Dates, times, and timestamps in Teradata (V2R4.1) can be a little tricky. This document explains how to do various things that you
may want to do.
Preliminaries
The examples below use Teradata syntax, so they assume that you are running in BTEQ. If you are using Queryman, be sure to
uncheck the option box "Allow use of ODBC SQL Extensions in queries".
datecol means a column defined as DATE.
Defining columns and formats
Times and timestamps can be defined with any number of decimal places from 0 to 6 (time(0), timestamp(6), etc.). As it turns out,
however, the Teradata hardware doesn't keep track of anything beyond 2 decimal places (hundredths of seconds), so it is useless to
define columns with more than 2 decimal places. The formats for all columns are described in the SQL Reference, Volume
3, Chapter 8. For dates, various combinations of YYYY, MM, DD, and so on are recognized, because those elements can be
arranged in different ways. For instance, you can ask for a format of 'YYYYMMDD' or 'MMMbDD,bYYYY'. Time and timestamp fields have
default formats that cannot be changed, because the order of the elements is fixed: hours always appear first, then minutes, and so
on. Here are some guidelines:
Date Type 
Length after Formatting (characters) 
time(0)  8 
time(2)  11 
timestamp(0)  19 
timestamp(2)  22 
Dates
Number of days between two dates
datecol  datecol
will return the number of days between two dates.
select date '20030815'  date '20030101';
(2003081520030101)

226
Adding or subtracting months
Generally speaking, you should use the ADD_MONTHS function to add months to a date (or to subtract
months). Your project may require adding a number of days, but if calendar months are required,
ADD_MONTHS is the way to go.
select add_months(current_date, 3);
ADD_MONTHS(Date, 3)

20030722
select add_months(current_date, 2);
ADD_MONTHS(Date, 2)

20030222
/*** Last day of the month is still the last day ***/
select add_months(date '20020131', 1);
ADD_MONTHS(20020131, 1)

20020228
Computing the day of the week
Computing the day of the week for a given date is not easy in SQL. If you need a weekday, I recommend
that you look it up in the view sys_calendar.calendar (or join to it), thus:
select day_of_week
from sys_calendar.calendar
where calendar_date = date '20030501';
day_of_week

5 [i.e. Thursday]
Computing the first day of a month
select datecol  extract(day from datecol) + 1
This subtracts the number of days since the beginning of the month, taking you to "day 0", or the day
before the first of the month; then adds 1.
Computing the last day of a month
select add_months((datecol  extract(day from datecol)+1),1)1
Same idea, but this computes the first day of the following month and then subtracts 1.
Special calendars
Your business may have special requirements, such as a retail calendar that is always a multiple of
weeks. Such data must be stored in a table and joined to in your queries.
Times
Changing floats to times
Some Data Dictionary tables have time columns that are defined as FLOAT rather than TIME. Here's how to
convert them to TIMEs, believe it or not:
select cast(cast(cast(TimeFld as format '99:99:99.99')
as char(11)) as time(6))
from DBC.EventLog ...
Time differences
First we shall address the case where your time data is defined as a number (FLOAT or DECIMAL, perhaps)
in hhmmss form — e.g., 194328 for 7:43:28 p.m. If you have two such times and can be sure that they both
fall within the same day, and you want to compute the difference between them in seconds, you will have to
break them up into their parts like so:
select ((time02 / 10000) * 3600 +
(time02 / 100 MOD 100) * 60 +
(time02 MOD 100)) 
((time01 / 10000) * 3600 +
(time01 / 100 MOD 100) * 60 +
(time01 MOD 100)) as time_diff
from dttest2;
time_diff

6432. [in seconds  about 1.8 hours]
If the earlier time could fall on one day and the later time on the next day, you may have to add 86,400
(the number of seconds in one day) to the later time, like so:
select case
when time02 >= time01 then
((time02 / 10000) * 3600 +
(time02 / 100 MOD 100) * 60 +
(time02 MOD 100)) 
((time01 / 10000) * 3600 +
(time01 / 100 MOD 100) * 60 +
(time01 MOD 100))
else /*** Midnight has passed ***/
(((time02 / 10000) * 3600 +
(time02 / 100 MOD 100) * 60 +
(time02 MOD 100)) + 86400) 
((time01 / 10000) * 3600 +
(time01 / 100 MOD 100) * 60 +
(time01 MOD 100))
end as time_diff
from dttest2;
time_diff

18094.
Next we consider the case where your time data is defined as TIME(n). The usual way to take
the difference of two times would be as follows:
select time02  time01 hour(2) to second
from dttest3;
(time02  time01) HOUR TO SECOND

5:02:40.000000
The above result has a data type of INTERVAL. If, however, you want to compute the difference in
seconds, as above, you again have to split the times up:
select (extract(hour from time02) * 3600 +
extract(minute from time02) * 60 +
extract(second from time02)) 
(extract(hour from time01) * 3600 +
extract(minute from time01) * 60 +
extract(second from time01)) as time_diff
from dttest3;
time_diff

18160
/*** After midnight ... ***/
select case
when time02 >= time01 then
(extract(hour from time02) * 3600 +
extract(minute from time02) * 60 +
extract(second from time02)) 
(extract(hour from time01) * 3600 +
extract(minute from time01) * 60 +
extract(second from time01))
else
(extract(hour from time02) * 3600 +
extract(minute from time02) * 60 +
extract(second from time02) + 86400) 
(extract(hour from time01) * 3600 +
extract(minute from time01) * 60 +
extract(second from time01))
end as time_diff
from dttest3;
time_diff

61360
Timestamps
Extracting the date or time portion of a timestamp
Extract the date or time portion of a timestamp thus:
select cast(ts01 as date) from dttest;
ts01

03/08/15
select cast(ts01 as time(0)) from dttest;
ts01

03:04:05
Length of time between two timestamps
You can subtract one timestamp from another. The result will be an interval, and you must specify
a precision for the interval, like so:
select ts01  ts04 day(4) to second(0)
from dttest;
(ts01  ts04) DAY TO SECOND

226 02:02:02
select ts04  ts01 day(4) to second(0)
from dttest;
(ts04  ts01) DAY TO SECOND

226 02:02:02
You can also convert this interval to seconds or minutes like so:
/*** Difference in seconds ***/
select (ts01  ts04 day(4) to second) as tsdiff,
(extract(day from tsdiff) * 86400)
+ (extract(hour from tsdiff) * 3600)
+ (extract(minute from tsdiff) * 60)
+ extract(second from tsdiff) as sec_diff
from dttest;
tsdiff sec_diff
 
2 07:11:24.000000 198684.000000
/*** Difference in minutes ***/
select (ts01  ts04 day(4) to minute) as tsdiff,
(extract(day from tsdiff) * 1440)
+ (extract(hour from tsdiff) * 60)
+ extract(minute from tsdiff) as min_diff
from dttest;
tsdiff min_diff
 
2 07:11 3311
Number of days between two timestamps
If you just want the number of days between two timestamps and wish to ignore the time
portion, either of the following two techniques will work, but note the differences. The
first technique lops off the time portion of each timestamp, so it will be equivalent to
subtracting the two days; the result is an integer. The second will take the time portion
into account and return an interval, so it will not count periods of time less than 24 hours.
Thus, the result could be one less than with the first technique. If you use the second
technique, be sure to allow enough digits for DAY.
sel cast(ts01 as date)  cast(ts04 as date)
from dttest;
(ts01ts04)

226  type of this result is INTEGER
sel ts01  ts04 day(4)
from dttest;
(ts01  ts04) DAY

226  type of this result is INTERVAL DAY
Other operations on timestamps
Operand 1 
Operator 
Operand 2 
Result Type 
Timestamp 
+ or  
Interval 
Timestamp 
Interval 
+ 
Timestamp 
Timestamp 
Interval 
+ or  
Interval 
Interval 
Interval 
* or / 
Numeric 
Interval 
Numeric 
* 
Interval 
Interval 
References
Teradata RDBMS SQL Reference, Volume 3: Data Types and Literals,
B0351101061A (June 2001).
American National Standard for Information Technology — Database Languages — SQLPart 2: Foundation,
ANSI/ISO/IEC 90752:1999, esp. Subclause 4.7.
