Hive's unix_timestamp and from_unixtime functions - hive

I am under the impression that unix_timestamp and from_unixtime Hive functions are 'reverse' of each other.
When I try to convert timestamp string to seconds in Hive:
SELECT unix_timestamp('10-Jun-15 10.00.00.000000 AM', 'dd-MMM-yy hh.mm.ss.MS a');
I get 1418176800.
When I try to convert 1418176800 to timestamp string:
SELECT from_unixtime(1418176800, 'dd-MMM-yy hh.mm.ss.MS a');
I get 10-Dec-14 10.00.00.120 AM, which is obviously not equal to the original.
Can someone explain what's going on? Thanks.

From the language manual:
Convert time string with given pattern to Unix time stamp (in seconds)
The result of this function is in seconds.
Your result changes with the milliseconds portion of the date, but the unix functions only support seconds. For example:
SELECT unix_timestamp('10-Jun-15 10.00.00 AM', 'dd-MMM-yy hh.mm.ss a');
1433930400
SELECT from_unixtime(1433930400, 'dd-MMM-yy hh.mm.ss a');
10-Jun-15 10.00.00 AM

Don't use from unix_timestamp as you have in your second query. Additionally your statement has a formatting in it that gives the result where DEC is used in steads of 12. See dd-MM-yy. Don't specify a format and it should work. See the examples below.
You are however correct that from_unixtime() and unix_timestamp() are used to convert back and forth from time string.
select unix_timestamp('2015-04-09 03:04:26') from dual;
results in "1428566666"
select from_unixtime(1428566666) from dual;
results in "2015-04-09 03:04:26"

Related

correct to_char date syntax to have trailing zeroes after milliseconds

My current query in oracle sql for getting a timestamp format is TO_CHAR(c2.start_on,'DD-MM-YY HH:MI:SS.FF PM'), it outputs the timestamp like this 25-11-20 07:00:13.36 PM
However I want it to display the date in this way 25-11-20 07:00:13.360000000 PM
What should I add in the timestamp format for this to be possible ?
I have tried doing it like this HH:MI:SS.FM00000 as suggested here
but it gives me the error. ORA-01821: date format not recognized
what is the correct way to get the date in the desired format ?
If you want fractional seconds, you don't want a DATE, you want a TIMESTAMP. So here's a timestamp formatted with 6 digits of precision
select to_char(systimestamp, 'HH:MI:SS.FF6') from dual;
If you have a date, you could convert it to a TIMESTAMP (using CAST AS TIMESTAMP), but better to look at updating your data model to use the proper type for the source column as starters.

In SQL How to convert time into UNIX timestamp

In hive there is some data I have. Now I want to convert the start_timestamp into unix_timestamp in second. How to do that? Because the start_timestamp has two formats:
First format:
2018-03-22 02:54:35
Second format:
May 15 2018 5:15PM
First format is 'yyyy-MM-dd HH:mm:ss', second is 'MMM dd yyyy hh:mm:aa'. If the format is wrong, unix_timestamp function will return NULL. Try to convert using one format, if NULL, try to convert using the other format. This can be done using coalesce function:
select
coalesce(unix_timestamp(start_timestamp ,'yyyy-MM-dd HH:mm:ss'),
unix_timestamp(start_timestamp ,'MMM dd yyyy hh:mm:aa')
) as UnixTimestamp
from my_table;
Use from_unixtime() to convert it back to given format if necessary, like in this answer.
See patterns examples here: SimpleDateFormat

Oracle SQL how to convert time zone string to date

I have following 2015-06-17T00:00:00.000+05:00 string.
I want to convert this string to Date using oracle sql.
I tried lot of format mask but none works for me :
SELECT TO_DATE('2015-06-17T00:00:00.000+05:00','yyyy-mm-dd HH24:MI:SS TZR') FROM DUAL;
Any idea which format mask should i apply for above conversion.
Also please note that i only need date information i.e (mm-dd-yyyy). So its also ok if the conversion results in date information only (i.e skipping time information)
This should work:
SELECT TO_DATE(SUBSTR('2015-06-17T00:00:00.000+05:00',1,10),'yyyy-mm-dd') from dual
If you need to keep track of the time zone you should probably look at something like this:
SELECT CAST(TO_TIMESTAMP_TZ('2015-06-17T00:00:00.000+05:00','yyyy-mm-dd"T"HH24:MI:SS.FFTZH:TZM') AT TIME ZONE 'UTC' AS DATE) FROM DUAL;

ORA-01722 INVALID NUMBER in oracle

I am getting invalid number error message while executing the below select statement.Can any one have an idea about the issue..Please let me know.
select TO_DATE(TO_CHAR('2015/01/22 00:00:00','YYYY/MM/DD'),'YYYY/MM/DD')
actually i want oracle standard date format without time stamp for this date '2015/01/22 00:00:00'
select to_date('2015/01/22 00:00:00','YYYY/MM/DD HH24:MI:SS') as dt
from dual
Fiddle - http://sqlfiddle.com/#!4/6a3a6/1/0
As an FYI, the Oracle DATE data type does include the time component (just not down to fractional seconds, as is the case with the TIMESTAMP data type).
If you are converting values and want to bring all the time values to zero you can use the trunc function like this (which changes 12:07:00 to 00:00:00):
select trunc(to_date('2015/01/22 12:07:00','YYYY/MM/DD HH24:MI:SS'),'DD') as dt_with_time_zerod
from dual
Fiddle - http://sqlfiddle.com/#!4/6a3a6/2/0
If the source is itself a date and you want to convert the date to a string in the Oracle default date format ('DD-MON-RR') you can achieve that by running:
select to_char(trunc(to_date('2015/01/22 12:07:00','YYYY/MM/DD HH24:MI:SS'),'DD'),'DD-MON-RR') as dt_with_time_zerod
from dual
Fiddle - http://sqlfiddle.com/#!4/6a3a6/3/0
If it's a date field, to_char without a mask will give you what you say you want.
actually i want oracle standard date format without time stamp for this date '2015/01/22 00:00:00'
I'm not sure what you mean by "Oracle standard date format." The format in which a date would appear would be based on your NLS settings (in particular, NLS_DATE_FORMAT). If you are just trying to format this string representing a date, then you might want something like the following:
SELECT TO_CHAR(TO_DATE('2015/01/22 00:00:00','YYYY/MM/DD HH:MI:SS'), 'YYYY/MM/DD')
FROM dual;
That is, you have the TO_CHAR() and TO_DATE() functions in the wrong order, and an incomplete date mask for the call to TO_DATE().
Try using date literals with the standard ISO 8601 format.
date '2015-01-22'
I suggest you not to give hour-minute-second if you do not want to show the time.
This is my simplest answer :
SELECT TO_DATE('2015/01/22','YYYY/MM/DD') FROM dual

EXTRACT the date and time - (Teradata)

I am trying to extract the date and time from a field in Teradata.
The field in question is:
VwNIMEventFct.EVENT_GMT_TIMESTAMP
Here is what the data look like:
01/02/2012 12:18:59.306000
I'd like the date and time only.
I have tried using EXTRACT(Date, EXTRACT(DAY_HOUR and a few others with no success.
DATE_FORMAT() does not appear to work since I'm on Teradata.
How would I select the date and time from VwNIMEventFct.EVENT_GMT_TIMESTAMP?
If the datatype of EVENT_GMT_TIMESTAMP is a TIMESTAMP, it's simple Standard SQL:
CAST(EVENT_GMT_TIMESTAMP AS DATE)
CAST(EVENT_GMT_TIMESTAMP AS TIME)
If it's a CHAR you need to apply a FORMAT, too:
CAST(CAST(EVENT_GMT_TIMESTAMP AS TIMESTAMP FORMAT 'dd/mm/yyyyBhh:mi:SS.s(6)') AS DATE)
CAST(CAST(EVENT_GMT_TIMESTAMP AS TIMESTAMP FORMAT 'dd/mm/yyyyBhh:mi:SS.s(6)') AS TIME)
Edit:
For simply changing the display format you need to add a FORMAT and a CAST to a string:
CAST(CAST(EVENT_GMT_TIMESTAMP AS FORMAT 'YYYYMMDDHHMI') AS CHAR(12))
or
CAST(CAST(EVENT_GMT_TIMESTAMP AS FORMAT 'YYYYMMDDHHMISS') AS CHAR(14))
If you don't care about display, just want to truncate the seconds:
EVENT_GMT_TIMESTAMP - (EXTRACT(SECOND FROM EVENT_GMT_TIMESTAMP) * INTERVAL '1.000000' SECOND)
Working with timestamps is a bit tricky :-)
I know this is an old topic, but I've struggled with this too. Try:
CAST(EVENT_GMT_TIMESTAMP AS TIMESTAMP(0))
The result will be
01/02/2012 12:18:59
The datatype will still be timestamp, but it will just be the date and time with no microseconds (looks just like a datetime object in Microsoft SQL).