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));
Related
I want to query postgres database table which has the column with timestamp in long milliseconds. But I have the time in date format "yyyy-MM-dd HH:mm:ssZ" like this. How can I convert this date format to long milliseconds to run the query?
You can either convert your long value to a proper timestamp:
select *
from the_table
where to_timestamp(the_millisecond_column / 1000) = timestamp '2020-10-05 07:42'
Or extract the seconds from the timestamp value :
select *
from the_table
where the_millisecond_column = extract(epoch from timestamp '2020-10-05 07:42') * 1000
The better solution is however to convert that column to a proper timestamp column to avoid the constant conversion between (milliseconds) and proper timestamp values
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 have column of type date time and values are getting stored in format 10-29-2011 08:25.
I would like to find out the rows only which are less then current date-time. What will be the condition for date comparison for current date and this date-time column field?
Thanks.
you could use the datetime function
SELECT * FROM mytable
WHERE mydate > datetime('now')
you can even make date operations
SELECT * FROM mytable
WHERE mydate > datetime('now','-15 days')
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;
let's say that I have a SmallDateTime column in my table. How to update hours in each row in T-SQL ?
Get the target value's hour part.
Find the difference between the hour you want and the found hour.
Add the difference of hours to the target value.
The script:
UPDATE atable
SET datetimevalue = DATEADD(hour, #hour - DATEPART(hour, datetimevalue),
datetimevalue)
WHERE ...
UPDATE YourTable SET YourDateColumn = DateAdd(hh, 1, YourDateColumn)
will add 1 hour to the time value in YourDateColumn in every single row.
See this page for more info.
UPDATE YourTableName
SET YourSmallDateTimeColumn = DATEADD(HH, 1, YourSmallDateTimeColumn)
Will add any number of hours to your column. So if you get rid of the time component first:
SELECT CAST(CONVERT(char(8), YourSmallDateTimeColumn, 112) AS smalldatetime)
and add hour component to it later it should work.
From here
To add or subtract hours to a datetime
or smalldatetime value, you will use
the DATEADD date function. The
DATEADD date function returns a new
datetime value based on adding an
interval to the specified date. The
syntax of the DATEADD date function is
as follows:
DATEADD ( datepart , number, date )
datepart is the parameter that
specifies on which part of the date to
return a new value. For hours, you
can use either HOUR or HH. number is
the value used to increment datepart.
date is an expression that returns a
datetime or smalldatetime value, or a
character string in a date format.
Here's an example on how to use the
DATEADD date function to increase or
decrease a datetime value by a certain
number of hours:
SELECT DATEADD(HOUR, -12, GETDATE())AS [TwelveHoursAgo]
SELECT DATEADD(HH, 6, GETDATE()) AS [SixHoursLater]