I know Quicksight could help parse the time into DAY, WEEK, MONTH and YEAR in Field Wells, but how could I customize it into 3 days(72 hrs)? Is there any windowsFunction I could use?
You can introduce a calculated field with formula looking something like that:
addDateTime((extract("DD", {create_time})/3)*3, "DD", truncDate("DD", {create_time}))
And use it for grouping. Depends on your desired time interval it possibly will not do exactly what you need - cause it will group days in month by three, but you can use dateDiff and either some fixed date as starting point. For example:
addDateTime((dateDiff(parseDate("2020-01-01"), truncDate("DD", {create_time}))/3)*3, "DD", parseDate("2020-01-01"))
Related
I need to get the difference between dates, but I just need to get the whole months that have passed. So for example between "1990-05-24" and "1990-05-27" it should say 0. It would also be 0 for "1990-05-02" and "1990-05-29" because the month has not finished.
I already got the difference in months using MONTHS_BETWEEN(), but I get months with decimals, and ROUNDing is not an option since sometimes it should be up and sometimes down.
I thought about setting al dates to day 01. In both colums Closing_date and Opening_date. But can't figure out how to do it.
I think you want to count boundaries between months. If so, you can use months_between() after truncating to the first of the month:
months_between(trunc(date1, 'MON'), trunc(date2, 'MON')
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).
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.
So what I'm looking to do is create a report that shows how many sales a company had on a weekly basis.
So we have a time field called created that looks like this:
2016-04-06 20:58:06 UTC
This field represents when the sale takes place.
Now lets say I wanted to create a report that gives you how many sales you had on a weekly basis. So the above example will fall into something like Week of 2016-04-03 (it doesn't have to exactly say that, I'm just going for the simplest way to do this)
Anyone have any advice? I imagine it involves using the UTEC_TO_xxxxxx functions.
The documentation advises to use standard SQL functions, like DATE_TRUNC():
SELECT DATE_TRUNC(DATE '2019-12-25', WEEK) as week;
you can use WEEK() function - it gives you week number
SELECT WEEK('2016-04-06 20:58:06 UTC')
if you need first day of the week - you can try something like
STRFTIME_UTC_USEC((UTC_USEC_TO_WEEK(TIMESTAMP_TO_USEC(TIMESTAMP('2016-05-02 20:58:06 UTC')), 0)),'%Y-%m-%d')
I had to add parentheses:
SELECT DATE_TRUNC(DATE('2016-04-06 20:58:06 UTC'), WEEK) as week;
This is quite an old question and things have moved on since.
In my case, I found that the old WEEK function is no longer recognised, so I had to instead use the EXTRACT function. The doc for it can be found here.
For me it was enough to just extract the ISOWEEK from the timestamp, which results in the week of the year (the ISOYEAR) as a number.
ISOWEEK: Returns the ISO 8601 week number of the datetime_expression. ISOWEEKs begin on Monday. Return values are in the range [1, 53]. The first ISOWEEK of each ISO year begins on the Monday before the first Thursday of the Gregorian calendar year.
So I did this:
SELECT EXTRACT(ISOWEEK FROM created) as week
And if you want to see the week's last day, rather than the week's number in a year, then:
SELECT last_day(datetime(created), isoweek) as week