how to extract day of month from date, and select only odd days in Informix SQL?
Day of month is simply the DAY(date_or_datetime_column) function, related to the MONTH() and YEAR() functions. Getting only odd numbers is done with a simple modulo 2 expression.
So I think you only need:
SELECT date_col, DAY(date_col)
FROM table
WHERE MOD(DAY(date_col), 2) = 1
Hope that's useful.
Related
How do I select the date with a specific month?
For example I have in my table:
1-mar-2015
16-mar-2013
12-feb-2016
14-apr-2014
And I want to get only the dates from march.
The result should be:
1-mar-2015
16-mar-2013
Databases have ways of extracting date parts. The ANSI standard method is:
where extract(month from date) = 3
Other databases support functions such as month() or to_char() to achieve the same purpose.
I'm pretty new to SQL so I hope this isn't a dumb question, tried to google but couldn't find anything.
I'm summing sales of departments per week in SQL and am using TD_SYSFNLIB.WEEKNUMBER_OF_YEAR (trans_dt) to get the week number.
I think everything is working except I'd like to change the format of the weeks to the start date of the week, e.g. week 1 = 1/4/15
Also, i'm not sure how to handle the very first of the year week 0 since I think that should be grouped up with week 52 of last year.
The following date math trick should get you Beginning of Week as an actual date without having to join to the SYS_CALENDAR view or using a function:
SELECT CURRENT_DATE - ((CURRENT_DATE - DATE '0001-01-07) MOD 7) AS BOW;
Starting with TD14 there's NEXT_DAY which returns the following weekday, if you subtract 7 days you get the previous day:
next_day(trans_dt - 7, 'sunday')
I am using Hive (which is similar to SQL, but the syntax can be little different for the SQL users). I have looked at the other stackoverflow, but they seems to be in the SQL with different syntax.
I am trying to the get the first day of the month through this query. This one gives me today's day. For example, if today is 2015-04-30, then result would be 2015-04-01. Thanks!
select
cust_id,
FROM_UNIXTIME(UNIX_TIMESTAMP(),'yyyy-MM-dd') as first_day_of_month_transaction
--DATEADD(MONTH, DATEDIFF(MONTH, 0, GETDATE()), 0) as first_day_of_month_transaction --SQL format. Not compatible in Hive.
from
customers;
Try this
date_format(current_date,'yyyy-MM-01')
To get the first day of the month, you can use:
date_add(<date>,
1 - day(<date>) )
Applied to your expression:
date_add(FROM_UNIXTIME(UNIX_TIMESTAMP(), 'yyyy-MM-dd'),
1 - day(FROM_UNIXTIME(UNIX_TIMESTAMP(), 'yyyy-MM-dd'))
)
But this will work for any column in the right format.
SELECT TRUNC(rpt.statement_date,'MM') will give you the first day of month.
You can use this query to calculate the First day of the current month
Select date_add(last_day(add_months(current_date, -1)),1);
Instead of current_date, you can use the date field name.
last_day => Gives the last day of current month
add_months with -1 => Gives previous month
date_add with 1 => Add one day
Another way - select concat(substring(current_date,1,7),'-01');
I want to split a datetime column so that the year and the month both have their own column in a select statement output. I also want to have a column by week of the year, as opposed to specific date.
Basically, I want separate year, month, and week columns to show up in my select statement output.
Try using the DatePart function as shown in the following:
select
datepart(year,Mydate),
datepart(month,Mydate),
datepart(week,Mydate)
From
MyTable
Note: If you need to calculate the week number by ISO 8601 standards then you'll need to use datepart(iso_week,Mydate)
You could also look at the DateName function
select
datename(month,Mydate)
From
MyTable
Here is another way. Use SQL Servers YEAR() and MONTH() functions. For week, I use datepart(week,Mydate) as noted by #DMK.
SELECT YEAR(MyDate), MONTH(MyDate), DATEPART(WEEK,Mydate) From YourTable
check this
select datepart(Month,DateColumn) Mnth,datename(Month,DateColumn) MnthName,datepart(Year,DateColumn) Year1,((day(DateColumn)-1) / 7) + 1 week from dbo.Table
This should be doable, but how can I extract the day of the week from a field containing data in date format with Netezza SQL? I can write the following query:
SELECT date_part('day',a.report_dt) as report_dt
FROM table as a
but that gives me the day of the month.
thanks for any help
The below queries give day numbers for any week,month,year for a particular date.
--Day of Week
SELECT EXTRACT(dow FROM report_dt) FROM table;
--Day of Month
SELECT DATE_PART('day', report_dt) FROM table;
--Day of Year
SELECT EXTRACT(doy FROM report_dt) FROM table;
Netezza is just ANSI SQL, originally derived from PostgreSQL. I'd expect this to work.
select extract(dow from a.report_dt) as report_dt
from table as a
Returns values should range from 0 to 6; 0 is Sunday. You might expect that to be an integer, but in PostgreSQL at least, the returned value is a double-precision floating point.
If you want to extract directly the day name :
Select to_char(date, 'Day') as Day_Name From table;
In Netezza SQL, SELECT EXTRACT(dow FROM report_dt) would return values 1 to 7. 1 is Sunday, 7 is Saturday.