I'd like to change a specific column in my PostgreSQL database from character_varying type to type date. Date is in the format yyyy:mm:dd
I tried to do:
alter table table_name
alter column date_time type date using (date_time::text::date);
But I received an error message:
date/time field value out of range: "2011:06:15"
When you cast text or varchar to date, the default date format of your installation is expected - depending on the datestyle setting of your session, typically set in your postgresql.conf. If in doubt, check with:
SHOW datestyle;
Generally, colon (:) is a time separator, In a simple cast, PostgreSQL will probably try to interpret '2011:06:15' as time - and fail.
To remove ambiguity use to_date() with a matching pattern for your dates:
ALTER TABLE table_name
ALTER COLUMN date_time type date
USING to_date(date_time, 'YYYY:MM:DD'); -- pattern for your example
Related
I'm trying to run a replace and update on a date column in Postgresql.
UPDATE table SET date_time = REPLACE(date_time, '12:', '00:');
I've tried casting as text. But still errors.
The error I get is
function replace(timestamp without time zone, unknown, unknown) does not exist
I'm guessing it's not picking it up due to the column's data type. But I'm not quite sure how to achieve what I'm after now.
Cast date_time to text, do the replacement and then cast the result to timestamp again. BTW, what are you trying to achieve? Isn't it a presentation format issue?
UPDATE "table"
SET date_time = replace(date_time::text, ' 12:', ' 00:')::timestamp;
I am dealing with the table which has date information in CHar(20) type. This date is in dd.mm.yyyy HH.MM.SS format but my pgadmin has Month first format. I tried editing posgres config file to change the date format. I tried to use SET timezone and then tried to convert type to timestamp but nothing is working. How can I convert following column into timestamp format? I followed miost of the answers here on stackoverflow but getting out of range error even after using set function or editing config file.
Use to_timestamp:
to_timestamp(stringcol, 'DD/MM/YYYY HH24:MI')
To change the data type of the column, which is highly commendable:
ALTER TABLE mytable ALTER date1 TYPE timestamp
USING CAST (to_timestamp(date1, 'DD/MM/YYYY HH24:MI') AS timestamp);
We are making use of tRedshiftOutputBulk exec and we have set it to 'Drop table if exists and create' as an action on table. The problem is that a Date field of with a pattern of 'yyyy-MM-dd HH:mm:ssZ' is being created as Timestamp rather than TimestampTZ on Redshift.
#mark
What pattern have you given in Date format and Time format field of the component?
you can use the change time pattern 'yyyy-MM-dd HH:mm:ssTZ'
This question already has answers here:
in postgres, can you set the default formatting for a timestamp, by session or globally?
(3 answers)
Closed 6 years ago.
INSERT into Group (Name,CreatedDate) VALUES ('Test',UTC_TIMESTAMP(), 1);
This is the query I have used for mysql to insert current date time.
When I am using this in postgresql, I am getting below error.
HINT: No function matches the given name and argument types. You might need to add explicit type casts.
********** Error **********
ERROR: function utc_timestamp() does not exist
SQL state: 42883
I have tried like below using now(), however it is inserting like "2016-07-07 17:01:18.410677". I need to insert in 'yyyymmdd hh:mi:ss tt' format.
INSERT into Group (Name,CreatedDate) VALUES ('Test',UTC_TIMESTAMP(), 1);
How to insert current date time in insert query of postgresql in above format ?
timestamp (or date or time columns) do NOT have "a format".
Any formatting you see is applied by the SQL client you are using.
To insert the current time use current_timestamp as documented in the manual:
INSERT into "Group" (name,createddate)
VALUES ('Test', current_timestamp);
To display that value in a different format change the configuration of your SQL client or format the value when SELECTing the data:
select name, to_char(createddate, 'yyyymmdd hh:mi:ss tt') as created_date
from "Group"
For psql (the default command line client) you can configure the display format through the configuration parameter DateStyle: https://www.postgresql.org/docs/current/static/runtime-config-client.html#GUC-DATESTYLE
For current datetime, you can use now() function in postgresql insert query.
You can also refer following link.
insert statement in postgres for data type timestamp without time zone NOT NULL,.
You can of course format the result of current_timestamp().
Please have a look at the various formatting functions in the official documentation.
I have the following function:
CREATE function SEMANA_ISO (fecha date) returns text as $$
select to_char(fecha, 'mm-dd-yyyy');
$$ LANGUAGE sql;
It works with:
Select SEMANA_ISO ('28/12/2014');
Select SEMANA_ISO ('01/01/2015');
Select SEMANA_ISO ('01/07/2015');
As you may see below
But not with:
Select SEMANA_ISO ('12/31/2014');
It shows:
********** Error **********
ERROR: The value of time / date is out of range "12/31/2014"
SQL state: 22008
Hint: You may need a different configuration of "dateStyle".
Character: 20
Do you have any suggestion without having to change the datestyle so I can enter
Select SEMANA_ISO ('12/31/2014');
And get an output of:
12-31-2014
using just one function to "parse" all dates?
Your function is declared to get a parameter of type date so you also need to pass such a value. '12/31/2014' is a character value, not a date.
When you pass a character literal (aka string) to the function Postgres is forced to do an implicit data type conversion based on the current datestyle - not something you should rely on.
If you want to call the function independently of the datestyle you need to pass a proper date literal, e.g. DATE '2014-12-31'
For more details on specifying date values, please see the manual: http://www.postgresql.org/docs/current/static/datatype-datetime.html#DATATYPE-DATETIME-INPUT
If you want 12/31/2014 accepted as a date set datestyle to iso, dmy;
if you want 31/12/2014 accepted as a date set datestyle to iso, mdy;
you can't have both at the same time, else 4/1/2015 is ambiguos, and must be rejected.
There is to_date() to input date literals with an arbitrary (given) format:
SELECT to_date('12/31/2014', 'MM/DD/YYYY') AS date1
, to_date('31/12/2014', 'DD/MM/YYYY') AS date2;
That's "without changing any datestyle". Obviously, you need to provide a matching format pattern, though.
To output the same in any desired format, use to_char():
SELECT to_char(to_date('12/31/2014', 'MM/DD/YYYY'), 'DD/MM/YYYY') AS date_as_text