I have a postgres database carrying date/time information in a text format. There is no way of changing it, but I have to retrieve those values as milisecons since epoch.
I managed to make a query, converting those date-time records to timestamps so that I get a correct "max" function behaviour like so:
SELECT max(TO_TIMESTAMP(column_name, 'YYYY/MM/DD HH24:MI:SS'))
FROM table_name;
But converting other results into miliseconds does not seem to work. And all the examples int the documentation and forums showcase only the usage for some literal value, not a value selected from a database. So lines like these don't work:
SELECT EXTRACT(EPOCH FROM TIMESTAMP
(select max(TO_TIMESTAMP(column_name, 'YYYY/MM/DD HH24:MI:SS'))
FROM table_name));
SELECT EXTRACT(EPOCH FROM TIMESTAMP
(select TO_TIMESTAMP(column_name,'YYYY/MM/DD HH24:MI:SS')
FROM table_name));
SELECT EXTRACT(EPOCH FROM TIMESTAMP WITH TIME ZONE(
SELECT TO_TIMESTAMP(column_name, 'YYYY/MM/DD HH24:MI:SS')
FROM table_name));
SELECT EXTRACT(EPOCH FROM TIMESTAMP WITH TIME ZONE TO_TIMESTAMP
(column_name, 'YYYY/MM/DD HH24:MI:SS'))
FROM table_name;
Is there an actual way to accomplish what I want by using a query, or I have to do something more complicated?
P.S.
Of course I can just retrieve all the infomation as text and use Qt (QDateTime) to convert it to miliseconds, but It would be more expensive and I was wondering if there is a way to ask the database to do it for me.
The timestamp keyword is only needed for literals (constants), not if you have a proper timestamp value available:
SELECT extract(epoch from max(TO_TIMESTAMP(column_name, 'YYYY/MM/DD HH24:MI:SS')))
FROM table_name;
Note that epoch represents seconds, not milliseconds.
Related
I have create a datetime with type timestamp. datetime timestamp NOT NULL I am not sure why the output is like this:
I want to extract the day part. I have tried these different approach but in both cases I am getting an error. How can I fix it?
extract(DAY FROM TIMESTAMP min(to_char(u.datetime ,'YYYY-MM-DD HH24:MI'))::timestamp)
EXTRACT(DAY FROM TIMESTAMP min(to_char(u.datetime ,'YYYY-MM-DD HH24:MI')))
date_part('day', min(to_char(u.datetime ,'YYYY-MM-DD HH24:MI')))
As mentioned in response I modified query to be like below and it does work.
extract(day from MIN(datetime)) as Day
All you need is:
select *, extract(day from activated_at) as Day from yourTable;
What you are seeing is a timestamp formatted as text for the display. Underlying data is timestamp as you said, directly use it.
Lets do the following:
SELECT to_timestamp(1453336500)::date
Then i get a date 2016-01-21
How does the function work backwards. With the date as input and the number (i guess seconds from 1970) as result?
You use extract:
SELECT extract(epoch FROM current_timestamp);
I need to convert the value stored in a bigint column to a date field. The first step of the conversion involves converting it to timestamp, and subsequently use the TRUNC method to convert this column to a date value.
However, my query is failing while converting the bigint value to timestamp.
The error that I'm getting is:-
Amazon Invalid operation: cannot cast type bigint to timestamp without time zone;
The query I'm trying for now is something like this:-
select ts::timestamp from events limit 1;
I was able to avoid the time zone error by using the method described in this thread: https://stackoverflow.com/a/36399361
My dates are based on epochs, and I was able to do the following:
SELECT
(TIMESTAMP 'epoch' + contract_start_date * INTERVAL '1 Second ')
FROM
table_name
SELECT TIMESTAMP 'epoch' + {column of bigint}/1000 * INTERVAL '1 second' as adate FROM tbl
If you are starting with a POSIX timestamp, and trying to get a timezone aware datetime value, you will need to supply a timezone - even if you later want to truncate the time part away. I'm not familiar with redshift, but perhaps there is a way to specify you mean UTC.
I'm trying to format the timestamps in my Postgres database to a certain format:
YYYY-MM-DD HH24:MI:SS
By doing:
update myTable set tds = to_char(tds, 'YYYY-MM-DD HH24:MI:SS')::timestamp;
I managed to set all the previously stored tds to this format. However, any newly added entry goes back to: YYYY-MM-DD HH24:MI:SS.MS since the default is set to now().
How do I change this so that newly added entries also have the format: YYYY-MM-DD HH24:MI:SS?
There is no format stored in a timestamp type. You can set its default to a timestamp truncated to the second at creation time
create table t (
tds timestamp default date_trunc('second', now())
)
Or alter the table
alter table t
alter column tds
set default date_trunc('second', now());
insert into t values (default);
INSERT 0 1
select * from t;
tds
---------------------
2014-03-11 19:24:11
If you just don't want to show the milliseconds part format the output
select to_char(now(), 'YYYY-MM-DD HH24:MI:SS');
to_char
---------------------
2014-03-11 19:39:40
The types timestamp or timestamptz optionally take a precision modifier p: timestamp(p).
To round to full seconds, set the default to:
now()::timestamp(0)
or:
now()::timestamptz(0)
Standard SQL functions CURRENT_TIMESTAMP (returns timestamptz) or LOCALTIMESTAMP (returns timestamp) allow the same precision modifier:
CURRENT_TIMESTAMP(0)
LOCALTIMESTAMP(0)
That's a bit shorter than calling date_trunc() - which truncates fractional seconds (may be what you really want!)
date_trunc('second', now())
Store timestamps as timestamptz (or timestamp), not as character type.
Finally, to make sure that ...
newly added entries also have the format: YYYY-MM-DD HH24:MI:SS
you could define your column as type timestamptz(0). This covers all values entered into that column, not just the default. But the rounding may introduce timestamps up to half a second in the future. If that can be an issue in any way, rather use date_trunc().
See #Clodoaldo's answer for instructions on to_char() and how to ALTER TABLE.
This related answer for in-depth information on timestamps and time zone handling:
Ignoring time zones altogether in Rails and PostgreSQL
When we use the query like this in oracle to get the the total number of hours in round figure,
hours= select round(out_time-in_time)*24 from table_name;
what is the datatype of hours here?
out_time and in_time are column names
You can take NUMBER as the datatype in your case.
EDIT:- This will return you an integer
select 24 * round((to_date('2013-07-07 22:00', 'YYYY-MM-DD hh24:mi')
- to_date('2013-07-07 19:30', 'YYYY-MM-DD hh24:mi'))) diff_hours
from table_name;
Maybe a datatype called "INTERVAL" will suit you. It's called:
INTERVAL DAY [(day_precision)] TO SECOND
Stores a period of time in days, hours, minutes, and seconds
Interval is something that can we added (or substrated) to a DATE or TIMESTAMP.
see: http://docs.oracle.com/cd/E11882_01/server.112/e26088/sql_elements001.htm