I am not a SQL user, it happens that I need to know what this type of SQL code means, can you help me please translating it to me?
CASE
WHEN DATEDIFF(to_date(B.DT_CAREPACK_END),
LAST_DAY(date_sub(add_months(current_date, 3), 20))) > 0
CASE statements let you introduce some conditional logic.
A full CASE sttatement would have a syntax like:
CASE
WHEN some logic is true THEN x
WHEN some other logic is true THEN y
ELSE THEN z
END as column_title
In your example, it doesn't look like you've provided the full statement as their is no END keyword.
Basically, this logic is checking when the difference between two dates (date-x and date-y) is positive or not. DATE_DIFF looks at the different between a start date (date-x) and an end date (date-y).
If date-x, the start date, is before date-y, the end date, then the result is positive. If the start date is after the end date, then the result is negative.
date-x is a date representation of the column DT_CAREPACK_END
date-y is taking the current_date, adding on 3 months (e.g. 4th September becomes 4th December), is is then subtracting 20 units (presumably days) and then setting that date to the last date of that month.
So, imagine DT_CAREPACK_END (presumably a date when something ends) is in the future and is 2022-10-02.
The inner logic here will take the current date (2022-09-04) and add 3 months to that date, making it 2022-12-04. Then, we are subtracting 20 days which is 2022-11-14. Then, we find the last day in that month, which would be 2022-11-30.
Finally, we look at the difference between 2022-10-02 (start date) and 2022-11-30 (end date). If that is a positive number, then the logic is satisfied. In this case, 2nd October is before 30th November, resulting in a positive logic and therefore the case logic is satisfied.
If the DT_CAREPACK_END is before the current_date logic, then it would be negative.
*N.B. I thought that date_add, date_sub and date_diff functions needed an interval unit to be explicitly stated (e.g. INTERVAL 20 DAY). I'm guessing the default here is days but that's an assumption on my part. I'm working in good-faith that the code snip is syntatically correct. *
Resources:-
Add Months:
https://docs.oracle.com/cd/B19306_01/server.102/b14200/functions004.htm
Date Sub:
https://docs.oracle.com/cd/E17952_01/mysql-5.7-en/date-and-time-functions.html#function_date-add
Last Day:
https://docs.oracle.com/cd/E17952_01/mysql-5.7-en/date-and-time-functions.html#function_last-day
Related
I need to round up a month-date based on certain parameters. For example: If I have a parameter where if a day in a given month is between the 6th and the 4th of the next month, I need my query to return the next months date. Is there a way to round up the month given these parameters without hard coding case whens for every single month ever?
SELECT case when date_trunc('day',li.created_at between '2019-03-06 00:00:00' and '2019-04-06 00:00:00' then '2019-04-01' end)
FROM line_items li
If you want the beginning of the month, but offset by 4 days, you can use date_trunc() and subtract some number of days (or add some number of days). You seem to want something like this:
select dateadd(month, 1, date_trunc('month', li.created_at - interval '4 day'))
Another approach is to create a canonical "dates" table that precomputes the mapping from a given date to a new date using your rounding scheme. The mapping could be done outside of redshift in a script and the table loaded in (or within redshift using a user defined function).
I need a WHERE statement where the date of the record is the previous day. I have the below code which will do this
WHERE DOC_DATE = dateadd(day,datediff(day,1,GETDATE()),0)
However I need this statement to get Friday's record when the current day is Monday. I have the below code but it will not work for me. No errors come back on SQL although no records results come back either. I have the below code for this
WHERE DOC_DATE = DATEADD(day, CASE WHEN datepart(dw,(GETDATE())) IN (2) then -3 ELSE -1 END ,0)
Important to add that this needs to be in a WHERE clause. This is for a Docuware administrative view I am creating. I have no control on how to write the SELECT statement, I only have access to edit the WHERE clause:
Here's a slightly "magical" way to compute the value that doesn't depend on any particular server settings such as datefirst. It's probably not immediately obvious how it works:
WHERE DOC_DATE = dateadd(day,datediff(day,'20150316',getdate()),
CASE WHEN DATEPART(weekday,getdate()) = DATEPART(weekday,'20150330')
THEN '20150313'
ELSE '20150315' END)
In the first line, we're computing the number of days which have elapsed since some arbitrary date. I picked a day in March 2015 to use as my base date.
The second line asks what today's day of the week is and if it's the same as some arbitrary "Known good" Monday. Just taking one value and comparing it to 2 depends on what your DATEFIRST setting is so I prefer not to write that.
In the third line, we decide what to do if it's a monday - I give a date that is 3 days before my arbitrary date above. If it wasn't a monday, we pick the day before.
Adding it all together, when we add the days difference from the arbitrary date back to one of these two dates from lines 3 and 4, it has the effect of shifting the date backwards 1 or 3 days.
It's can be an odd structure to see if you're not familiar with it - but combining dateadd/datediff and exploiting relationships between an arbitrary date and other dates computed from it can be useful for performing all kinds of calculations. A similar structure can be used for computing e.g. the last day of the month 15 months ago using just dateadd/datediff, an arbitrary date and another date with the right offset from the first:
SELECT DATEADD(month,DATEDIFF(month,'20010101',GETDATE()),'19991031')
As I said in a comment though, usually doing this sort of thing is only a short step away from needing to properly model your organisation's business days, at which point you'd typically want to introduce a calendar table. At one row per day, 20 years worth of pre-calculated calendar (adjusted as necessary as the business changes) is still less than 10000 rows.
You can try this.
WHERE DOC_DATE = DATEADD(DAY, CASE WHEN datepart(dw, GETDATE()) = 2 THEN -3 ELSE -1 END, CAST(GETDATE() AS DATE))
Select MOD (RUNC (MONTHS_BETWEEN (TO_DATE ('28-Mar-2017', 'DD-MON-YYYY'),
TO_DATE ('28-Feb-2017', 'DD-MON-YYYY'))), 12) Months from dual
Why is this function returning 1 month even if not completed the month? Its supposed to be 0
You are wrong. MONTHS_BETWEEN treats the time span from date '2017-02-28' to date '2017-03-28' as exactly one month (same day in two adjacent months). This can be read in the docs: https://docs.oracle.com/cd/B19306_01/server.102/b14200/functions089.htm
MONTHS_BETWEEN returns number of months between dates date1 and date2. If date1 is later than date2, then the result is positive. If date1 is earlier than date2, then the result is negative. If date1 and date2 are either the same days of the month or both last days of months, then the result is always an integer. Otherwise Oracle Database calculates the fractional portion of the result based on a 31-day month and considers the difference in time components date1 and date2.
One date is bigger than the other so you get a positive or negative number (i.e. not zero) depending on which is first and which is second parameter, and as the days equal, the result will be an integer. That is 1 or -1 for adjacent months.
Month calculation is a strange thing anyway, as a month is not a defined time span. You seem to have a certain definition in mind, which simply is different from how MONTHS_BETWEEN defines it. Nothing wrong about that. I would agree with MONTHS_BETWEEN in this case; from Feb 28 to Mar 28 is "exactly" one month. If you want different rules, then apply your own math :-)
Because it is working to its documented specification:
MONTHS_BETWEEN returns number of months between dates date1 and date2.
The month and the last day of the month are defined by the parameter
NLS_CALENDAR. If date1 is later than date2, then the result is
positive. If date1 is earlier than date2, then the result is negative.
If date1 and date2 are either the same days of the month or both last
days of months, then the result is always an integer. Otherwise Oracle
Database calculates the fractional portion of the result based on a
31-day month and considers the difference in time components date1 and
date2.
If you have requirements that are different to this then you will need to write your own function.
In one of my SQL queries, I am using
[... and z.READ_TIMESTAMP > TIMESTAMP_TO_EPOCH(TRUNC(SYSDATE-3)]
If I want the date to be exactly 5/31/2017, will I use 'SYSDATE' (date-n) function or some other expression? or how can a modify my query for 5/31/2017
If you want the date to be exactly 5/31/2017 then use TO_DATE() or TO_TIMESTAMP() depending on which data type you need (date or timestamp). As you are using SYSDATE already the the date data type should work.
-- e.g.
select
to_date('5/31/2017','mm/dd/yyyy')
, to_timestamp('5/31/2017','mm/dd/yyyy')
from dual
...
and z.READ_TIMESTAMP > TIMESTAMP_TO_EPOCH(to_date('5/31/2017','mm/dd/yyyy'))
HOWEVER
I suspect you may want more than just a way to establish a fixed date. For example are you asking for "how do I get that last day of the previous month?" which perhaps can be satisfied by using >= and the first day of current month like this:
...
and z.READ_TIMESTAMP >= TIMESTAMP_TO_EPOCH(trunc(sysdate,'MM'))
or if it really is the last day of the previous month can be achieved with a combination of LAST_DAY() and ADD_MONTHS()
and z.READ_TIMESTAMP >
TIMESTAMP_TO_EPOCH( last_day(add_months(trunc(sysdate,'MM'),-1)) )
Without knowing a great deal more about the nature of your data and query purpose please do note that each date you use when "truncated" also has the time set to 00:00:000 - so IF you data contains time within a day other than 00:00:00 then these 2 queries might NOT produce the same result
.... datetimecolumn > to_date('05/31/2017','mm/dd/yyyy') -- "a"
.... datetimecolumn >= to_date('06/01/2017','mm/dd/yyyy') -- "b"
For example "a" the entire 24 hour duration of 05/31/2017 would be included in the results, but for example "b" that same 24 hour duration would be excluded from results. In my experience the last day of any month isn't really the best method for locating date/time based data, instead usually it is the first day of the next month that produces the correct result.
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.