How do I return a character string that represents the specified datepart of the specified date using PostgreSQL?
Note: In SQL Server we use the following syntax.
Example: (Using SQL Server)
SELECT DATENAME(DW,'2014-05-22'); /*For Weekday */
As documented in the manual the to_char() function can be used for this:
select to_char(DATE '2014-05-22', 'Day');
returns Thursday
date_part(text, timestamp)
Example for day of the week:
select date_part('dow', timestamp '2014-05-22 00:00:00')
Related
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 tried to ruh this query in postgres :
Select to_char((select add_months (to_date ('10/10/2019', 'dd/mm/yyyy'), '11/11/2019') ) , 'dd/mm/yyyy') as temp_date
I got an error :
Function add_months (date, unknown) does not exist
Hint: no function matches the given name and argument types. You might need to add explicit type casts.
Please help
As documented in the manual there is no add_months function in Postgres
But you can simply add an interval:
select to_date('10/10/2019', 'dd/mm/yyyy') + interval '10 months'
If you need to format that date value to something:
select to_char(to_date('10/10/2019', 'dd/mm/yyyy') + interval '10 months', 'yyyy-mm-dd')
No one, even running on Oracle, has run the original query- at least not successfully. It appears that query is expecting to add two months together (in this case Oct and Nov). That is not what the function does. It adds an integer number of months to the specified date and returns the resulting date. As indicated in Postgres just adding the desired interval. However, if you have many occurrences ( like converting) of this the following implements a Postgres version.
create or replace function add_months(
date_in date
, n_months_in integer)
returns date
language sql immutable strict
as
$$
-- given a date and an integer for number of months return the calendar date for the specified number of months away.
select (date_in + n_months_in * interval '1 month')::date
$$ ;
-- test
-- +/- 6 months from today.
select current_date "today"
, add_months(current_date,6) "6 months from now"
, add_months(current_date,-6) "6 months ago"
;
I have a 'timestamp without time zone' value in a column example:
PunchIn = "2019-06-10 09:10:00"
How can I extract "2019-06-10 00:00:00" from this column?
demo:db<>fiddle
SELECT my_date::date::timestamp
This converts the date column my_date to date (cuts the time part) and if you cast it back into timestamp it gets the 0 time.
Alternatively you can use the date_trunc function:
SELECT date_trunc('day', my_date)
SELECT SUBSTRING("PunchIn = '2019-06-10 09:10:00'" FROM 11 FOR 21);
should return the string you're looking for (if I count the index number right:)
You can use the DATE function to extract date:
SELECT DATE(SUBSTRING("PunchIn = '2019-06-10 09:10:00'" FROM 11 FOR 21));
In the view I have a text column which contains a timestamp in this format '20/03/2018 00:00' and I'm trying to make a selection with a between clause but it's not working
SELECT id,entry_date
FROM v_view
WHERE entrada BETWEEN to_timestamp('20/03/2018 00:00','DD/MM/YYYY')::timestamp and to_timestamp('22/03/2018 00:00')::timestamp
order entry_date
with this error message
ERROR: el operador no existe: text >= timestamp without time zone
LINE 3: WHERE entry_date BETWEEN to_timestamp('20/03/2018 00:00','DD/MM.
you need to convert the entrada column value to a timestamp.
Also: casting the result of to_timestamp() to a timestamp is useless because to_timestamp() already returns a timestamp
SELECT id,entry_date
FROM v_view
WHERE to_timestamp(entrada, 'dd/mm/yyyy hh24:mi')
BETWEEN to_timestamp('20/03/2018', 'DD/MM/YYYY')
and to_timestamp('22/03/2018', 'dd/mm/yyyy')
order entry_date;
I prefer to use ANSI SQL timestamp literals over the to_timestamp function:
SELECT id,entry_date
FROM v_view
WHERE to_timestamp(entrada, 'dd/mm/yyyy hh24:mi')
BETWEEN timestamp '2018-03-20 00:00:00'
and timestamp '2018-03-22 00:00:00'
order entry_date
Do not store date, time or timestamp values in a text or varchar column. You should define that column as timestamp then you don't need to convert anything and you don't need to deal with invalid timestamp values in that column.
I'm trying to convert value for DIM_DT_ID to MMddYY. I'm successful in doinf that. However, query fails because ultimately I'm comparing a character value to date here. Is there a way by which I can get value for DIM_DT_ID in MMddyy format and its data type still remains DATE ?
Here DIM_DT_ID
SELECT DIM_DT_ID
DIM_DT_ID >= FORMATDATE('MMddyy',ADDDAY(TO_date('yyyy-MM-dd','2016-12-21'), -25)); from abc;
Regards,
Ajay
In Denodo, to convert a string to a date field, use "to_date()" (which returns a date).
Then, don't convert back to a string, leave that field as a date (so don't use "Formatdate()", which returns a string).
So:
SELECT *
FROM MyTable
WHERE now() >= to_date('yyyy-MM-dd',myStringFieldThatLooksLikeADate)
In my example, "now()" is a date, and so is the output of the to_date function... so you can do a comparison.
If you try to convert the date back to a string using formatdate, it won't work:
#This doesn't work:
SELECT *
FROM MyTable
WHERE now() >= formatdate('MMddyy',to_date('yyyy-MM-dd',myStringFieldThatLooksLikeADate))
It doesn't work because we are comparing a date ("now()") to a string.