I'm trying to find the best way to subtract 1 year and also one month from the current date in a Hive query. Using the following, I don't believe it will take into account leap years or if the fact that months have different amounts of days so eventually the code will break. Any help would be greatly appreciated!
set my_date = from_unixtime(unix_timestamp()-365*60*60*24, 'yyyy-MM-dd');
set my_date = from_unixtime(unix_timestamp()-30*60*60*24, 'yyyy-MM-dd');
Thank!
-Rebecca
If you have date format like yyyy-MM-dd hh:mm:ss in Hive, it is easy to implement using following functions
concat((year(date_field)-1),'-', (month(date_field)-1), '-', day(date_field))
Use IF and CASE functions to implement your logic to find whether it is a leap year or not(by dividing year by 4)
Related
I'm working with Big Query and I need to get first day of week.
For example if date = '2022-08-26' I want to have '2022-08-22' where 22 is Monday.
Any solutions please ?
Thanks in advance.
To achieve this you'll want to use the DATE_TRUNC function as follows:
select date_trunc(date('2022-08-26'), WEEK(MONDAY))
You can change the parameter for WEEK to be any day of the week, default is SUNDAY.
Documentation can be found here:
https://cloud.google.com/bigquery/docs/reference/standard-sql/date_functions#date_trunc
I am trying to compare yesterday's data to the same day the year before. For example, yesterday is 11 November 2018. I want to compare to 12 November 2017 (same day but the year before). I am wanting this to be applied automatically on the filter so all I need to do is open the file and verify the numbers are correct before sending off the report.
Any help would be appreciated.
Thanks
There are many Tableau functions that manipulate dates. A couple in particular are relevant to your problem:
Today() - returns the current date
DateAdd() - adds or subtracts an interval from a date. For instance, DateAdd('year', Today(), -1) gives the date one year prior to today. The first argument to DateAdd is the level of granularity or date part.
DateDiff() - determines the difference of the interval between two dates. DateDiff('day', [Start Date], [End Date]) returns the number of days separating the two date arguments.
The functions are well documented in the online help. Construct the formulas you need and filter accordingly.
Isolate yesterday's date as its own field. For instance if that is the max date in your data, then {max([Date])} would create an LOD of the maximum date.
Then make a calculation that will display the same date last year:
year([Date]) = year([max_date])-1
and datepart('week',[Date]) = datepart('week',[max_date])
and datepart('weekday',[Date]) = datepart('weekday',[max_date])
I have a condition here in which I will have total experience in terms of month and year. For example, two drop down will be there for asking total number of experience in month and year. So if I am working from 1 Jan 2012, then I will write total experience as 3 year and 11 months. Now I have to convert this 3 year and 11 months into date format so that I can save this into database
You could use java.util.Calendar:
Calendar calendar = Calendar.getInstance();
calendar.add(Calendar.MONTH, month);
calendar.add(Calendar.YEAR, year);
Date date = calendar.getTime();
As a word of caution, the day field would be set to today's date. Check the intended behaviour if the current day is outside of the bounds for the target month. For example, setting the month to February when calendar has a day field of 30. It might be wise to set the day to a known, valid value for every month (eg: 1) before setting the month and year.
Use DATE_SUB() function:
Try this:
SELECT DATE_SUB(DATE_SUB(CURRENT_DATE(), INTERVAL 3 YEAR), INTERVAL 11 MONTH);
You can use mysql's date_sub() function or <date> - interval <expression> unit syntax to subtract an interval from a date.
select date_sub(curdate(),interval '3-11' YEAR_MONTH) as start_date
UPDATE:
Following the conversation between the OP and #eggyal, the OP need to replace the period in the incoming data with - and construct an insert statement as follows:
insert into mytable (...,join_date,...) values (...,date_sub(curdate(),interval '3-11' YEAR_MONTH),...)
I have to calculate the difference in days between two dates and I figured out that
there is no such thing as a DATEDIFF() function in DB2.
I tried doing it like that:
(DAYOFYEAR(date1)-DAYOFYEAR(date2)+(YEAR(date1)-YEAR(date2))*365)
This is obviously not working for leap years, but I do not have to deal with that.
I know that date1 is always later than date2.
Do I have any flaws in my logic? It is not working (it's an exercise and I have a function to test my results). Is there an easier way to do that?
Thank you.
It would possibly help if we knew what version of DB2 you were using and what platform it was running on. But it seems likely that you can do something like this:
select
days(my1stdate) - days(my2nddate) as myDuration
from mySchema.myTable
The DAYS() function converts a DATE value into the number of days between Jan 1, 0001, and the supplied DATE value. Once both DATEs are converted, the subtraction (difference) is straightforward.
Essentially I may have any date and I want to get the proper quarter-end date for the previous quarter.
Three examples:
19/07/2013 -> 30/06/2013
30/06/2013 -> 31/03/2013
28/02/2013 -> 31/12/2012
In Excel VBA it seems using DateAdd to subtract a quarter just subtracts three months from the date, e.g. 19/07/2013 -> 19/04/2013. This is no good for me.
What do you think would be the best way of doing this?
Perhaps extracting the month part and then comparing it to a list of the
possible quarters?
Or maybe having some sort of null date
01/01/2000 and then adding the number of quarters the given date
shows as to this null date and then subtracting a day to get a
previous quarter-end position (although this might cause problems
when the date flips under a year to December 31)?
Please consider:
=IF(MOD(MONTH(A1),3)=0,EOMONTH(A1,-3),
IF(MOD(MONTH(A1),3)=1,EOMONTH(A1,-1),
EOMONTH(A1,-2)))
Edit: There is a better answer in the comments.