I need to change the date format from yyyy-mm-dd to dd/mm/yyyy, but I need to keep the date type for the column. So far, I have used this method that changes the format correctly but transforms the column type from date to text.
TO_CHAR(mydate::TIMESTAMP WITH TIME ZONE, 'dd/mm/yyyy'::TEXT) ;
How can I do it with keeping the date type?
This is a misunderstanding.
If your data type is date, then it is stored without format. It's just a 4-byte integer counting seconds since 2000.
You can format it any way when displaying to the client.
SELECT to_char(mydate, 'yyyy-mm-dd') AS one_way
,to_char(mydate, 'dd/mm/yyyy') AS other_way
'yyyy-mm-dd' happens to be ISO 8601 format, which is the default text representation in many locales.
You can always create a VIEW with a text representation of the date:
CREATE VIEW v_tbl_with_date_foramt AS
SELECT id, some_column, to_char(mydate, 'dd/mm/yyyy') AS mydate_text
FROM tbl;
But now it's a text column, not a date column.
datestyle is responsible for how input date literals are interpreted.
LC_TIME regulates how date/time functions behave.
Formatting Date(YY:MM:DD:Time) in Excel
The default display is the ISO 8601: 'yyyy-mm-dd'.
Related
The Postgres Query select date('22-02-2022')
is showing Error message.
I need to save date in a column of a table, but the input date can be any format. Please suggest how to save date in YYYY-MM-DD format.
Following is the Error screenshot:
A date has no format at all. But if you specify a date literal (constant) the way you tried, it has to be in yyyy-mm-dd format:
select date '2022-02-22'
If you want to specify the value in a different format, use the to_date() function:
select to_date('22-02-2022', 'dd-mm-yyyyy');
If the column in the table is defined with the date data type, the way you specify the actually value is irrelevant as it will be stored as a binary value without any format.
If you need a specific format when selecting (displaying) those values, you can use the to_char() function.
If all your dates and timestamps are going to come in this way then as the HINT suggests you need to change the DateStyle.
Run SHOW DateStyle, I'm guessing it will return something like ISO, MDY.
Change it SET DateStyle = 'ISO, DMY', then your example will work.
Example:
show datestyle ;
DateStyle
-----------
ISO, MDY
select date('22-02-2022');
ERROR: date/time field value out of range: "22-02-2022"
LINE 1: select date('22-02-2022');
^
HINT: Perhaps you need a different "datestyle" setting.
set datestyle = 'ISO,DMY';
select date '22-02-2022';
date
------------
2022-02-22
I need to execute a INSERT statement writing a date with a YYYY-MM-DD format.
Would to_date('2021-09-28','YYYY-MM-DD') work?
YYYY-MM-DD is the the ISO 8601 standard date format and unambiguous default in Postgres. Just insert your date literally.
The type date is stored as a 4-byte integer quantity internally, which does not preserve any format. You can format any way you like on output with some basic locale settings or settings of your client, or explicitly with to_char().
Input with to_date('2021-09-28','YYYY-MM-DD') works, too. But you don't need to_date() while operating with ISO format.
I have a column which contians date as string but in many formats like - dd/MM/yy, dd/MMM/yyy .. etc etc. And I am using the following code to convert all strings to one specific date format (yyyy-MM-dd) in hive :
select
from_unixtime(unix_timestamp('31/02/2021','dd/MM/yyyy'),'yyyy-MM-dd')
but this gives me 2021-03-03 in HIVE.
Is there any other way to identify such invalid dates and give null.
Assume, you recognized format correctly and it is exactly 'dd/MM/yyyy' and date is invalid one '31/02/2021'.
unix_timestamp function in such case will move date to the next month and there is no way to change it's behavior. But you can check if the date double-converted from original string to timestamp and back to original format is the same. In case it is not the same, then the date is invalid one.
case
-- check double-converted date is the same as original string
when from_unixtime(unix_timestamp(date_col,'dd/MM/yyyy'),'dd/MM/yyyy') = date_col
--convert to yyyy-MM-dd if the date is valid
then from_unixtime(unix_timestamp('31/02/2021','dd/MM/yyyy'),'yyyy-MM-dd')
else null -- null if invalid date
end as date_converted
I'm trying to add only the current date in "DD/MM/YYYY" format in a field of type ' timestamp in PostgreSQL' .
I try:
select to_char(now(),'DD/MM/YYYY') as date;
But PostgreSQL return me:
TIP : You will need to rewrite the expression or apply a type conversion.
There is no such thing as "only the date" in a timestamp field. A timestamp field will store timestamps.
Try using the date type instead. Please read about this here.
Also, please consider using the ISO 8601 format instead. Getting used to it helps in a lot of cases.
Human-readable formats like "DD/MM/YYYY" should only be used for presentation.
If you want to use timestamp fields and insert a human-readable formatted dates, then you are looking for:
to_timestamp('05/04/2016', 'DD/MM/YYYY')
If it is about the current date, then Postgres provides the CURRENT_DATE function, which you may use:
SELECT CURRENT_DATE;
INSERT INTO t (timestamp_field) VALUES (CURRENT_DATE);
So I have one big file (13 million rows) and date formatted as:
2009-04-08T01:57:47Z. Now I would like to split it into 2 columns now,
one with just date as dd-MM-yyyy and other with time only hh:MM.
How do I do it?
You can simply use tMap and parseDate/formatDate to do what you want. It is neither necessary nor recommended to implement your own date parsing logic with regexes.
First of all, parse the timestamp using the format yyyy-MM-dd'T'HH:mm:ss'Z'. Then you can use the parsed Date to output the formatted date and time information you want:
dd-MM-yyyy for the date
HH:mm for the time (Note: you mixed up the case in your question, MM stands for the month)
If you put that logic into a tMap:
you will get the following:
Input:
timestamp 2009-04-08T01:57:47Z
Output:
date 08-04-2009
time 01:57
NOTE
Note that when you parse the timestamp with the mentioned format string (yyyy-MM-dd'T'HH:mm:ss'Z'), the time zone information is not parsed (having 'Z' as a literal). Since many applications do not properly set the time zone information anyway but always use 'Z' instead, so this can be safely ignored in most cases.
If you need proper time zone handling and by any chance are able to use Java 7, you may use yyyy-MM-dd'T'HH:mm:ssXXX instead to parse your timestamp.
I'm guessing Talend is falling over on the T and Z part of your date time stamp but this is easily resolved.
As your date time stamp is in a regular pattern we can easily extract the date and time from it with a tExtractRegexFields component.
You'll want to use "^([0-9]{4}-[0-9]{2}-[0-9]{2})T([0-9]{2}:[0-9]{2}):[0-9]{2}Z" as your regex which will capture the date in yyyy-MM-dd format and the time as mm:HH (you'll want to replace the date time field with a date field and a time field in the schema).
Then to format your date to your required format you'll want to use a tMap and use TalendDate.formatDate("dd-MM-yyyy",TalendDate.parseDate("yyyy-MM-dd",row7.date)) to return a string in the dd-MM-yyyy format.