Can anyone tell please tell me how to change 2018-01-15T08:54:45.000Z to 2018-01-15 08:54:45 in PostgreSQL.
Here my timestamp: 2018-01-15T08:54:45.000Z is in text format.
I need to split it into two different columns like one for the only date:2018-01-15 and another is for only time:08:54:45
You should be able to directly cast your text to a timestamp and then cast again to date or time to get two different columns:
SELECT
('2018-01-15T08:54:45.000Z'::timestamp)::time AS time,
('2018-01-15T08:54:45.000Z'::timestamp)::date AS date
;
Related
I am beginner in Backend technology and I am developing one query in Typeorm QueryBuilder + PostgreSQL.
My query is look like this :
But I can't convert my timestamp into following format. Can anyone who has expertise in it, please help me to select all records with timestamp in specific format.
I am suffering this things from last two days but still not find any solution. Actually I want to compare this date format in having clause. so I want to first select that date format then I can use that same method to compare column with current date and previous dates.
I have table ABC in which I have column Z of datatype Date. The format of the data is YYYYMMDD. Now I am looking to convert the above format to YYYY-MON-DD format. Can someone help?
You can use to_char
TO_CHAR(Z,'YYYY-MON-DD')
Depending on what the purpose of the reformatting is, you can either explicitly cast it to a VARCHAR/CHAR and define the format, or you can change your display format to however you'd like to see all dates:
ALTER SESSION SET DATE_OUTPUT_FORMAT = 'YYYY-MON-DD';
It's important to understand that if the data is in a DATE field, then it is stored as a date, and the format of the date is dependent on your viewing preferences, not how it is stored.
Since the value of the date field is stored as a number, you have to convert it to date.
ALTER SESSION SET DATE_OUTPUT_FORMAT = 'YYYY-MON-DD';
select to_date(to_char( z ), 'YYYYMMDD');
(adding this answer to summarize and resolve the question - since the clues and answers are scattered through comments)
The question stated that column Z is of type DATE, but it really seems to be a NUMBER.
Then before parsing a number like 20201017 to a date, first you need to transform it to a STRING.
Once the original number is parsed to a date, it can be represented as a new string formatted as desired.
WITH data AS (
SELECT 20201017 AS z
)
SELECT TO_CHAR(TO_DATE(TO_CHAR(z), 'YYYYMMDD'), 'YYYY-MON-DD')
FROM data;
# 2020-Oct-17
I am extracting three values (server, region, max(date)) from my postgresql> But I want to extract an additional 4th field which should be the numerical addition of 1 to 3rd field. I am unable to use date add function as in the database date field is defined as an integer.
date type in DB
date|integer|not null
tried using cast and date add function
MAX(s.date)::date + cast('1 day' as interval)
Error Received
ERROR: cannot cast type integer to date
Required output
select server, region, max(alarm_date), next date from table .....
testserver, europe, 20190901, 20190902
testserver2, europe, 20191001, 20191002
next date value should be the addition to alarm_date
To convert an integer like 20190901 to a date, use something like
to_date(CAST(s.date AS text), 'YYYYMMDD')
It is a bad idea to store dates as integers like that. Using the date data type will prevent corrupted data from entering the database, and it will make all operations natural.
First solution that came to my mind:
select (20190901::varchar)::date + 1
Which output 2019-09-02 as type date.
Other solutions can be found here.
I am building a map in CartoDB which uses Postgres. I'm simply trying to display my dates as: 10-16-2014 but, haven't been able to because Postgres includes an unneeded timestamp in every date column.
Should I alter the column to remove the timestamp or, is it simply a matter of a (correct) SELECT query? I can SELECT records from a date range no problem with:
SELECT * FROM mytable
WHERE myTableDate >= '2014-01-01' AND myTableDate < '2014-12-31'
However, my dates appear in my CartoDB maps as: 2014-10-16T00:00:00Z and I'm just trying to get the popups on my maps to read: 10-16-2014.
Any help would be appreciated - Thank you!
You are confusing storage with display.
Store a timestamp or date, depending on whethether you need time or not.
If you want formatted output, ask the database for formatted output with to_char, e.g.
SELECT col1, col2, to_char(col3, 'DD-MM-YY'), ... FROM ...;
See the PostgreSQL manual.
There is no way to set a user-specified date output format. Dates are always output in ISO format. If PostgreSQL let you specify other formats without changing the SQL query text it'd really confuse client drivers and applications that expect the date format the protocol specifies and get something entirely different.
You have two basic options.
1 Change the column from a timestamp to a date column.
2 Cast to date in your SQL query (i.e. mytimestamp::date works).
In general if this is a presentation issue, I don't usually think that is a good reason to muck around with the database structure. That's better handled by client-side processing or casting in an SQL query. On the other hand if the issue is a semantic one, then you may want to revisit your database structure.
Please pardon my ignorance if I have missed any documentation/solution for the same. But I searched the web and could not find an answer.
I have a simple question. In the DB2 table,I have a column of type date and the with data of format 04/25/2013 12:00:00AM . When I query the DB2 database, I want to obtain just the date and not the timestamp i.e to obtain "04/25/2013" and not "04/25/2013 12:00:00AM". I tried DATE(column name) and just gave back the complete value including the time stamp.
This looks like a TIMESTAMP and not a DATE column. If it is indeed a TIMESTAMP column try this:
select varchar_format(current timestamp, 'MM/DD/YYYY') from sysibm.sysdummy1 ;
Just replace the current timestamp in the above example with your column and sysibm.sysdummy1 with your table.
The good thing about varchar_format is that it lets you easily format the timestamp. Just change the 'MM/DD/YYYY' part to 'YYYY.MM.DD' to get a format like '2017.08.18'.