PostgreSQL now() function not returning time - postgresql

When inserting now() in a table, the written value only contains the date (e.g. 2017-12-20), but not the date and time as specified in the documentation.
See this SQLfiddle.
create table timetest (
id serial primary key,
mydate date
);
insert into timetest (mydate) values (
now());
Is there some specific command that should be passed to either write or retrieve also the time information?

You created mydate date as a date type column. date only represents the date fraction (unsurprisingly).
If you need both date and time use timestamp type instead.
References:
https://www.postgresql.org/docs/current/static/datatype-datetime.html

Related

How do I convert timestamp and offset columns to a single timestamptz column?

Image I have a table containing the following two columns:
timestampwithouttimezone (of type TIMESTAMP WITHOUT TIME ZONE)
utcoffset (of type INTEGER)
I want to convert those two column to a single one of type TIMESTAMP WITH TIME ZONE. Can this be achieved using a ALTER TABLE ALTER COLUMN [column] SET DATE TYPE TIMESTAMP WITH TIME ZONE query and the additional USING clause?
Or do I need a separate UPDATE query that takes the offset and sets the timezone of the timestamps? If that's the case, what would that query be? I can't find any examples that show how to update the timezone using an integer.
You could do that like this, assuming the offset is in hours:
ALTER TABLE mytab
ALTER timestampwithouttimezone
TYPE timestamp with time zone
USING CAST (timestampwithouttimezone::text || ' '
|| to_char(utcoffset, 'S00FM')
AS timestamp with time zone),
DROP utcoffset;

what will be data type to store date time in cassandra

what will be data type to store date time in cassandra
CREATE TABLE testTable (
dateValue date,
time timestamp
)
n my insert staements would be like this ,
insert into caliper.log_per_day ( timeStampValue,dateValue ) values ('2015-12-30 16:10:31','2015-12-30');
i wanted to store date & time both in one column like this '2015-12-30 16:10:31'.
but if i used timestamp it would be store like this '2015-12-30 04:10:31+0530'
Notes : primary key n other things are skip here... ignore it.
Cqlsh will display timestamps in the following format by default:
yyyy-mm-dd HH:mm:ssZ
The Z in these formats refers to an RFC-822 4-digit time zone,
If no time zone is supplied, the current time zone for the Cassandra
server node will be used.
so if you don't want to store in this way you can store it as varchar.

Convert bigint data type to timestamp (and subsequently to date) in redshift

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.

how to insert the current system date and time in oracle10g database

I have created a table with a column date_time type (varchar2 (40) ) but when i try to insert the current system date and time the doesnt work it gives error (too many values). please tell me what's wrong with the insert statement.
create table HR (type varchar2 (20), raised_by number (6), complaint varchar2 (500), date_time varchar2(40))
insert into HR values ('request',6785,'good morning',sysdate,'YYYY/MM/DD:HH:MI:SSAM')
The immediate cause of the error is that you have too many values, as the message says; that is, more elements in your values clause than there are columns. It is better to explicitly list the column names to avoid future problems and confusion, so you're really doing this:
insert into HR (type, raised_by, complaint, date_time)
values ('request',6785,'good morning',sysdate,'YYYY/MM/DD:HH:MI:SSAM')
... sp you have four columns, but five values. You're trying to insert the current date/time as a string so you would need to use the to_char() function:
insert into HR (type, raised_by, complaint, date_time)
values ('request',6785,'good morning',
to_char(sysdate,'YYYY/MM/DD:HH:MI:SSAM'))
But it is bad practice to store a date (or any other structured data, such as a number) as a string. As the documentation notes:
Each value manipulated by Oracle Database has a data type. The data
type of a value associates a fixed set of properties with the value.
These properties cause Oracle to treat values of one data type
differently from values of another. For example, you can add values of
NUMBER data type, but not values of RAW data type.
If you use a string then you can put invalid values in. If you use a proper DATE data type then you cannot accidentally put an invalid or confusing value in. Oracle will also be able to optimise the use of the column, and will be able to compare values safely and efficiently. Although the format you're using is better than some, using string comparison you still can't easily compare two values to see which is earlier, so you can't properly order by the date_time column for example.
Say you inserted two rows with values 2013/11/15:09:00:00AM and 2013/11/15:08:00:00PM - which is earlier? You need to look at the AM/PM marker to realise the first one is earlier; with a string comparison you'd get it wrong because 8 would be sorted before 9. Using HH24 instead of HH and AM avoids that, but would still be less efficient than a true date.
If you need to store a date with a time component you can use the DATE data type, which has precision down to the second; or if you need fractional seconds too then you can use TIMESTAMP. Then your table and insert would be:
create table HR (type varchar2 (20), raised_by number (6),
complaint varchar2 (500), date_time date);
insert into HR (type, raised_by, complaint, date_time)
values ('request',6785,'good morning',sysdate);
You can still get the value in the format you wanted for display purposes as part of a query:
select type, raised_by, complaint,
to_char(date_time, 'YYYY/MM/DD:HH:MI:SSAM') as date_time
from HR
order by date_time;
TYPE RAISED_BY COMPLAINT DATE_TIME
-------------------- ---------- -------------------- ---------------------
request 6785 good morning 2013/11/15:08:44:35AM
Only treat a date as a string for display.
You can use TO_DATE() or TO_TIMESTAMP or To_char() function,
insert into HR values ('request',6785,'good morning',TO_DATE(sysdate, 'yyyy/mm/dd hh24:mi:ss'))
insert into HR values ('request',6785,'good morning',TO_TIMESTAMP(systimestamp, 'yyyy/mm/dd hh24:mi:ss'))
sysdate - It will give date with time.
systimestamp - It will give datetime with milliseconds.
To_date() - Used to convert string to date.
To_char() - Used to convert date to string.
Probably here you have to use To_char() because your table definition have varchar type for date_time column.
Use TIMESTAMP datatype for date_time. And while inserting use the current timestamp.
create table HR (type varchar2(20), raised_by number(6), complaint varchar2(500), date_time timestamp);
insert into HR values ('request',6785,'good morning', systimestamp);
For other options: http://psoug.org/reference/timestamp.html

how to insert a time in oracle 10g database

I want to insert date and time in oracle database, I have created the table with columns
create table myadmin
( employe_id number(5),
supervisor Varchar2(20),
department Varchar2(20),
action Varchar2(20),
sdate date,
stime date)
While inserting the values below it gives an error. Please tell me how to insert the time ?
insert into myadmin
( employe_id,supervisor,department,action,sdate,stime) values
(83,'gaurav','helpdesk','pick','23-jan-2013','09:43:00');
You have to use keyword to_date for date insert in oracle like this.
to_date('23-01-2013','dd-mm-yyyy')
Basically you have to use keyword to_date('your date','your date format').
You can also add date and time together if you want and it would be something like this
to_date('23-01-2013 09:43:00','dd-mm-yyyy hh24:mi:ss')
A date in Oracle always has a date part and a time part. Having date and time in two separate columns only makes sense, if it can occur that date is null and time is not. (And still, you could set date to an improbable value like 1.1.0001 then.)
However, if you want to stick to those two separate fields, then make your string a datetime with the to_date function specifying the format used:
insert into myadmin
( employe_id,supervisor,department,action,sdate,stime) values
(83,'gaurav','helpdesk','pick',to_date('23-01-2013','dd-mm-yyyy'), to_date('09:43:00', 'hh24:mi:ss'));