TSQL: Function to display month in words not numbers - tsql

MONTH(CMS.App_Received_Date) as 'App Month'
will return 4 for April when date is like 2012-04-01
will return 5 for May when date is like 2012-05-02 etc
Is there a TSQL function to return April, May, June instead?
i.e. MONTHNAME(CMS.App_Received_Date) as 'App Month'

DATENAME(month, CMS.App_Received_Date) as 'App Month'

SELECT DATENAME(month, '2012-04-01') AS 'App Month'
Returns a character string that represents the specified datepart of the specified date.
Note that the return value depends on the language environment, in my case it returns Januar (german).
http://msdn.microsoft.com/en-us/library/ms174395.aspx

Related

'3rd Friday of the Month' to a timestamp in PLPGSQL?

I have a database column giving me information on how often a file comes in.
Frequency_month
-------------
3rd Friday of the month
2nd Tuesday of the month
3rd Thursday of the month
I need to update this column and have it be a timestamp. e.g.
Frequency_month
-------------
2020-05-21 00:00:00
2020-05-11 00:00:00
2020-05-20 00:00:00
How can I accomplish this using postgres PLPGSQL language?
The following yields what your looking for. As far a parsing the Frequency_month it imposes the following restrictions:
The first character in the string is a digit indicating the relative
number.
This is followed 2 characters ordinal spec (st, nd, etc) and a space.
Actually any 3 characters, they are not checked.
Position 5 - 7 con the first 3 characters of the English day of week (dow).
If any of those are not satisfied you will need to change the S1 subquery.
Further it requires you to provide a date of reference. This may be any date in the month of interest. See comment by #sddk.
It proceeds as follows:
Parse the above extracting the week number, day of week, and last
day of the prior month. (S1).
Determine the ISODOW id numbers for the day of week specified and
DOW for last of prior month. (S2).
Using the ISODOW id numbers Determine, determine the first
occurrence of the target day in the target month. (S3).
Adjust the date from #3 by the additional weeks. (S4).
Finally, if the resulting date in #4 in still in the target month
return the date form #4. If it is not the same month then return
null. This occurs when there in no nth dow in the month or the dow
is incorrectly specified.
I have wrapped the above into a SQL function making parameterization easy. See Demo.
create or replace
function frequency_month( frequency_string text
, target_month date
)
returns date
language sql
as $$
with day_names( l_days) as
( values (array['mon','tue','wed','thu','fri','sat','sun']) )
select -- if the calculated date in still in the target month return that date else return null
-- covers invalid week in frequency 6th Friday or 0th Monday
case when extract(month from target_date) = extract (month from target_month)
then target_date
else null
end
from ( -- Advance from first dow in month the number of weeks to desirded dates
--select (first_of_mon + (7*(rel_num-1)) * interval '1 day')::date target_date
select (first_of_mon + (rel_num-1) * interval '1 week')::date target_date
from ( -- with last day of prior month get first DOW week of target month
select case when dow_day_nbr <= from_day_nbr
then (from_date + (dow_day_nbr-from_day_nbr+7) * interval '1 days' )::date
else (from_date + (dow_day_nbr-from_day_nbr) * interval '1 days' )::date
end first_of_mon
, rel_num
from ( -- Pick up ISODOW numbers
select array_position(l_days, (substring(to_char(from_date, 'day'),1,3))) as from_day_nbr
, array_position(l_days, lower(substring(rel_dow,1,3))) as dow_day_nbr
, from_date
, rel_num
from day_names
cross join ( -- get last day of prior month, desired relative day, relative dow
select substr(frequency_string,1,1)::integer rel_num
, lower(substr(frequency_string,5,3)) rel_dow
, (date_trunc('month',target_month) - interval '1 day')::date from_date
) s1
) s2
) s3
) s4;
$$;
Note: The demo also includes a standalone version if a function is not desired.

How to write the query to get the first and last date of a January and other month's in postgresql

How to get the first and last date of the particular month i.e if i pass the particular month name say March it should return output as 01/03/2019 and 31/03/2019.( For current year)
If you want to pass value March you would have to modify the code to understand every month. I'm not sure it's worth the trouble. Anyways, here's a code to return two values (start and end of month) based on current_date. Should you wish to change the day, you could put for example '2019-04-13' in that place.
SELECT
date_trunc('month', current_date) as month_start
, (date_trunc('month', current_date) + interval '1 month' - interval '1 day')::date as month_end
DATE_TRUNC function truncates the date to the precision specified in first argument, thus making the date as of first day of given month (taken from current_date in above example).
For end of month you need a bit more computation. I've always used this in production and what it does is it first truncates your date to first day of month, then adds one month and goes back one day, so that you have your end of month date (whether it's 30, 31, or special case for February during leap years).
for any month, the first day must be 1st,
so it is:
make_date(2019, 3, 1)
and for any month, the last day is 1 day before the first day of next month,
so it is:
make_date(2019, 4, 1) - integer '1'
sorry, I don't have a PostgreSQL environment to test if it is correct,
so please test it yourself.
and, BTW,
you can find more details about date/time operators and functions here:
https://www.postgresql.org/docs/current/functions-datetime.html
One straightforward approach, which would also work on most other databases, would be to truncate the incoming date by month to obtain the first day of that month. Then, truncate the date with one month added to it, and subtract one day, to obtain the last day of the month.
SELECT
DATE_TRUNC('month', '2019-03-15'::date) AS date_start,
DATE_TRUNC('month', '2019-03-15'::date + INTERVAL '1 MONTH')
- INTERVAL '1 DAY' AS date_end;
Demo
From here Date LastDay
SELECT date_trunc('MONTH', dtCol)::DATE;
CREATE OR REPLACE FUNCTION last_day(DATE)
RETURNS DATE AS
$$
SELECT (date_trunc('MONTH', $1) + INTERVAL '1 MONTH - 1 day')::DATE;
$$ LANGUAGE 'sql' IMMUTABLE STRICT;
The conversion from month name parameter is actually rather simple. Create an array with the month names and find the position in the array of the parameter, that result becomes the month value into the make_date function with year extracted from current date and day 1. The below contains an overloaded function providing for either date or month name with optional year.
create type first_last_date as ( first_of date, last_of date);
create or replace function first_last_of_month(date_in date)
returns first_last_date
language sql immutable strict leakproof
as $$
select (date_trunc('month', date_in))::date, (date_trunc('month', date_in) + interval '1 month' - interval '1 day')::date ;
$$;
create or replace function first_last_of_month( month_name_in text
, year_in integer default null
)
returns first_last_date
language sql immutable leakproof
as $$
select first_last_of_month ( make_date ( coalesce (year_in, extract ('year' from now())::integer)
, array_position(ARRAY['jan','feb','mar','apr','may','jun','jul','aug','sep','nov','dec']
, lower(substring(month_name_in,1,3)))
,1 ) );
$$;
-- test
Select first_last_of_month('March');
Select first_last_of_month('February') y2019
, first_last_of_month('February', 2020) y2020;
Select first_last_of_month(now()::date);

How to calculate end of the month in Postgres?

How to calculate end of the month in Postgres? I have table with column date datatype. I want to calculate end of the month of every date. For Eg. In the table there values like "2015-07-10 17:52:51","2015-05-30 11:30:19" then end of the month should be like 31 July 2015,31 May 2015.
Please guide me in this.
How about truncating to the beginning of this month, jumping forward one month, then back one day?
=# select (date_trunc('month', now()) + interval '1 month - 1 day')::date;
date
------------
2015-07-31
(1 row)
Change now() to your date variable, which must be a timestamp, per the docs. You can then manipulate this output (with strftime, etc.) to any format you need.
Source
SELECT TO_CHAR(
DATE_TRUNC('month', CURRENT_DATE)
+ INTERVAL '1 month'
- INTERVAL '1 day',
'YYYY-MM-DD HH-MM-SS'
) endOfTheMonth
Hi I tried like this and it worked
Date(to_char(date_trunc('month'::text, msm013.msa011) + '1 mon - 1 day '::interval , 'DD-MON-YYYY') )
Thanks a lot!!

How to get the number of days in a month?

I am trying to get the following in Postgres:
select day_in_month(2);
Expected output:
28
Is there any built-in way in Postgres to do that?
SELECT
DATE_PART('days',
DATE_TRUNC('month', NOW())
+ '1 MONTH'::INTERVAL
- '1 DAY'::INTERVAL
)
Substitute NOW() with any other date.
Using the smart "trick" to extract the day part from the last date of the month, as demonstrated by Quassnoi. But it can be a bit simpler / faster:
SELECT extract(days FROM date_trunc('month', now()) + interval '1 month - 1 day');
Rationale
extract is standard SQL, so maybe preferable, but it resolves to the same function internally as date_part(). The manual:
The date_part function is modeled on the traditional Ingres equivalent to the SQL-standard function extract:
But we only need to add a single interval. Postgres allows multiple time units at once. The manual:
interval values can be written using the following verbose syntax:
[#] quantity unit[quantity unit...] [direction]
where quantity is a number (possibly signed); unit is microsecond,
millisecond, second, minute, hour, day, week, month, year, decade,
century, millennium, or abbreviations or plurals of these units;
ISO 8601 or standard SQL format are also accepted. Either way, the manual again:
Internally interval values are stored as months, days, and seconds.
This is done because the number of days in a month varies, and a day
can have 23 or 25 hours if a daylight savings time adjustment is
involved. The months and days fields are integers while the seconds
field can store fractions.
(Output / display depends on the setting of IntervalStyle.)
The above example uses default Postgres format: interval '1 month - 1 day'. These are also valid (while less readable):
interval '1 mon - 1 d' -- unambiguous abbreviations of time units are allowed
IS0 8601 format:
interval '0-1 -1 0:0'
Standard SQL format:
interval 'P1M-1D';
All the same.
Note that expected output for day_in_month(2) can be 29 because of leap years. You might want to pass a date instead of an int.
Also, beware of daylight saving : remove the timezone or else some monthes calculations could be wrong (next example in CET / CEST) :
SELECT DATE_TRUNC('month', '2016-03-12'::timestamptz) + '1 MONTH'::INTERVAL
- DATE_TRUNC('month', '2016-03-12'::timestamptz) ;
------------------
30 days 23:00:00
SELECT DATE_TRUNC('month', '2016-03-12'::timestamp) + '1 MONTH'::INTERVAL
- DATE_TRUNC('month', '2016-03-12'::timestamp) ;
----------
31 days
This works as well.
WITH date_ AS (SELECT your_date AS d)
SELECT d + INTERVAL '1 month' - d FROM date_;
Or just:
SELECT your_date + INTERVAL '1 month' - your_date;
These two return interval, not integer.
SELECT cnt_dayofmonth(2016, 2); -- 29
create or replace function cnt_dayofmonth(_year int, _month int)
returns int2 as
$BODY$
-- ZU 2017.09.15, returns the count of days in mounth, inputs are year and month
declare
datetime_start date := ('01.01.'||_year::char(4))::date;
datetime_month date := ('01.'||_month||'.'||_year)::date;
cnt int2;
begin
select extract(day from (select (datetime_month + INTERVAL '1 month -1 day'))) into cnt;
return cnt;
end;
$BODY$
language plpgsql;
You can write a function:
CREATE OR REPLACE FUNCTION get_total_days_in_month(timestamp)
RETURNS decimal
IMMUTABLE
AS $$
select cast(datediff(day, date_trunc('mon', $1), last_day($1) + 1) as decimal)
$$ LANGUAGE sql;

How to get DateTime formatted

In sql server how can I get current date but time 12:00 AM.
With GETDATE() i get current date and time.I need to have datetime formated like this:
2011-02-04 12:00 AM OR 2011-02-04 00:01
you first need to know how to get the time you want. It is called "flooring" date time. see this example:
--Floor a datetime
SELECT '0 None', GETDATE() -- none 2008-09-17 12:56:53.430
UNION SELECT '1 Second',DATEADD(second,DATEDIFF(second,'2000-01-01',GETDATE()),'2000-01-01') -- Second: 2008-09-17 12:56:53.000
UNION SELECT '2 Minute',DATEADD(minute,DATEDIFF(minute,0,GETDATE()),0) -- Minute: 2008-09-17 12:56:00.000
UNION SELECT '3 Hour', DATEADD(hour,DATEDIFF(hour,0,GETDATE()),0) -- Hour: 2008-09-17 12:00:00.000
UNION SELECT '4 Day', DATEADD(day,DATEDIFF(day,0,GETDATE()),0) -- Day: 2008-09-17 00:00:00.000
UNION SELECT '5 Month', DATEADD(month,DATEDIFF(month,0,GETDATE()),0) -- Month: 2008-09-01 00:00:00.000
UNION SELECT '6 Year', DATEADD(year,DATEDIFF(year,0,GETDATE()),0) -- Year: 2008-01-01 00:00:00.000
ORDER BY 1
PRINT' '
PRINT 'Note that when you are flooring by the second, you will often get an arithmetic overflow if you use 0. So pick a known value that is guaranteed to be lower than the datetime you are attempting to floor'
PRINT 'this always uses a date less than the given date, so there will be no arithmetic overflow'
SELECT '1 Second',DATEADD(second,DATEDIFF(second,DATEADD(day,DATEDIFF(day,0,GETDATE()),0)-1,GETDATE()),DATEADD(day,DATEDIFF(day,0,GETDATE()),0)-1) -- Second: 2008-09-17 12:56:53.000
once you floor it, use one of the flavors of CONVERT() to format it as you would like:
this does the format you want, but without changing the time:
select CONVERT(char(10), GETDATE(), 121)+' '+LTRIM(RIGHT(CONVERT(varchar(30), GETDATE(), 100),7))
OUTPUT:
------------------
2011-02-04 7:19AM
to format and set the time to what you want:
select CONVERT(char(10),DATEADD(day,DATEDIFF(day,0,GETDATE()),0), 121)+' '+LTRIM(RIGHT(CONVERT(varchar(30), DATEADD(day,DATEDIFF(day,0,GETDATE()),0), 100),7))
OUTPUT:
------------------
2011-02-04 12:00AM
Check out the CAST and CONVERT functions in T-SQL - they allow you to format your DATETIME values in various ways.
There are a number of Date functions you can use in SQL Server - See here.
Hopefully these will help!
To simply set the time to 12:00 AM but maintain the datetime datatype try:
SELECT DATEADD(DAY,0,DATEDIFF(DAY,0,GETDATE()))