select MONTH (birth_date) ...
....
..
However this gives numeric values.
I need Jan , Sep ...
How can I do this ? Case birth_date when 1 then 'jan' when 2 then ... is too long
Any other efficient ways ?
You use several functions that provide the information that you need:
VARCHAR_FORMAT - http://pic.dhe.ibm.com/infocenter/db2luw/v10r5/topic/com.ibm.db2.luw.sql.ref.doc/doc/r0007110.html
SUBSTR - http://pic.dhe.ibm.com/infocenter/db2luw/v10r5/topic/com.ibm.db2.luw.sql.ref.doc/doc/r0000854.html
LCASE - http://pic.dhe.ibm.com/infocenter/db2luw/v10r5/topic/com.ibm.db2.luw.sql.ref.doc/doc/r0052924.html
Example
Select LCASE(SUBSTR(VARCHAR_FORMAT(birth_date,'Month'),1,4))
From ...
This works with Timestamp or date
If the values of that column are not a time stamp you will have to do a CASE statement. However, if it is a timestamp than maybe you can use the MONTHNAME function
Related
Is there a way to find the number of days in a month in DB2. For example I have a datetime field which I display as Jan-2020, Feb-2020 and so on. Based on this field I need to fetch the number of days for that month. The output should be something like below table,
I'm using the below query
select reportdate, TO_CHAR(reportdate, 'Mon-YYYY') as textmonth from mytable
Expected output
ReportDate textMonth No of Days
1-1-2020 08:00 Jan-2020 31
1-2-2020 09:00 Feb-2020 29
12-03-2020 07:00 Mar-2020 31
Try this:
/*
WITH MYTABLE (reportdate) AS
(
VALUES
TIMESTAMP('2020-01-01 08:00:00')
, TIMESTAMP('2020-02-01 09:00:00')
, TIMESTAMP('2020-03-12 07:00:00')
)
*/
SELECT reportdate, textMonth, DAYS(D + 1 MONTH) - DAYS(D) AS NO_OF_DAYS
FROM
(
SELECT
reportdate, TO_CHAR(reportdate, 'Mon-YYYY') textMonth
, DATE(TO_DATE('01-' || TO_CHAR(reportdate, 'Mon-YYYY'), 'dd-Mon-yyyy')) D
FROM MYTABLE
);
Db2 has the function DAYS_TO_END_OF_MONTH and several others which you could use. Based on your month input, construct the first day of the month. This should be something like 2020-01-01 for Jan-2020 or 2020-02-01 for Feb-2020. Follow the link for several other conversion functions which allow you to transform between formats and to perform date arithmetics.
convert your column to a proper date and try this: day(last_day(date_column))
Below is a script i am trying to run in Presto; Subtracting today's date from an integer field I am attempting to convert to date. To get the exacts days between. Unfortunately, it seems the highlighted block does not always convert the date correctly and my final answer is not correct. Please does anyone know another way around this or a standard method on presto of converting integer values to date.
Interger value in the column is in the format '20191123' for year-month-date
select ms, activ_dt, current_date, date_diff('day',act_dt,current_date) from
(
select ms,activ_dt, **CAST(parse_datetime(CAST(activ_dt AS varchar), 'YYYYMMDD') AS date) as act_dt**, nov19
from h.A_Subs_1 where msisdn_key=23480320012
) limit 19
You can convert "date as a number" (eg. 20180527 for May 27, 2018) using the following:
cast to varchar
parse_datetime with appropriate format
cast to date (since parse_datetime returns a timestamp)
Example:
presto> SELECT CAST(parse_datetime(CAST(20180527 AS varchar), 'yyyyMMdd') AS date);
_col0
------------
2018-05-27
You can use below sample query for your requirement:
select date_diff('day', date_parse('20191209', '%Y%m%d'), current_timestamp);
I am trying to use the so called DateDiff function to subtract the End Date from the Start Date and obtain the numbers of days apart.For example:
10/11/1995 - 7/11/1995 = 3 (extract the 'dd' from DD/MM/YYYY format)
As date values are double with the integer part counting for a day, you can use this simple expression:
[Due Date]-[Start Date]
or, for integer days only:
Fix([Due Date]-[Start Date])
That said, you should a query for tasks like this.
table has DEPNO DATE it contains 2 records as :
1 2004-05-02
2 03-APR-04
I want to process above 2 records dates in in yyyy-mm-dd format in hive
You can try in following way:
select TO_DATE(from_unixtime(UNIX_TIMESTAMP(recordDate, 'yyyy-mm-dd'))) from table_name
Refer Hive Date Functions for more details
couple of options
1. Use UDF to take the column and return what ever format you want.
2. Use case when statement like
select CASE WHEN length(col) = 10 then unix_timestamp(col, 'yyyy-MM-dd')
else unix_timestamp(col, 'yy-MMM-dd') from whatever.
Let's say that I have a range of SQL tables that are named name_YYYY_WW where YYYY = year and WW = week number. If I call upon a function that guides a user defined date to the right table.
If the date entered is "20110101":
SELECT EXTRACT (WEEK FROM DATE '20110101') returns 52 and
SELECT EXTRACT (YEAR FROM DATE '20110101') returns 2011.
While is nothing wrong with these results I want "20110101" to either point to table name_2010_52 or name_2011_01, not name_2011_52 as it does now when I concanate the results to form the query for the table.
Any elegant solutions to this problem?
The function to_char() will allow you to format a date or timestamp to output correct the iso week and iso year.
SELECT to_char('2011-01-01'::date, 'IYYY_IW') as iso_year_week;
will produce:
iso_year_week
---------------
2010_52
(1 row)
You could use a CASE:
WITH sub(field) AS (
SELECT CAST('20110101' AS date) -- just to test
)
SELECT
CASE
WHEN EXTRACT (WEEK FROM field ) > 1 AND EXTRACT (MONTH FROM field) = 1 AND EXTRACT (DAY FROM field) < 3 THEN 1
ELSE
EXTRACT (WEEK FROM field)
END
FROM
sub;