I have this anonymous block:
DECLARE
V_DATA DATE;
BEGIN
V_DATA := '01-GEN-2000';
HR.STATISTICHE.RATINGOPERATORI (V_DATA);
COMMIT;
END;
but I would to generate the date in a random way. How can I do?
You can generate random dates between two dates ,as displayed in the query below .Random Dates are generated between 1-jan-2000 and 31-dec-9999
SELECT TO_DATE(
TRUNC(
DBMS_RANDOM.VALUE(TO_CHAR(DATE '2000-01-01','J')
,TO_CHAR(DATE '9999-12-31','J')
)
),'J'
) FROM DUAL;
OR you can use
SELECT TO_DATE (
TRUNC (
DBMS_RANDOM.VALUE (2451545, 5373484)
)
, 'J'
)
FROM DUAL
In the above example ,the first value is 01-Jan-2000 and the second value id 31-dec-9999
To generate random date you can use
select to_date('2010-01-01', 'yyyy-mm-dd')+trunc(dbms_random.value(1,1000)) from dual
or for random datetime
select to_date('2010-01-01', 'yyyy-mm-dd')+dbms_random.value(1,1000) from dual
If you want to see it's logic, you can also use this code.
create or replace procedure genDate(result out nvarchar2) IS
year number;
month number;
day number;
Begin
year:=FLOOR(DBMS_RANDOM.value(2000,2100));
month:=FLOOR(DBMS_RANDOM.value(1,12));
IF month=2 and (year/4)=0 and (year/100)!=0 then
day:=FLOOR(DBMS_RANDOM.value(1,29));
ELSIF month=2 or (year/100)=0 then
day:=FLOOR(DBMS_RANDOM.value(1,28));
ELSIF MOD(month,2)=1 then
day:=FLOOR(DBMS_RANDOM.value(1,31));
ELSIF MOD(month,2)=0 and month!=2 then
day:=FLOOR(DBMS_RANDOM.value(1,30));
END IF;
result:=month||'-'||day||'-'||year;
End;
here is one more option to generate date going back from now where 365 - days quanitity to move back from today, 'DD.MM.YYYY'- mask
to_char(sysdate-dbms_random.value()*365, 'DD.MM.YYYY')
I needed to generate employee data for testing. Each employee needed a date of birth that put them between 16 and 65 years of age, and a date of hire sometime between their 16th birthday and SYSDATE. Here's how...
FUNCTION randomDateInRange(alpha IN DATE, omega IN DATE) RETURN DATE IS
BEGIN
RETURN alpha + DBMS_RANDOM.VALUE(0, omega - alpha);
END;
...and then, to use this function...
-- an employee can be any age from 16 to 65 years of age
DoB := randomDateInRange(
SYSDATE - INTERVAL '65' YEAR,
SYSDATE - INTERVAL '16' YEAR
);
-- an employee could have been hired any date since their sixteenth birthday
DoH := randomDateInRange(
DoB + INTERVAL '16' YEAR,
SYSDATE
);
Related
Im trying to add hours to midnight of today eg: like 27 hours
I have tried various methods from the internet but am getting the trunc of the dated expected. eg 23-nov-2022 not 23-nov-2022 03:00. when i run it outside my pl/sql procedure/block i get the desired output
the select:
select to_char(to_date(sysdate,'DD-MON-RRRR HH:MI')+hours/24,'DD-MON-RRRR HH:MI') into v_from from dual;
I need some expert assistance
Add an INTERVAL DAY TO SECOND data type to SYSDATE TRUNCated back to midnight:
DECLARE
v_from DATE;
BEGIN
SELECT TRUNC(sysdate) + INTERVAL '27' HOUR
INTO v_from
FROM DUAL;
DBMS_OUTPUT.PUT_LINE(TO_CHAR(v_from,'DD-MON-RRRR HH:MI'));
END;
/
or, more simply:
DECLARE
v_from DATE;
BEGIN
v_from := TRUNC(sysdate) + INTERVAL '27' HOUR;
DBMS_OUTPUT.PUT_LINE(TO_CHAR(v_from,'DD-MON-RRRR HH:MI'));
END;
/
or, dynamically, with the NUMTODSINTERVAL function:
DECLARE
v_from DATE;
v_hours NUMBER(3,0) := 27;
BEGIN
v_from := TRUNC(sysdate) + NUMTODSINTERVAL(v_hours,'HOUR');
DBMS_OUTPUT.PUT_LINE(TO_CHAR(v_from,'DD-MON-RRRR HH:MI'));
END;
/
fiddle
Trunc SYSDATE to get midnight. Then add hours: 1 hour is 1/24 day so you'd add the number of hours divided by 24. Example.
koen>DECLARE
2 l_date DATE;
3 l_hours NUMBER := 27;
4 BEGIN
5 l_date := TRUNC(SYSDATE) + 27/24;
6 dbms_output.put_line('l_date is: '||TO_CHAR(l_date,'DD-MON-YYYY HH24:MI'));
7 END;
8* /
l_date is: 23-NOV-2022 03:00
PL/SQL procedure successfully completed.
koen>
The select from dual is not advised, you can just assign a variable in pl/sql using the assignment operator :=. The select from dual requires an additional context switch (invoke the sql engine from within pl/sql).
I need a help on this function. The code has no error but it keeps returning the same result, that the value from of the second condition/statement.
This is how it suppose to work:
If employees age at hire date (DOFA) is less than or equals to 25 the retirement date is 35 years from hire date. Otherwise retirement date is when employees age is 60
create or replace function EDOR_FUNCTION
(DOFA in date, DOB in date)
return date
is
new_edor_date date;
begin
if
DOFA - DOB <= 25 then new_edor_date := add_months(DOFA, 35*12);
else
new_edor_date := add_months(DOB, 60*12);
end if;
return new_edor_date;
end;
Your condition subtracts one date from another. This gives the number of days between the two, not the number of years.
months_between() gives the number of months between two dates. Multiply by twelve to get number of years
if months_between(DOFA , DOB) <= (25*12) then
new_edor_date := add_months(DOFA, 35*12);
else
new_edor_date := add_months(DOB, 60*12);
end if;
I got parameter
:dateFrom
which gonna be used as an argument in a function as a TIMESTAMP. I need to add to :dateFrom + 7 hours, how can I do that?
If your parameter is not already a timestamp, use to_timestamp or to_date to convert it:
to_timestamp(dateFrom,'mm/dd/yyyy hh24:mi:ss')
(substitute the appropriate mask based on the format of your input parameter)
Then just add 7/24.
to_timestamp(dateFrom,'mm/dd/yyyy hh24:mi:ss') + 7/24;
Adding 1 adds a full day, so adding 1/24 adds 1 hour.
This can also be done with the INTERVAL operator:
to_timestamp(dateFrom,'mm/dd/yyyy hh24:mi:ss') + INTERVAL '7' hour
Here is some PL/SQL that will:
DECLARE
dateFrom TIMESTAMP;
BEGIN
dateFrom := SYSTIMESTAMP;
DBMS_OUTPUT.PUT_LINE('BEFORE :: ' || dateFrom);
dateFrom := dateFrom + INTERVAL '2' HOUR;
DBMS_OUTPUT.PUT_LINE('AFTER :: ' || dateFrom);
END;
/
Look into the INTERVAL operator.
You can do like this,
select dateFrom + interval '7' hours from dual
In the arithmetic of the dates, in Oracle, when you add a number, it is intended as NUMBER OF DAYS.
7 hours are 7/24 days, so you have simply to add 7/24.
i want to make a function in oracle database. in witch i want to get 1 day previous date from selected date, if selected date is "Monday" then i want to select date of "Friday" as previous date as it is last working day of week.
Try this:
CREATE OR REPLACE FUNCTION PREVIOUS_WORKING_DAY(pDate IN DATE)
RETURN DATE
IS
strDay_of_week VARCHAR2(50);
dtPrev DATE;
BEGIN
strDay_of_week := TRIM(TO_CHAR(pDate, 'DAY'));
CASE strDay_of_week
WHEN 'MONDAY' THEN
dtPrev := TRUNC(pDate) - INTERVAL '3' DAY;
WHEN 'SUNDAY' THEN
dtPrev := TRUNC(pDate) - INTERVAL '2' DAY;
ELSE
dtPrev := TRUNC(pDATE) - INTERVAL '1' DAY;
END CASE;
RETURN dtPrev;
END PREVIOUS_WORKING_DAY;
Share and enjoy.
can i use to_number(to_char()) function in order to exclude all the weekends from a range of dates?
For instance, there are two date columns in my table such as start and finish (in the form of '06/06/2011 10:00:00 am'), and i want to estimate the duration of finish-start excluding Saturdays and Sundays.
If I understand you right you want to calculate the difference between two dates, but exclude the 2 days of each weekend in the range from the result. Is that correct?
If this is what you want the below code should work with the following assumptions:
I am assume start and end will not be on weekends.
I am not validating that end is before start.
Basically its just a matter of working out how many weekends are in the date range. So obviously there's one weekend per 7 days. Then we just have to check if the range wraps around a weekend, and if so add one more.
FUNCION dateDiff( dt_start DATE, dt_end DATE ) RETURN NUMBER
IS
raw_diff NUMBER;
weekends NUMBER;
BEGIN
raw_diff := dt_end - dt_start;
weekends := TRUNC( raw_diff / 7 );
IF( ( dt_start - TRUNC( dt_start, 'DAY' ) )
> ( dt_end - TRUNC( dt_end , 'DAY' ) ) )
THEN
weekends := weekends + 1;
END IF;
RETURN raw_diff - ( weekends * 2 );
END;