I have approximately the same table (excluding count column). I want to calculate the number of working days (Mon-Fri) and exclude public holidays.
I tried to try the following query
SELECT count(distinct(date)) from MYDB where dummy <> 1
However, it gives the only total number of days including weekends. Additionally, if use this command it counts distinct dates, however, my dates do not show a full month, so another logic should've used. Could you help to figure out which code is better to use?
there should be a function in Vertica that extracts weekday from date, so to exclude weekends you'll need to add another condition like
extract(dow from date) not in (6,0)
(6 is Sat, 0 is Sun in this case)
Table format for the date column is "yyyyMMdd" and I'm using the following functions to convert into standard format so that HIVE day, months and year can be performed to get the respective values.
(from_unixtime(unix_timestamp(cast(created_day as STRING) ,'yyyyMMdd'), 'yyyy-MM-dd'))
To get the current year data, I would subtract the year obtained from all the records with the year returned by the current date and if it return zero, then it falls in this year.
(year(current_date()) - year(from_unixtime(unix_timestamp(cast(created_day as STRING) ,'yyyyMMdd'), 'yyyy-MM-dd'))) = 0
Problem: If the current date falls in January, I would get only January data month, but i need to get the data from February(last year) to January(current year)?
Also I need to scale this to obtain the last 24 months.
I always set my date range parameters outside of Hive and pass them as arguments as this lends itself to reproducibility and testability.
select <fields> from <table> where created_day between ${hiveconf:start_day} and ${hiveconf:end_day}
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 know this one is pretty easy but I've always had a nightmare when it comes to comparing dates in SQL please can someone help me out with this, thanks.
I need to get the month and year of now then compare it to a date stored in a DB.
Time Format in the DB:
2015-08-17 11:10:14.000
I need to compare the month and year with now and if its > 12 months old I will increment a count. I just need the number of rows where this argument is true.
I assume you have a datetime field.
You can use the DATEDIFF function, which takes the kind of "crossed boundaries", the start date and the end date.
Your boundary is the month because you are only interested in year and month, not days, so you can use the month macro.
Your start time is the value stored in the table's row.
Your end time is now. You can get system time selecting SYSDATETIME function.
So, assuming your table is called mtable and the datetime object is stored in its date field, you simply have to query:
SELECT COUNT(*) FROM mtable where DATEDIFF(month, mtable.date, (SELECT SYSDATETIME())) > 12
I have a table with two dates, "Start_Date" and "End_Date". In DB2 SQL, is there a way to determine if a leap day falls between these two dates?
Thank you!
Sure, you can do this using some date math and the DAYS function, by comparing the number of days between the the start and end date to the number of days between the start date and end date when they've both been shifted by 1 year.
If the number of days between the two dates is the same in both cases, then no leap day has occurred. If the number of days differs, then there has been at least 1 leap day.
This expression will return the number of leap days:
select
( DAYS(end_date + 1 year) - DAYS(start_date + 1 year) ) -
( DAYS(end_date) - DAYS(start_date) )
from
sysibm.sysdummy1
This should work as long as end_date >= start_date.
It's trivial to encapsulate this into a scalar User Defined Function.