I can I 52 previous weeks listed from todays date? Or if simpler get the first day of the previous 52 weeks?
Thanks
E
I believe what you are looking for is using the BETWEEN inside your WHERE clause in conjunction with using the DATEADD() function....
For example if you have a dbo.Date table with a DATETIME or DATE column...
SELECT DATEPART(week,Date) as Week
FROM dbo.Date
WHERE Date BETWEEN DATEADD(Week,-52,GETDATE()) AND GETDATE()
Related
I want to find out the previous weeks's last day in Teradata SQL using Sunday as the last day of the week. For example, today is Friday 1/27, so the last week ended on Sunday (1/22) and I would want to return 2023-01-22.
Other examples:
If current date is '2023-01-02', then the output I require is '2023-01-01'
If current date is '2023-01-18', then the output I require is '2023-01-15'
With Hive query I would use:
date_sub(current_date, cast(date_format(current_date, 'u') as int));
What would the equivalent be in Teradata? I've tried using the code below but it seems to return the date of the closest Sunday instead of the date of the previous Sunday.
SELECT ROUND(current_date, 'd') (FORMAT 'yyyy-mm-dd');
There are several ways:
Probably the best one is one of the built-in functions to return the previous xxxday <= the input date:
Td_Sunday(Current_Date - 1)
Or the function to return the next xxxday > input date:
Next_Day(Current_Date - 8, 'sun')
Truncating is least understandable:
Trunc(Current_Date, 'IW') -1
TRUNC supports three variations, only IW is usable, but restricted to Monday as week start:
IW: the Monday of the ISO week
WW: the same day of the week as January 1st of the year
W: the same day of the week as the first day of the month
You can use the trunc function to return the first day of the a week, month, ect.
select trunc(current_date -7 ,'IW')
Current date today is 2023-01-27. This will return 2023-01-15, the previous Sunday.
EDIT: Sorry, meant to use the ISO week. As Dnoeth points out, the regular week option doesn't work consistently (which I didn't know, never used it for this before). Anyhoo, his answer is better than mine...
I've been having a look around but I can't seem to find anything that answers this question. I want to calculate the date for the last week of every month. For example, the date for the last week of April 2021 is 26-04-2021. I want a date, not a week number.
I use Google Big Query, I do have a calendar table I could use to extract year and month.
Thanks,
Emily
Try date_trunc:
SELECT date_trunc(last_day(month1st, month), week(monday))
from unnest(GENERATE_DATE_ARRAY('2021-01-01', '2021-12-01', interval 1 month)) AS month1st;
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 am new to crystal reports. I have a report on which there are two date fields : Benefit Start & Benefit End Dates.
I have to compare the age of the employees.
If age>65 years, Benefit Start date should be 20160101.
If age<65 years, Benefit Start date should be first day of the month effective.
eg: if the date is '09/21/2015', the output should be'09/01/2015'.
Similarly for benefit End Date,
If age<65 years, Benefit End date should be last day of the month effective.
eg: if the date is '09/02/2015', the output should be 09/30/2015.
How do I hard code the days in dates so that I get the first day & last day of the months for the two fields?
Please help me
Thanks in advance.
date(year({effectivedate})+1,1,1) -- returns Jan 1 of following year
date(year({effectivedate}),month({effectivedate}),1) -- returns first day of month for effective date
date(year(dateadd('m',1,{effectivedate})),month(dateadd('m',1,{effectivedate})),1)-1 -- returns last day of effective month
I hope... I didnt test
try this:
Start date:
DateSerial (YEAR(Cdate(datefield)),MONTH(Cdate(datefield)) ,1 );
End Date:
DateADD("d",-1,DateAdd ("m",1 ,Cdate(datefield) ))
I have tried the following:
add_months(to_Date('04/01/ind.birth_dte','MM/DD/YYYY'), 864) >= to_date('&StartDt','MM/DD/YYYY')
Is there a better way to pull April first of the participant's 72nd birth date?
You could use an interval calculation instead, but not sure how you're defining 'better'. Assuming you do want April 1st of the year in which their 72 birthday falls:
trunc(ind.birth_dte, 'YYYY') + interval '72-3' year to month
The trunc() function goes to the first day of their birth year, and the interval adds 72 years and 3 months to that, which will be April 1st.
SQL Fiddle with some sample dates, including a leap day to show that isn't a problem.
Or to compare that adjusted date with a fixed date as a filter:
where trunc(ind.birth_dte, 'YYYY') + interval '72-3' year to month
> to_date('&StartDt','MM/DD/YYYY');
SQL Fiddle.
You can use the trunc() method with your version as well to save building up a string and calling to_date, adding an additional three months to the add_months call (though I'd suggest you at least need a comment indicating where '867' comes from):
where add_months(trunc(ind.birth_dte, 'YYYY'), 867)
> to_date('&StartDt','MM/DD/YYYY');