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}
Related
I am trying to dynamically get the dates, starting from begging of month to current date using SSAS Tabular - DAX, I have tried many date functions but I couldn't end up with solution, so is there any idea to share pls ?
The following DAX function will return a table of DATES for each day from the start of the current month through today:
DatesMTD = CALENDAR(DATE(YEAR(TODAY()), MONTH(TODAY()), 1), TODAY())
To be able to filter the dates for the first 10 days, create a calculated column that identifies as a date in this range
1st to 10th Date = If(Day([Date]) <11,1,0)
This can then be used to either filter out the dates past the 10th of each month
I have a date field in Hive 2018-06-10, from which i need to get WeekOfMonth
WEEKOFYEAR(order_time)
I need output for 2018-06-10 as 3 (which is 3rd week. assuming week starts from Sunday)
Is there any built in function in Hive to retrieve WeekofMonth. I couldn't find any. I tried below to convert based on minutes and seconds but
from_unixtime(unix_timestamp(CURRENT_DATE())+7200)
But the above is not giving correct value
For the week of the month, you can get the day part of the month and divide by 7.
select case
when DAYOFMONTH(order_time)%7 = 0
then DAYOFMONTH(order_time)/7
else DAYOFMONTH(order_time)/7 + 1
end
Also you can use date_format function:
select date_format('2018-06-10','W');
See more format patterns here: SimpleDateFormat
SELECT strftime('%W', 'Week'), sum(income) FROM tableOne GROUP BY Week;
Format for date is a simple date: YYYY-MM-DD
PROBLEM: When run no value for the Week column is provided. Any suggestions?
There is data in the table and when the query is run the income is summarized by the date in the week column. Thing is, this column contains a date that may be any day of the week and often multiple different days of the same week. I need to summarize the income by week.
In SQL, 'Week' is a string containing four characters.
To reference the value in a column named "Week", remove the quotes: Week.
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
How to get the difference between two dates (informix) in integer format like that
day = 15
mon = 2
year = 1
There are two sets of date/time values in Informix: DATE and DATETIME.
The DATE type is oldest (it was in the precursor to the SQL-based Informix), and represents the integer number of days since a reference date (where day 0 is 1899-12-31, so day 1 was 1900-01-01).
You get the difference between two DATE values in days by subtracting one from the other.
The DATETIME system is newer (but still old — circa 1990). You can take the difference between two DATETIME YEAR TO DAY values and get a result that is an INTERVAL DAY TO DAY (essentially the number of days).
You could also take the difference between two DATETIME YEAR TO MONTH values and get a result that is an INTERVAL YEAR TO MONTH.
However, there is no way to get a difference in years, months and days because there is no simple way to deduce that value. In fact, ISO SQL recognizes two classes of INTERVAL: those in the YEAR-MONTH group, and those in the DAY-SECOND group. You can't have an INTERVAL that crosses the MONTH/DAY barrier.
Use the MDY function :
select mdy(2,15,2014) - mdy(1,15,2014) from sysmaster:sysdual