I'm using Postgres with Rails. I've two string fields which store date like this, field_1 = "2020.06.09", I 'm comparing them with TO_DATE function and they are working finr but now I want to compare them after adding 5 months in one field. What I've tried so far is:
select * from users as u where TO_DATE(u.field_1, 'YYYY MM DD') > DATEADD(MONTH, 5, TO_DATE(u.field_2, 'YYYY MM DD'))
can anyone help me on this?
Thanks in advance.
There is no dateadd() in Postgres (or standard SQL), you need to add an interval:
select *
from users u
where TO_DATE(u.field_1, 'YYYY MM DD') > to_date(u.field_2, 'yyyy mm dd') + interval '5 month';
Related
Lets say we have the dates
'2017-01-01'
and
'2017-01-15'
and I would like to get a series of exactly N timestamps in between these dates, in this case 7 dates:
SELECT * FROM
generate_series_n(
'2017-01-01'::timestamp,
'2017-01-04'::timestamp,
7
)
Which I would like to return something like this:
2017-01-01-00:00:00
2017-01-01-12:00:00
2017-01-02-00:00:00
2017-01-02-12:00:00
2017-01-03-00:00:00
2017-01-03-12:00:00
2017-01-04-00:00:00
How can I do this in postgres?
Possibly this can be useful, using the generate series, and doing the math in the select
select '2022-01-01'::date + generate_series *('2022-05-31'::date - '2022-01-01'::date)/15
FROM generate_series(1, 15)
;
output
?column?
------------
2022-01-11
2022-01-21
2022-01-31
2022-02-10
2022-02-20
2022-03-02
2022-03-12
2022-03-22
2022-04-01
2022-04-11
2022-04-21
2022-05-01
2022-05-11
2022-05-21
2022-05-31
(15 rows)
WITH seconds AS
(
SELECT EXTRACT(epoch FROM('2017-01-04'::timestamp - '2017-01-01'::timestamp))::integer AS sec
),
step_seconds AS
(
SELECT sec / 7 AS step FROM seconds
)
SELECT generate_series('2017-01-01'::timestamp, '2017-01-04'::timestamp, (step || 'S')::interval)
FROM step_seconds
Conversion to function is easy, let me know if have trouble with it.
One problem with this solution is that extract epoch always assumes 30-days months. If this is problem for your use case (long intervals), you can tweak the logic for getting seconds from interval.
You can divide the difference between the end and the start value by the number of values you want:
SELECT *
FROM generate_series('2017-01-01'::timestamp,
'2017-01-04'::timestamp,
('2017-01-04'::timestamp - '2017-01-01'::timestamp) / 7)
This could be wrapped into a function if you want to avoid repeating the start and end value.
I have issue change ADD_MONTHS Oracle to PostgreSQL.
I have Oracle query like this :
ADD_MONTHS (to_date(to_char(start_billdate,'DD-MM-YYYY'),'DD-MM-YYYY'),
(processed_num*periodvalue)
)
So how to implement that query to PostgreSQL?
One option might be to multiply number of months (processed_num * periodvalue) with the interval of 1 month and add that to start_billdate:
start_billdate + (interval '1 month' * processed_num * periodvalue);
You can construct an interval with the contents of the column:
start_billdate + make_interval(months => processed_num*periodvalue)
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))
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;
I get to dust off my VBScript hat and write some classic ASP to query a SQL Server 2000 database.
Here's the scenario:
I have two datetime fields called fieldA and fieldB.
fieldB will never have a year value that's greater than the year of fieldA
It is possible the that two fields will have the same year.
What I want is all records where fieldA >= fieldB, independent of the year. Just pretend that each field is just a month & day.
How can I get this? My knowledge of T-SQL date/time functions is spotty at best.
You may want to use the built in time functions such as DAY and MONTH. e.g.
SELECT * from table where
MONTH(fieldA) > MONTH(fieldB) OR(
MONTH(fieldA) = MONTH(fieldB) AND DAY(fieldA) >= DAY(fieldB))
Selecting all rows where either the fieldA's month is greater or the months are the same and fieldA's day is greater.
select *
from t
where datepart(month,t.fieldA) >= datepart(month,t.fieldB)
or (datepart(month,t.fieldA) = datepart(month,t.fieldB)
and datepart(day,t.fieldA) >= datepart(day,t.fieldB))
If you care about hours, minutes, seconds, you'll need to extend this to cover the cases, although it may be faster to cast to a suitable string, remove the year and compare.
select *
from t
where substring(convert(varchar,t.fieldA,21),5,20)
>= substring(convert(varchar,t.fieldB,21),5,20)
SELECT *
FROM SOME_TABLE
WHERE MONTH(fieldA) > MONTH(fieldB)
OR ( MONTH(fieldA) = MONTH(fieldB) AND DAY(fieldA) >= DAY(fieldB) )
I would approach this from a Julian date perspective, convert each field into the Julian date (number of days after the first of year), then compare those values.
This may or may not produce desired results with respect to leap years.
If you were worried about hours, minutes, seconds, etc., you could adjust the DateDiff functions to calculate the number of hours (or minutes or seconds) since the beginning of the year.
SELECT *
FROM SOME_Table
WHERE DateDiff(d, '1/01/' + Cast(DatePart(yy, fieldA) AS VarChar(5)), fieldA) >=
DateDiff(d, '1/01/' + Cast(DatePart(yy, fieldB) AS VarChar(5)), fieldB)
Temp table for testing
Create table #t (calDate date)
Declare #curDate date = '2010-01-01'
while #curDate < '2021-01-01'
begin
insert into #t values (#curDate)
Set #curDate = dateadd(dd,1,#curDate)
end
Example of any date greater than or equal to today
Declare #testDate date = getdate()
SELECT *
FROM #t
WHERE datediff(dd,dateadd(yy,1900 - year(#testDate),#testDate),dateadd(yy,1900 - year(calDate),calDate)) >= 0
One more example with any day less than today
Declare #testDate date = getdate()
SELECT *
FROM #t
WHERE datediff(dd,dateadd(yy,1900 - year(#testDate),#testDate),dateadd(yy,1900 - year(calDate),calDate)) < 0