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Dates, times, and timestamps in Teradata (V2R5.1) 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 SQL Assistant (Queryman), be sure to uncheck the option box "Allow use of ODBC SQL Extensions in queries". Changing the Date Time Format in your ODBC data source to AAA or IIA may also help.

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, MP-RAS on Intel doesn't keep track of anything beyond 2 decimal places (hundredths of seconds), so in many cases it is useless to define columns with more than 2 decimal places. If, however, you are loading data from another server, that data could have digits down to the microsecond.

The formats for all columns are described in the SQL Reference: Data Types and Literals;, Chapter 8. In V2R5, you can specify many different formats for time and timestamp fields.

Dates

Literals

Date literals can be specified in many ways:

-- in the same format as the column they are being compared to

where infamy_date = '12/07/1941'

-- with an explicit format

where infamy_date = '07Dec1941' (date, format 'DDMMMYYYY')

-- in ANSI standard form (must be preceded by the keyword DATE)

where infamy_date = date '1941-12-07'

-- in numeric form (not recommended)

where infamy_date = 411207

Observe in the examples above that the word DATE must be used to specify the data type. To get the system date, therefore, CURRENT_DATE is a better choice than DATE.

Number of days between two dates

datecol - datecol will return the number of days between two dates.

select date '2003-08-15' - date '2003-01-01';

(2003-08-15-2003-01-01)
-----------------------
                    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)
-------------------
         2005-10-19

select add_months(current_date, -2);

ADD_MONTHS(Date, -2)
--------------------
          2005-05-19

/*** Last day of the month is still the last day ***/
select add_months(date '2002-01-31', 1);

ADD_MONTHS(2002-01-31, 1)
-------------------------
               2002-02-28

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 '2003-05-01';

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.

Times

Literals

The easiest way to specify a time literal is in ANSI form with a 24-hour clock:

time '15:09:17'

But in V2R5 you can use another format as long as you tell Teradata what it is:

'08-09-17PM' (time, format 'HH-MI-SST')

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.

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
--------------------------------
                 10:39:23.000000

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
-----------
      38363

/*** 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
-----------
      38363

Timestamps

Literals

The easiest way to specify a timestamp literal is in ANSI form:

timestamp '1994-12-25 23:46:29'

But in V2R5 you can use another format as long as you tell Teradata what it is:

'12/25/1994 11:46:29PM' (timestamp, format 'MM/DD/YYYYBHH:MI:SST')

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
--------
04/07/27

select cast(ts01 as time(6)) from dttest;

           ts01
---------------
10:24:37.739920

Combining date and time to make a timestamp

Thanks to Dieter Noeth for this tip. The most efficient way to combine a date and a time is to cast the date to a timestamp and add the time. But you can't simply add a time, because TIME is a point in time and not a duration. So you have to transform it into an interval:

cast(Date_column as TimeStamp(6))
+ ((Time_column - time '00:00:00') hour to second(6))

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 ts02 - ts01 day(4) to second(6)
   from dttest;

(ts02 - ts01) DAY TO SECOND
---------------------------
        371 05:26:35.639649

select ts01 - ts02 day(4) to second(6)
   from dttest;

(ts01 - ts02) DAY TO SECOND
---------------------------
       -371 05:26:35.639649

You can also convert this interval to seconds or minutes like so:

/*** Difference in seconds ***/
select (ts02 - ts01 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
---------------------   -----------------
  371 05:26:35.639649     32073995.639649

/*** Difference in minutes ***/
select (ts02 - ts01 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;
 *** Query completed. One row found. 2 columns returned.
 *** Total elapsed time was 1 second.

     tsdiff      min_diff
-----------   -----------
  371 05:26        534566

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(ts02 as date) - cast(ts01 as date)
   from dttest;

(ts02-ts01)
-----------
        371  -- [type of this result is INTEGER]

sel ts02 - ts01 day(4)
   from dttest;

(ts02 - ts01) DAY
-----------------
              371  -- [type of this result is INTERVAL DAY]

Other operations on timestamps

References

Teradata Database SQL Reference: Data Types and Literals,
    B035-1143-083A (Nov. 2003).

Information Technology — Database Languages — SQL-Part 2: Foundation,
    ISO/IEC 9075-2:2003, esp. Subclause 4.6.