I am trying to round upwards to the nearest month. So far, I have:
SELECT ROUND(CURRENT_DATE, 'MM') FROM DUAL
Which rounds to the closest month, which is upwards in this case. At the time of posting, the output is 03/01/2019 in MM/DD/YYYY format.
But what if it's the first of a month for example?
SELECT ROUND(TO_DATE('01-03-19','DD-MM-YY'), 'MM') FROM DUAL
This produces the same output as above. But I am expecting 04/01/2019.
I could do something like:
SELECT TRUNC(ADD_MONTHS(TO_DATE('30-03-19','DD-MM-YY'),1), 'MM') - 1 FROM DUAL
Which produced the output 03/31/2019, which is as expected.
I take the 30th March, add one month onto it. Truncate that, to get the first day of that month, then just subtract one for the last day of the previous month.
Now this works, but it seems long and tedious. Surely there is a better way?
if you are trying to find last day of the current month
LAST_DAY(SYSDATE)
if you are trying to find first day of the next month
LAST_DAY(SYSDATE) + 1
Related
I have 70.000 rows of data, including a date time column (YYYY-MM-DD HH24-MM-SS.).
I want to split this data into 3 separate columns; Hour, day and Week number.
The date time column name is 'REGISTRATIONDATE' from the table 'CONTRACTS'.
This is what I have so far for the day and hour columns:
SELECT substr(REGISTRATIONDATE, 0, 10) AS "Date",
substr(REGISTRATIONDATE, 11, 9) AS "Hour"
FROM CONTRACTS;
I have seen the options to get a week number for specific dates, this assignment concerns 70.000 dates so this is not an option.
You (the OP) still have to explain what week number to assign to the first few days in a year, until the first Monday of the year. Do you assign a week number for the prior calendar year? In a Comment I asked about January 1, 2017, as an example; that was a Sunday. The week from January 2 to January 8 of 2017 is "week 1" according to your definition; what week number do you assign to Sunday, January 1, 2017?
The straightforward calculation below assigns to it week number 0. Other than that, the computation is trivial.
Notes: To find the Monday of the week for any given date dt, we can use trunc(dt, 'iw'). iw stands for ISO Week, standard week which starts on Monday and ends on Sunday.
Then: To find the first Monday of the year, we can start with the date January 7 and ask for the Monday of the week in which January 7 falls. (I won't explain that one - it's easy logic and it has nothing to do with programming.)
To input a fixed date, the best way is with the date literal syntax: date '2017-01-07' for January 7. Please check the Oracle documentation for "date literals" if you are not familiar with it.
So: to find the week number for any date dt, compute
1 + ( trunc(dt, 'iw') - trunc(date '2017-01-07', 'iw') ) / 7
This formula finds the Monday of the ISO Week of dt and subtracts the first Monday of the year - using Oracle date arithmetic, where the difference between two dates is the number of days between them. So to find the number of weeks we divide by 7; and to have the first Monday be assigned the number 1, instead of 0, we need to add 1 to the result of dividing by 7.
The other issue you will have to address is to convert your strings into dates. The best solution would be to fix the data model itself (change the data type of the column so that it is DATE instead of VARCHAR2); then all the bits of data you need could be extracted more easily, you would make sure you don't have dates like '2017-02-29 12:30:00' in your data (currently, if you do, you will have a very hard time making any date calculations work), queries will be a lot faster, etc. Anyway, that's an entirely different issue so I'll leave it out of this discussion.
Assuming your REGISTRATIONDATE if formatted as 'MM/DD/YYYY'
the simples (and the faster ) query is based ond to to_char(to_date(REGISTRATIONDATE,'MM/DD/YYYY'),'WW')
(otherwise convert you column in a proper date and perform the conversio to week number)
SELECT substr(REGISTRATIONDATE, 0, 10) AS "Date",
substr(REGISTRATIONDATE, 11, 9) AS "Hour",
to_char(to_date(REGISTRATIONDATE,'MM/DD/YYYY'),'WW') as "Week"
FROM CONTRACTS;
This is messy, but it looks like it works:
to_char(
to_date(RegistrationDate,'YYYY-MM-DD HH24-MI-SS') +
to_number(to_char(trunc(to_date(RegistrationDate,'YYYY-MM-DD HH24-MI-SS'),'YEAR'),'D'))
- 2,
'WW')
On the outside you have the solution previous given by others but using the correct date format. In the middle there is an adjustment of a certain number of days to adjust for where the 1st Jan falls. The trunc part gets the first of Jan from the date, the 'D' gets the weekday of 1st Jan. Since 1 represents Sunday, we have to use -2 to get what we need.
EDIT: I may delete this answer later, but it looks to me that the one from #mathguy is the best. See also the comments on that answer for how to extend to a general solution.
But first you need to:
Decide what to do dates in Jan before the first Monday, and
Resolve the underlying problems in the date which prevent it being converted to dates.
On point 1, if assigning week 0 is not acceptable (you want week 52/53) it gets a bit more complicated, but we'll still be able to help.
As I see it, on point 2, either there is something systematically wrong (perhaps they are timestamps and include fractions of a second) or there are isolated cases of invalid data.
Either the length, or the format, or the specific values don't compute. The error message you got suggests that at least some of the data is "too long", and the code in my comment should help you locate that.
I am brand new to Oracle. I have figured out most of what I need but one field is driving me absolutely crazy. Seems like it should be simple but I think my brain is fried and I just can't get my head around it. I am trying to produce a Sales report. I am doing all kinds of crazy things based on the Invoice Date. The last thing I need to do is to be able to create a Week Number so I can report on weekly sales year vs year. For purposes of this report my fiscal year starts exactly on December 1 (regardless of day of week it falls on) every year. For example, Dec 1-7 will be week 1, etc. I can get the week number using various functions but all of them are based on either calendar year or ISO weeks. How can I easily generate a field that will give me the number of the week since December 1? Thanks so much for your help.
Forget about the default week number formats as that won't work for this specific requirement. I'd probably subtract the previous 1 December from invoice date and divide that by 7. Round down, add 1 and you should be fine.
select floor(
(
trunc(invoiceDate) -
case
-- if December is current month, than use 1st of this month
when to_char(invoiceDate, 'MM') = '12' then trunc(invoiceDate, 'MM')
-- else, use 1st December of previous year
else add_months(trunc(invoiceDate, 'YYYY'), -1)
end
) / 7
) + 1
from dual;
I have a date in sql which will always fall on a Monday and I'm subtracting 13 weeks to get a weekly report. I am trying to get the same 13 week report but for last year's figures as well.
At the moment, I'm using the following:
calendar_date >= TRUNC(sysdate) - 91
which is working fine.
I need the same for last year.
However, when I split this into calendar weeks, there will also be a partially complete week as it will include 1 or 2 days from the previous week. I need only whole weeks.
e.g. the dates that will be returned for last year will be 14-Feb-2015 to 16-May-2015. I need it to start on the Monday and be 16-Feb-2015. This will change each week as I am only interested in complete weeks...
I would do this:
Get the date by substracting 91 days as you're already doing.
Get the number of the day of the week with TO_CHAR(DATE,'d')
Add the number of days until the next monday to the date.
Something like this:
SELECT TO_DATE(TO_DATE('16/05/2015','DD/MM/YYYY'),'DD/MM/YYYY')-91 + MOD(7 - TO_NUMBER(TO_CHAR(TO_DATE(TO_DATE('16/05/2015','DD/MM/YYYY'),'DD/MM/RRRR')-91,'d'))+1,7) d
FROM dual
next_day - returns date of first weekday named by char.
with dates as (select to_date('16/05/2015','DD/MM/YYYY') d from dual)
select
trunc(next_day( trunc(d-91) - interval '1' second,'MONDAY'))
from dates;
I want to get next monday from calculated date. In situation when calculated date is monday i have to move back to previous week ( -1 second).
I'm pretty new to SQL so I hope this isn't a dumb question, tried to google but couldn't find anything.
I'm summing sales of departments per week in SQL and am using TD_SYSFNLIB.WEEKNUMBER_OF_YEAR (trans_dt) to get the week number.
I think everything is working except I'd like to change the format of the weeks to the start date of the week, e.g. week 1 = 1/4/15
Also, i'm not sure how to handle the very first of the year week 0 since I think that should be grouped up with week 52 of last year.
The following date math trick should get you Beginning of Week as an actual date without having to join to the SYS_CALENDAR view or using a function:
SELECT CURRENT_DATE - ((CURRENT_DATE - DATE '0001-01-07) MOD 7) AS BOW;
Starting with TD14 there's NEXT_DAY which returns the following weekday, if you subtract 7 days you get the previous day:
next_day(trans_dt - 7, 'sunday')
I could not find the meaning of the following SQL command:
where date between to_date('2013-03-01', 'yyyy-mm-dd') and trunc(sysdate, 'mm') -1
What does the "-1" mean / does?
The other example is
trunc(months_between(date1, date2))+1
I have searched for this, but could not find a thing.
Thank you for advice!
As others have answered, "date - 1" subtracts one day from the date. Here's more detail on your specific SQL snippets:
where date between to_date('2013-03-01', 'yyyy-mm-dd') and trunc(sysdate, 'mm') -1`
This evaluates to "date between 3/1/2013 and the end of last month"
TRUNC(some date, 'MM') chops the date to the beginning of the month
TRUNC(SYSDATE, 'MM') returns the beginning of the current month
TRUNC(SYSDATE, 'MM')-1 returns the last day of the previous month
trunc(months_between(date1, date2))+1
This is giving the number of full months between date1 and date2, treating any fraction of a month as a whole month. For example, if you gave it the dates 7/28/2013 and 7/29/2013 it would report one month, and it would also report one month if you gave it 7/1/2013 and 7/31/2013.
The MONTHS_BETWEEN function returns, as it implies, the number of months between two dates. The return value will have decimal places - for example a return value of 1.5 means one and a half months.
The TRUNC function, when applied against a numeric, will chop off all its decimals, so TRUNC(1.9999999) will return 1.
+1 is way to add a day to the date
-1 is way to remove a day to the date
In your specific case:
the instruction trunc(sysdate, 'mm') -1 remove one month to the date, in this case is one month before the current date.
the instruction trunc(months_between(date1, date2))+1 compute the difference in month between the two dates and then adds one.
Give a look at this SQLFiddle
it is a lazy way of adding or subtracting day(s)