR')
+ LEVEL - 1 date_extraction FROM DUAL CONNECT BY LEVEL <
(TO_DATE (p_date_end, 'DD-MON-RRRR')- TO_DATE (p_date_start,'DD-MON-RRRR'))+ 2)
WHERE TRIM (TO_CHAR (date_extraction, 'DAY')) NOT IN ('SATURDAY', 'SUNDAY'));
RETURN l_no_of_days;
EXCEPTION
WHEN OTHERS
THEN
RETURN 0;
END exlude_weekends;

Answer 1

As you mention in the comments, the key of the function is the hierarchical sub-query:
SELECT TO_DATE(p_date_start,'DD-MON-RRRR') + LEVEL - 1 date_extraction
FROM DUAL
CONNECT BY LEVEL <
(TO_DATE(p_date_end,'DD-MON-RRRR')-
TO_DATE(p_date_start,'DD-MON-RRRR'))+ 2
Hierarchical queries try to traverse a tree (CONNECT BY clause specifies how parents and children are related). In this example we find a tricky use (or abuse) of the connect by.
This sub-query generates date from p_date_start to p_date_end (both inclusive). How it does it?
Note that the expression being compare with LEVEL in the CONNECT BY is a constant, and it is the number of days between start and the day after the end date (why the day after the end date? because it is using < and the day after the end date is the first day out of the interval):
(TO_DATE(p_date_end,'DD-MON-RRRR')-TO_DATE(p_date_start,'DD-MON-RRRR'))+2
The select get the DUAL row (it has only one row) this row has LEVEL 1 (hierarchical query use the pseudo-column LEVEL to indicate the depth from the root where it started to evaluate).
The CONNECT BY checks that this level (1) is in the range of days to be generated.
Evaluates the expression:
TO_DATE(p_date_start,'DD-MON-RRRR') + LEVEL - 1
This is the start date plus the level minus one: this is, the start date.
Now a new cycle in hierarchical evaluation starts: the row generated in the previous cycle (the start date) is evaluated again (the new row will have level 2).
If it is in the range of days to be generated (controlled by the CONNECT BY clause) a new date is generated (the day after the start date).
A new cycle start (level 3)....
And the process iterates until LEVEL is greater than the number of days to be generated (which is the same than the number of levels required to iterate from the start date to the end date).
The outer queries in the function only filter SATURDAYS and SUNDAYS and count the remaining days.
Although oracle is very efficient evaluating this query, this function uses a brute force solution.
A more elegant and mathematical solution can be used (with no iterations). We have an equation that computes the number of a particular day of week between two dates:
TRUNC(( END – START – DAYOFWEEK(END-DAYOFWEEKTOBECOUNTED) + 8) / 7)
where DAYOFWEEK is a function that returns 0-6 (0 Sunday, 1 Monday ... 6 Saturday). And DAYOFWEEKTOBECOUNTED is the number of the day to be counted in the same format.
Note that TO_CHAR(date, 'd') returns the day of week in 1..7 format we must rectify to 0..6 format (In my region monday is the first day of week, so i get sunday as 0 and saturday as 6 with the mod function as follows):
MOD(TO_NUMBER(TO_CHAR(p_date_end, 'd')), 7)
Finally we want the number of days in the interval minus the number of sundays (day 0) and saturdays (day 6). So the final procedure with the mathematical approach will be:
CREATE OR REPLACE FUNCTION exlude_weekends (p_date_start DATE,
p_date_end DATE)
RETURN NUMBER
AS
l_no_of_days NUMBER := NULL;
BEGIN
SELECT TRUNC(p_date_end - p_date_start) + 1 -
( TRUNC((p_date_end - p_date_start -
MOD(to_number(to_char(p_date_end - 0, 'd')), 7)+8)/7)
+ TRUNC((p_date_end - p_date_start -
MOD(to_number(to_char(p_date_end - 6, 'd')), 7)+8)/7)
)
INTO l_no_of_days
FROM DUAL;
RETURN l_no_of_days;
EXCEPTION
WHEN OTHERS
THEN
RETURN 0;
END exlude_weekends;