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.
Related
select datepart(year, '2017/08/25') as week;
I believe this is for mysql but does not work for oracle sql
Week of year (1-53) where week 1 starts on the first day of the year and continues to the seventh day of the year.
select to_char(trunc(sysdate),'WW') from dual;
Week of year (1-52 or 1-53) based on the ISO standard.
select to_char(trunc(sysdate),'IW') from dual;
Taken from Oracle docs
I have to generate year wise, weekly reports for some data. Now When I aggregate date on week number, and week number is calculated from extract from creation date.
Now the problem is these both queries return week number 52.
SELECT EXTRACT(WEEK FROM TIMESTAMP '2006-01-01');
SELECT EXTRACT(WEEK FROM TIMESTAMP '2006-12-31');
First query return 52 (52nd week of 2005) and 2nd query return 52 (52nd week of year 2006). thats documented behavior.
But I want to Calculate local week number, and results for first query should be 1 and other query would return 53.
You can't do this with the exctract() function, it only supports ISO weeks.
But the to_char() function has an option for this:
SELECT to_char(DATE '2006-01-01', 'WW')::int` --> 1
SELECT to_char(DATE '2006-12-31', 'WW')::int` --> 53
For date 2006-01-01 end week is start in 2005 year, that same problem is 1999 year.
Clausule EXTRACT(WEEK getting year where week is started not ending.
You can use this code:
SELECT floor(EXTRACT(doy FROM TIMESTAMP '2006-01-01')/7 + 1);
SELECT floor(EXTRACT(doy FROM TIMESTAMP '2006-12-31')/7 + 1);
I have a date column in the format YY-MON-DD, e.g. 25-JUN-05. Is it possible to isolate this into 3 separate columns for year, month and day? Where month is converted from text to numerical, e.g. Year: 25, Month: 06, Day: 05?
MS SQL SERVER
As Nebi suggested, you can use DATEPART and extract each part and store it into different columns.
SELECT DATEPART(DAY,'2008-10-22'); -- Returns DAY part i.e 22
SELECT DATEPART(MONTH,'2008-10-22'); -- Returns MONTH part i.e 10
SELECT DATEPART(YEAR,'2008-10-22'); -- Returns YEAR part i.e 2008
Try with the below script,if you are using SQL Server.
SELECT 'Year: '+CAST(LEFT(YourdateColumn,2) as VARCHAR(2))+', Month: ' +CAST(MONTH('01-'+SUBSTRING(YourdateColumn,4,3)+'-15')as VARCHAR(2))+', Day:'+CAST(RIGHT(YourdateColumn,2)as VARCHAR(2))
FROM Yourtable
sample output :
You didn't specify your DBMS.
The following is standard SQL assuming that column really is a DATE column
select extract(year from the_column) as the_year,
extract(month from the_column) as the_month,
extract(day from the_column) as the_day
from the_table;
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');
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.