is it possibile rounding time to nearest 5 minutes down?
Something like that:
select datetime('now')
return 2017-05-09 07:34:26
Select stuffcodesql
return 2017-05-09 05:00:00
Get current datetime as unix timestamp
strftime('%s', 'now')
Subtract the remainder of the division by 300 (300 seconds or 5 minutes)
strftime('%s', 'now') % 300
Convert to local time format
select datetime(strftime('%s', 'now') - strftime('%s', 'now') % 300, 'unixepoch', 'localtime') as [5min];
Convert the timestamp into a number of days;
multiply by 288 (the number of 5-minute intervals per day);
round down to the nearest integer;
divide by 288 again;
convert back into a string:
SELECT datetime(CAST(julianday('now') * 288 AS INTEGER) / 288.0);
Related
I got two different datetimes: 2020-05-18 12:30:01 and 2020-05-17 13:00:00.
I want to show the time difference between them in the format HH:MM:SS, which is 23:30:01.
If the difference is higher than 24 hours, let's say 28 hours, 12 minutes and 45 seconds, it would show like 28:12:45.
How can I do that in SQLite?
SQLite supports a limited number of functions for datetime manipulation.
One of these functions is strftime(), and
strftime('%s', somedate)
returns the number of seconds from '1970-01-01' up to somedate.
With the use of this function, arithmetic calculations, string padding and concatenations you can get what you want like this:
CASE WHEN ((strftime('%s', date1) - strftime('%s', date2)) / 3600) < 10 THEN '0' ELSE '' END ||
((strftime('%s', date1) - strftime('%s', date2)) / 3600) || ':' ||
SUBSTR('0' || (((strftime('%s', date1) - strftime('%s', date2)) / 60) % 60), -2) || ':' ||
SUBSTR('0' || ((strftime('%s', date1) - strftime('%s', date2)) % 60), -2)
Replace date1 and date2 with your dates.
See a simplified demo.
SELECT time(
(
julianday('2020-05-18 12:30:01')-
julianday('2020-05-17 13:00:00')
)*60*60*24, 'unixepoch'
);
answsers the question when time difference is lower than 24h...
In my Teradata table, I have the epoch timestamps under the column dhTimestamp
dhTimestamp
1435308067705
1434965874565
1434763800794
1434775876034
1434765207057
How can I convert the epoch timestamp to Human Date/Time format on Teradata?
This is a SQL UDF for standard unixtime:
/**********
Converting Unix/POSIX time to a Timestamp
Unix time: Number of seconds since 1970-01-01 00:00:00 UTC not counting leap seconds (currently 24 in 2011)
Also working for negative numbers.
The maximum range of Timestamps is based on the range of INTEGERs:
1901-12-13 20:45:52 (-2147483648) to 2038-01-19 03:14:07 (2147483647)
Can be changed to use BIGINT instead of INTEGER
20101211 initial version - Dieter Noeth
**********/
REPLACE FUNCTION Epoch2Timestamp (UnixTime INT)
RETURNS TimeStamp(0)
LANGUAGE SQL
CONTAINS SQL
DETERMINISTIC
RETURNS NULL ON NULL INPUT
SQL SECURITY DEFINER
COLLATION INVOKER
INLINE TYPE 1
RETURN
CAST(DATE '1970-01-01' + (UnixTime / 86400) AS TIMESTAMP(0))
+ ((UnixTime MOD 86400) * INTERVAL '00:00:01' HOUR TO SECOND)
;
SELECT
Epoch2Timestamp(-2147483648)
,Epoch2Timestamp(0)
,Epoch2Timestamp(2147483647)
;
But your values seem to include milliseconds, this needs a modified calculation:
CAST(DATE '1970-01-01' + (UnixTime / 86400000) AS TIMESTAMP(3))
+ ((UnixTime / 1000.000 MOD 86400) * INTERVAL '00:00:01' HOUR TO SECOND)
Edit 2016-07-01:
There was an issue with dayight saving time (see this thread on Teradata's on DevEx), this should fix it:
-- Unix time to Timestamp WITH TIME ZONE (+00:00)
REPLACE FUNCTION UnixTime_to_TimeStamp_TZ (UnixTime INT)
RETURNS TIMESTAMP(0) WITH TIME ZONE
LANGUAGE SQL
CONTAINS SQL
DETERMINISTIC
SQL SECURITY DEFINER
COLLATION INVOKER
INLINE TYPE 1
RETURN
((CAST(DATE '1970-01-01' + (UnixTime / 86400) AS TIMESTAMP(0) AT 0)) AT 0)
+ ((UnixTime MOD 86400) * INTERVAL '00:00:01' HOUR TO SECOND);
-- Unixtime to Timestamp, implicit TIME ZONE of the local session
REPLACE FUNCTION UnixTime_to_TimeStamp (UnixTime INT)
RETURNS TIMESTAMP(0)
LANGUAGE SQL
CONTAINS SQL
DETERMINISTIC
SQL SECURITY DEFINER
COLLATION INVOKER
INLINE TYPE 1
RETURN
CAST(((CAST(DATE '1970-01-01' + (UnixTime / 86400) AS TIMESTAMP(0) AT 0)) AT 0)
+ ((UnixTime MOD 86400) * INTERVAL '00:00:01' HOUR TO SECOND) AS TIMESTAMP(0));
This is the most simplied way to convert EPOCH TO TERADATA LOCAL.
SELECT
dhTimestamp as unix_epoc_time ,
to_timestamp(unix_epoc_time) utc,
cast(cast(utc as char(19))||'+00:00' as timestamp(0) with time zone) AT LOCAL
If you have epoch with more than 10 digit , then chop the numbers after 10th digit, It will just work fine.
remember , Unix time is in UTC.
your system will treat this utc as local. So, let us make it understand that it is UTC by adding '+00:00' and then convert it to your LOCAL using AT LOCAL OR using either of these "America Central" , "America Eastern" , "America Mountain" etc..
I was able to convert epoch column to timestamp using below query..
SELECT CAST((date '1970-01-01' + CAST(epochtimecolumn/1000 AS INTEGER)/86400) AS TIMESTAMP(6)) + (CAST(epochtimecolumn/1000 AS INTEGER) MOD 86400) * INTERVAL '00:00:01' HOUR TO SECOND + (epcho_time_column_with_milliseconds MOD 1000) * INTERVAL '00:00:00.001' HOUR TO SECOND from table_name
Just replace epochtimecolumn with your column in the above query to run it in teradata.
Hope it helps!!
I am trying to convert 24 hour Time values to minutes
From:
(TIME)
17:00:00
16:55:00
17:30:00
To:
(NUM)
1060
993
1038
Currently I am multiplying the time values by 60 17*60, 16.55*60, 17.30*60
How would I accomplish this? am I doing it right? and what am I doing wrong?
SAS time values are stored as seconds. If it's truly a 'time' value, anyway, and not a character string. Thus, you can DIVIDE by 60, rather than multiplying.
data want;
input timeval TIME9.;
minutes=timeval/60;
format minutes BEST12.;
format timeval TIME9.;
put timeval= minutes=;
datalines;
17:00:00
16:55:00
17:30:00
;;;;
run;
If it's not stored as a time value (numeric) but as a string, you need to INPUT(timeval,TIME9.) in order to do that; so
minutes = input(timeval,TIME9.)/60;
would work.
No idea what this has to do with SQL, but in general, assuming that hours is 0-23 rather than 0-12 with an am/pm indicator, this pseudocode
( 60.0 * hours ) // convert 0-23 hours to minutes
+ minutes // minutes don't need conversion
+ ( seconds / 60.0 ) // convert 0-59 seconds to fractional minute
should give you the offset from start of day (midnight, 00:00:00) as a fractional number of minutes. If you've got an am/pm indicator, you need to factor that in, something like this:
( 60.0 * ( hours + ( isPM ? 12 : 0 ) ) ) // convert 0-23 hours to minutes
+ minutes // minutes don't need conversion
+ ( seconds / 60.0 ) // convert 0-59 seconds to fractional minutes
I want to make a function call that hase efect in SQLite like TIMEDIFF in MySQL.
I made this:
select strftime('%s','2012-01-01 12:00:00') - strftime('%s','2004-01-01 02:34:56')
but this is just the number of seconds. So how can i make a str like %Y-%m-%d %H:%M:%S where %H:%M:%S is the hours, minutes and seconds difference, and when it is bigger then 24 hours then %d will show how much dais it is and so on with %Y and %m.
You cannot represent a time difference with %Y-%m-%d ..., at least not as a date format. How would you express less than a day of difference? (0000-00-00 ... is not a valid date). Also, what would a month be? 30 days? 31 days? 23423432 days?
I suggest you keep your difference in seconds, and when presenting it you adapt it as necessary.
On the other hand, if you really want to do as you asked, here's one way:
sqlite> select datetime(strftime('%s','2012-01-01 12:00:00')
- strftime('%s','2004-01-01 02:34:56') - 62167305600, 'unixepoch');
0007-12-31 09:25:04
Even if I feel the downvote by the OP wasn't justified, I can't stop myself from explaining why what I mentioned above as clearly not a very good option returns "incorrect" results when the time difference is less than 1 day: the reason is implied in what I wrote above: there is no such date as 0000-00-00 ... so instead the datetime returned goes in negative territory: -001-12-31 ...
Here's a way to obtain 438:53:45, but it's quite involved:
earlier date: d1
later date: d2
select
cast(
(strftime('%s', d2) - strftime('%s', d1)) / 86400 * 24
+ cast(strftime("%H", time(strftime('%s', d2)
- strftime('%s', d1), 'unixepoch'))
as int)
as text)
|| ":"
|| substr(time(strftime('%s', d2) - strftime('%s', d1), 'unixepoch'), 4);
Example:
d1 = '2004-01-01 02:34:56'
d2 = '2012-01-01 12:00:00'
sqlite> select cast((strftime('%s','2012-01-01 12:00:00') - strftime('%s','2004-01-01 02:34:56')) / 86400 *24 + cast(strftime("%H", time(strftime('%s','2012-01-01 12:00:00') - strftime('%s','2004-01-01 02:34:56'), 'unixepoch')) as int) as text)
|| ":"
|| substr(time(strftime('%s','2012-01-01 12:00:00') - strftime('%s','2004-01-01 02:34:56'), 'unixepoch'), 4);
70137:25:04
I am trying to express the difference of two given dates in days, hours, and minutes (like 1 day, 6 hours, 17 minutes.) as SQLite query output. I have entryin and entryout as datetime fields in a SQLitedatabase. I tried all combinations of julianday and strftime but still running into rough weather.
I tried strftime('%d %H:%M', julianday(entryout)-julianday(entryin)). For a row the values are 2011-11-10 11:46, and 2011-11-09 09:00. but the output is 25 14:46 instead of 01 02:46.
Can some one help me with this, or point me correct logic for this? Thanks in advance.
You can try something like this:
SELECT
CAST((strftime('%s', '2011-11-10 11:46') - strftime('%s', '2011-11-09 09:00')) / (60 * 60 * 24) AS TEXT) || ' ' ||
CAST(((strftime('%s', '2011-11-10 11:46') - strftime('%s', '2011-11-09 09:00')) % (60 * 60 * 24)) / (60 * 60) AS TEXT) || ':' ||
CAST((((strftime('%s', '2011-11-10 11:46') - strftime('%s', '2011-11-09 09:00')) % (60 * 60 * 24)) % (60 * 60)) / 60 AS TEXT);