DATE_ADD works just fine with different months. The problem is that you are adding six months to 2001-01-01 and July 1st is supposed to be there. This is what you want to do: SELECT * FROM mydb WHERE creationdate BETWEEN “2011-01-01” AND DATE_ADD(“2011-01-01”, INTERVAL 6 MONTH) – INTERVAL 1 DAY GROUP BY MONTH(creationdate) OR SELECT … Read more
Yes, the syntax is accurate and it should be fine. Here is the SQL Fiddle Demo I created for your particular case create table sample2 ( id int primary key, created_date date, data varchar(10) ) insert into sample2 values (1,’2012-01-01′,’testing’); And here is how to select the data SELECT Created_Date FROM sample2 WHERE Created_Date >= … Read more
Use DATEADD: SELECT DATEADD(mi, @increase, @shift_start_time); db<>fiddle demo
Try this: UPDATE table SET date_field = date_field + interval ‘1 year’ It appears that you were trying to use SQL Server’s DATEADD() function, which does not exist in Postgres.
SELECT DATEADD(minute, -15, ‘2000-01-01 08:30:00′); The second value (-15 in this case) must be numeric (i.e. not a string like ’00:15′). If you need to subtract hours and minutes I would recommend splitting the string on the : to get the hours and minutes and subtracting using something like SELECT DATEADD(minute, -60 * @h – … Read more
Just do- Select (Getdate()+360) As MyDate There is no need to use dateadd function for adding or subtracting days from a given date. For adding years, months, hours you need the dateadd function.
Use the following function: DATEADD(type, value, date) date is the date you want to manipulate value is the integere value you want to add (or subtract if you provide a negative number) type is one of: yy, yyyy: year qq, q: quarter mm, m: month dy, y: day of year dd, d: day wk, ww: … Read more
You may use something like the strtotime() function to add something to the current timestamp. $new_time = date(“Y-m-d H:i:s”, strtotime(‘+5 hours’)). If you need variables in the function, you must use double quotes then like strtotime(“+{$hours} hours”), however better you use strtotime(sprintf(“+%d hours”, $hours)) then.