F is a function counting the number of weekdays betwwen two dates (using the format to_date(a, 'DD/MM/YYYY') [NLS_DATE_LANGUAGE = "FRENCH"].
It is stored in package P.
With a is the beginning date of the period and b the end.
When I use the SQL developer to run the function i.e. Right click on package p, execute, choose f function and fill the arguments:
if a = to_date('02/08/2019', 'DD/MM/YYYY') and b = to_date('12/08/2019', 'DD/MM/YYYY'))
I get a number of days = 7, which is the correct result.
But when I run it in SQL:
select p.f('02/08/2019','12/08/2019') from DUAL
I get a number of days = 8 and this result is obviously wrong.
I cannot understand how these two process can return two different values since the same F function is called, and the arguments are exactly the same.
I have been careful about the date format, still the two ways of calling f won't return the same result.
For information, this is the content of this function (sorry, cariable are expressed in french)
FUNCTION getjoursouvres
(
i_debut IN DATE,
i_fin IN DATE
)
RETURN NUMBER IS o_result NUMBER;
v_jour date;
v_nbjours NUMBER;
v_testferie number;
v_testweekend number;
v_testglobal number;
-- This cursor browses all dates between i_debut and i_fin (included)
cursor cx is
select to_date(i_debut, 'DD/MM/YYYY') + rownum -1 dt
from dual
connect by level <= to_date(i_fin, 'DD/MM/YYYY') - to_date(i_debut, 'DD/MM/YYYY') + 1;
BEGIN
open cx;
v_nbjours := 0;
loop
-- Browses all the days in the interval (begining and end included)
fetch cx into v_jour;
exit when cx%NOTFOUND;
-- testferie return 1 if the day is NOT a holiday, 0 if it is (so not be be added)
v_testferie := testferie(v_jour);
-- testweekend return 1 if the day is a weekday, 0 if it is on weekend (so not be be
v_testweekend := testweekend(v_jour);
--
v_testglobal := v_testferie + v_testweekend;
-- If v_testglobal = 2 then the day is neither weekend nor holiday. Therefore it is aded to the sum of days
if v_test = 2 then
v_nbjours := v_nbjours + 1;
end if;
end loop;
o_result := v_nbjours;
close cx;
return o_result;
END;
After many calls of both way to use the function. I reach to the conclusion that the element inducing the anomaly comes from the ue of
to_date(v_date, 'DD/MM/YYYY')
where v_date is already declared as a date format behave in an inpredictable way.
I decided to keep the date variable as such, and only use to_date for different cariable format.
I am not able to explain why exactly, but it solved my problems concerning dates behavior. Now I can call a function with data parameter from DUAL or from the SQL developer interface and get identical and coherent results.
I am trying to write a Trigger for before insert to validate a case date.
The case date should be between 5 years before and 7 years after current date.
For example, in 2018 the case date should be from year 2013 to 2025. If the date is out of range the trigger should stop inserting data.
CREATE OR REPLACE TRIGGER ch
BEFORE INSERT
on CASE
FOR EACH ROW
DECLARE
CASN number;
BEGIN
SELECT COUNT(*)
INTO CASN
FROM CASE
WHERE :new.STARTDATE > SYSDATE;
IF (CASN > 0) THEN
RAISE_APPLICATION_ERROR(-20000,'Start DATE CANNOT Be GREATER than today's
date');
END IF;
END;
Here STARTDATE is column of CASE table
This trigger starts when start date is greater than today's date but I need it to run when it's out range as given above.
How do I add an specified interval to the sysdate, so that it could work for the above condition?
The logic you are using in your Trigger is completely wrong. You don't need to get the count from the table using :NEW.STARTDATE. This is something what you are looking for.
CREATE OR replace TRIGGER ch
BEFORE INSERT ON cases
FOR EACH ROW
BEGIN
IF ( :NEW.casedate < SYSDATE - INTERVAL '5' year
OR :NEW.casedate > SYSDATE + INTERVAL '7' year ) THEN
RAISE_APPLICATION_ERROR(-20000,
'CASE DATE should be in range: current date - 5 years and current date + 7 years')
;
END IF;
END;
/
EDIT : I have not added TRUNC on the dates because I'm not sure if you want to consider time component as well while considering date range.If you are ok with just considering days, you may use TRUNC(SYSDATE) in place of just SYSDATE. Modify it accordingly as per your business needs.
Another option is to use CHECK constraint. Although Oracle does not allow you to have
use SYSDATE in a check constraint definition, you may create another column( or reuse existing) that defaults to SYSDATE and apply check constraint on that.
ALTER TABLE CASES ADD ( CURR_DATE DATE DEFAULT SYSDATE );
ALTER TABLE CASES ADD CONSTRAINT
RANGE_CHECK CHECK( casedate > CURR_DATE - INTERVAL '5' YEAR
AND casedate < CURR_DATE + INTERVAL '7' YEAR) ENABLE;
Date arithmetic. Oracle Database enables you to perform arithmetic operations on dates and time stamps in several ways:
Add a numeric value to or subtract it from a date, as in SYSDATE + 7; Oracle Database treats the number as the number of days.
Add one date to or subtract it from another, as in l_hiredate - SYSDATE.
Use a built-in function to “move” a date by a specified number of months or to another date in a week.
Here are some examples of date arithmetic with a date and a number (assume in all cases that the l_date variable has been declared as DATE):
Set a local variable to tomorrow’s date:
l_date := SYSDATE + 1;
Move back one hour:
l_date := SYSDATE - 1/24;
Move ahead 10 seconds:
l_date := SYSDATE + 10 / (60 * 60 * 24);
When you add one date to or subtract it from another, the result is the number of days between the two. As a result, executing this block:
DECLARE
l_date1 DATE := SYSDATE;
l_date2 DATE := SYSDATE + 10;
BEGIN
DBMS_OUTPUT.put_line (
l_date2 - l_date1);
DBMS_OUTPUT.put_line (
l_date1 - l_date2);
END;
returns the following output:
10
-10
I would like to find the simplest way of calculating the difference in hours between 2 dates from Oracle datetime fields whilst ignoring the days, months & years portion. For example:
Datetime 1 (DATE variable) = 10/05/2017 16:00:00
Datetime 2 (DATE variable) = 15/05/2017 19:34:23
Required result (NUMBER output) = 3.576 hours
This is formula will be used in a PLSQL procedure, the output needs to be a number as above. I would imagine some combination of TO_DATE & TRUNC might work. Any help would be most appriciated and apologies if this is a duplicate question.
Use the sssss date mask to get just the time element as the number of seconds since midnight. Then it's just a matter of simple arithmentic:
select (to_number(to_char(datetime2, 'sssss'))
- to_number(to_char(datetime1, 'sssss')) / 3600 as diff_hours
from dual;
PL/SQL version is the same....
declare
Datetime1 DATE := to_date('10/05/2017 16:00:00', 'dd/mm/yyyy hh24:mi:ss');
Datetime2 DATE := to_date('15/05/2017 19:34:23', 'dd/mm/yyyy hh24:mi:ss');
hours_diff number;
begin
hours_diff := (to_number(to_char(datetime2, 'sssss'))
- to_number(to_char(datetime1, 'sssss'))) / 3600 ;
dbms_output.put_line(hours_diff);
end;
I need a PL/SQL function that takes a date and returns the year as a number, including a fractional part from the days. For example, 1995-01-01 would be 1995.000 and 2015-11-24 would be 2015.897. This is what I managed to produce:
CREATE OR REPLACE FUNCTION year_fraction (in_date DATE)
RETURN NUMBER
IS
year NUMBER;
days NUMBER;
total_days NUMBER;
BEGIN
year := EXTRACT(YEAR FROM in_date);
days := in_date - TRUNC(in_date, 'YEAR');
total_days := ADD_MONTHS(TRUNC(in_date, 'YEAR'), 12) - TRUNC(in_date, 'YEAR');
RETURN year + days / total_days;
END year_fraction;
It works, but it feels like to much work to solve quite a simple problem. Can this be done in a neater way?
So you want the decimal part as percentage of days passed in current year. It would be a simple date arithmetic to divide the day of the year with total days in that year.
SQL> SELECT EXTRACT(YEAR FROM SYSDATE)
2 ||'.'
3 ||TRUNC(TO_NUMBER(TO_CHAR(SYSDATE, 'DDD'))/
4 (add_months(TRUNC(SYSDATE,'year'), 12) - TRUNC(SYSDATE,'year'))*1000) my_format
5 FROM DUAL;
MY_FORMAT
--------------------------------------------------------------------------------
2015.898
Eventually I ended up using this solution. It is quite similar to what I started out with in the question, but involves fewer function calls and is a bit more condensed.
CREATE OR REPLACE FUNCTION fractional_year (in_date DATE)
RETURN NUMBER
IS
year DATE;
BEGIN
year := TRUNC(in_date, 'YEAR');
-- Year + Days passed so far / Total number of days in year
RETURN EXTRACT(YEAR FROM year) + (in_date - year) / (ADD_MONTHS(year, 12) - year);
END fractional_year;
In Oracle, is there a function that calculates the difference between two Dates? If not, is a way to display the difference between two dates in hours and minutes?
Query:
SELECT Round(max((EndDate - StartDate ) * 24), 2) as MaximumScheduleTime,
Round(min((EndDate - StartDate) * 24), 2) as MinimumScheduleTime,
Round(avg((EndDate - StartDate) * 24), 2) as AveragegScheduleTime
FROM table1
You can subtract two dates in Oracle. The result is a FLOAT which represents the number of days between the two dates. You can do simple arithmetic on the fractional part to calculate the hours, minutes and seconds.
Here's an example:
SELECT TO_DATE('2000/01/02:12:00:00PM', 'yyyy/mm/dd:hh:mi:ssam')-TO_DATE('2000/01/01:12:00:00AM', 'yyyy/mm/dd:hh:mi:ssam') DAYS FROM DUAL
Results in: 1.5
You can use these functions :
1) EXTRACT(element FROM temporal_value)
2) NUMTOYMINTERVAL (n, unit)
3) NUMTODSINTERVAL (n, unit).
For example :
SELECT EXTRACT(DAY FROM NUMTODSINTERVAL(end_time - start_time, 'DAY'))
|| ' days ' ||
EXTRACT(HOUR FROM NUMTODSINTERVAL(end_time - start_time, 'DAY'))
||':'||
EXTRACT(MINUTE FROM NUMTODSINTERVAL(end_time - start_time, 'DAY'))
||':'||
EXTRACT(SECOND FROM NUMTODSINTERVAL(end_time - start_time, 'DAY'))
"Lead Time"
FROM table;
With Oracle Dates, this is pretty
trivial, you can get either TOTAL
(days, hours, minutes, seconds)
between 2 dates simply by subtracting
them or with a little mod'ing you can
get Days/Hours/Minutes/Seconds
between.
http://asktom.oracle.com/tkyte/Misc/DateDiff.html
Also, from the above link:
If you really want 'datediff' in your
database, you can just do something
like this:
SQL> create or replace function datediff( p_what in varchar2,
2 p_d1 in date,
3 p_d2 in date ) return number
4 as
5 l_result number;
6 begin
7 select (p_d2-p_d1) *
8 decode( upper(p_what),
9 'SS', 24*60*60, 'MI', 24*60, 'HH', 24, NULL )
10 into l_result from dual;
11
11 return l_result;
12 end;
13 /
Function created
Q: In Oracle, is there a function that calculates the difference between two Dates?
Just subtract one date expression from another to get the difference expressed as a number of days. The integer portion is the number of whole days, the fractional portion is the fraction of a day. Simple arithmetic after that, multiply by 24 to get hours.
Q: If not, is a way to display the difference between two dates in hours and minutes?
It's just a matter of expressing the duration as whole hours and remainder minutes.
We can go "old school" to get durations in hhhh:mi format using a combination of simple builtin functions:
SELECT decode(sign(t.maxst),-1,'-','')||to_char(floor(abs(t.maxst)/60))||
decode(t.maxst,null,'',':')||to_char(mod(abs(t.maxst),60),'FM00')
as MaximumScheduleTime
, decode(sign(t.minst),-1,'-','')||to_char(floor(abs(t.minst)/60))||
decode(t.minst,null,'',':')||to_char(mod(abs(t.minst),60),'FM00')
as MinimumScheduleTime
, decode(sign(t.avgst),-1,'-','')||to_char(floor(abs(t.avgst)/60))
decode(t.avgst,null,'',':')||to_char(mod(abs(t.avgst),60),'FM00')
as AverageScheduleTime
FROM (
SELECT round(max((EndDate - StartDate) *1440),0) as maxst
, round(min((EndDate - StartDate) *1440),0) as minst
, round(avg((EndDate - StartDate) *1440),0) as avgst
FROM table1
) t
Yeah, it's fugly, but it's pretty fast. Here's a simpler case, that shows better what's going on:
select dur as "minutes"
, abs(dur) as "unsigned_minutes"
, floor(abs(dur)/60) as "unsigned_whole_hours"
, to_char(floor(abs(dur)/60)) as "hhhh"
, mod(abs(dur),60) as "unsigned_remainder_minutes"
, to_char(mod(abs(dur),60),'FM00') as "mi"
, decode(sign(dur),-1,'-','') as "leading_sign"
, decode(dur,null,'',':') as "colon_separator"
from (select round(( date_expr1 - date_expr2 )*24*60,0) as dur
from ...
)
(replace date_expr1 and date_expr2 with date expressions)
let's unpack this
date_expr1 - date_expr2 returns difference in number of days
multiply by 1440 (24*60) to get duration in minutes
round (or floor) to resolve fractional minutes into integer minutes
divide by 60, integer quotient is hours, remainder is minutes
abs function to get absolute value (change negative values to positive)
to_char format model FM00 give two digits (leading zeros)
use decode function to format a negative sign and a colon (if needed)
The SQL statement could be made less ugly using a PL/SQL function, one that takes two DATE arguments a duration in (fractional) days and returns formatted hhhh:mi
(untested)
create function hhhhmi(an_dur in number)
return varchar2 deterministic
is
begin
if an_dur is null then
return null;
end if;
return decode(sign(an_dur),-1,'-','')
|| to_char(floor(abs(an_dur)*24))
||':'||to_char(mod((abs(an_dur)*1440),60),'FM00');
end;
With the function defined:
SELECT hhhhmi(max(EndDate - StartDate)) as MaximumScheduleTime
, hhhhmi(min(EndDate - StartDate)) as MinimumScheduleTime
, hhhhmi(avg(EndDate - StartDate)) as AverageScheduleTime
FROM table1
You can use the months_between function to convert dates to the difference in years and then use between the decimal years you are interested:
CASE
WHEN ( ( MONTHS_BETWEEN( TO_DATE(date1, 'YYYYMMDD'),
TO_DATE(date1,'YYYYMMDD'))/12
)
BETWEEN Age1DecimalInYears AND Age2DecimalInYears
)
THEN 'It is between the two dates'
ELSE 'It is not between the two dates'
END;
You may need to change date format to match the a given date format and verify that 31 day months work for your specific scenarios.
References:
( found on www on 05/15/2015 )
1. Oracle/PLSQL: MONTHS_BETWEEN Function
2. Oracle Help Center - MONTHS_BETWEEN