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How to store only time; not date and time?
(5 answers)
Closed 8 years ago.
I would like to insert data to db as time, not date. If I use to_date('2012-08-31 07:39:33', 'YYYY-MM-DD HH24:MI:SS') it adds date too.
If I use to_date('09:34:00', 'HH24:MI:SS') it adds year, month, day as well, from nowhere :|
Later I need to get rows where time is between x and y, not taking in account the year, month or day. How do I do that?
thanks
As an alternative to the date solution Dave shows, you could use an interval data type for the column:
create table t42(id number, t interval day to second);
insert into t42 (id, t) values(123, to_dsinterval('0 07:39:33'));
insert into t42 (id, t) values(456, to_dsinterval('0 09:34:00'));
select id
from t42
where t between to_dsinterval('0 07:00:00') and to_dsinterval('0 07:59:59');
ID
----------
123
Displaying intervals is a little awkward as they don't have format models, but see this question for some ideas if needed. If you only use them for filtering then that may not be an issue at all.
A DATE type always includes the date component.
One option is to continue using DATE and write your code to ignore the date component. In order to make queries on the time efficient, you might want to create a function-based index on something like TO_CHAR( date_field, 'HH24:MI:SS' ) and use that expression in your queries.
Alternatively, you could use a NUMBER field to store the number of seconds since midnight, and write your queries in terms of that.
you can use number column type and insert value as
INSERT INTO table_name (nTime)
VALUES (date - trunc(date));
and then select values
select *
from table_name t
where t.nTime between (10 / 24 + 15 / 24 / 60) and (12 / 24 + 30 / 24 / 60) --between 10:15 and 12:30
Related
In Oracle SQL Developer, I have a table called t1 who have two columns col1 defined as NUMBER(19,0) and col2 defined as TIMESTAMP(3).
I have these rows
col1 col2
1 03/01/22 12:00:00,000000000
2 03/01/22 13:00:00,000000000
3 26/11/21 10:27:11,750000000
4 26/11/21 10:27:59,606000000
5 16/12/21 11:47:04,105000000
6 16/12/21 12:29:27,101000000
My sysdate looks like this:
select sysdate from dual;
SYSDATE
03/03/22
I want to create a stored procedure (SP) which will delete rows older than 2 months and displayed message n rows are deleted
But when i execute this statement
select * from t1 where to_date(TRUNC(col2), 'DD/MM/YY') < add_months(sysdate, -2);
I don't get the first 2 rows of my t1 table. I get more than 2 rows
1 03/01/22 12:00:00,000000000
2 03/01/22 13:00:00,000000000
How can i get these rows and deleted it please ?
In Oracle, a DATE data type is a binary data type consisting of 7 bytes (century, year-of-century, month, day, hour, minute and second). It ALWAYS has all of those components and it is NEVER stored with a particular formatting (such as DD/MM/RR).
Your client application (i.e. SQL Developer) may choose to DISPLAY the binary DATE value in a human readable manner by formatting it as DD/MM/RR but that is a function of the client application you are using and not the database.
When you show the entire value:
SELECT TO_CHAR(ADD_MONTHS(sysdate, -2), 'YYYY-MM-DD HH24:MI:SS') AS dt FROM DUAL;
Then it outputs (depending on time zone):
DT
2022-01-03 10:11:28
If you compare that to your values then you can see that 2022-01-03 12:00:00 is not "more than 2 months ago" so it will not be matched.
What you appear to want is not "more than 2 months ago" but "equal to or more than 2 months, ignoring the time component, ago"; which you can get using:
SELECT *
FROM t1
WHERE col2 < add_months(TRUNC(sysdate), -2) + INTERVAL '1' DAY;
or
SELECT *
FROM t1
WHERE TRUNC(col2) <= add_months(TRUNC(sysdate), -2);
(Note: the first query would use an index on col2 but the second query would not; it would require a function-based index on TRUNC(col2) instead.)
Also, don't use TO_DATE on a column that is already a DATE or TIMESTAMP data type. TO_DATE takes a string as the first argument and not a DATE or TIMESTAMP so Oracle will perform an implicit conversion using TO_CHAR and if the format models do not match then you will introduce errors (and since any user can set their own date format in their session parameters at any time then you may get errors for one user that are not present for other users and is very hard to debug).
db<>fiddle here
Perhaps just:
select *
from t1
where col2 < add_months(sysdate, -2);
I am trying to separate the time and date in one column to be independent off each other. I am new at writing scripts
this is my query:
select
*
from
[tablename]
where
to_date([column_name]) in ( '15-Jun-2021', '16-Jun-2021' )
and
to_char([column_name],'dd-Mon-yyyy HH:MM:ss') < '15-Jun-2021 19:54:30'
The way you put it, it would be
select *
from your_table
where date_column >= date '2021-06-15'
and date_column < to_date('15.06.2021 19:54:30', 'dd.mm.yyyy hh24:mi:ss')
because
date_column should be of date datatype. If it isn't, you'll have problems of many kinds in the future. Therefore,
don't to_date it, it is already a date
don't to_char it either, because you'd be comparing strings and get unexpected result. Use that function when you want to nicely display the result
the second condition you wrote makes the first one questionable. If date_column is less than value you wrote, then you can omit date '2021-06-16' from the first condition because you won't get any rows for that date anyway
date literal (date '2021-06-15') sets time to midnight, so condition I wrote should return rows you want
SQL> select date '2021-06-15' first,
2 to_date('15.06.2021 19:54:30', 'dd.mm.yyyy hh24:mi:ss') second
3 from dual;
FIRST SECOND
------------------- -------------------
15.06.2021 00:00:00 15.06.2021 19:54:30
SQL>
I want to find the day between two timestamps. The query must return an Integer value.
I have a column value and a fixed date (like t.movementdate and '2014-07-23 00:00:00.0').
You can try this
CREATE TABLE t (movementdate TIMESTAMP);
INSERT INTO t VALUES (TIMESTAMP '1014-07-21 03:23:02.0');
INSERT INTO t VALUES (TIMESTAMP '2014-07-22 10:54:02.0');
select actual_diff,extract (day from actual_diff)+
extract (hour from actual_diff)/24
+extract (minute from actual_diff)/(60*24)
+extract (second from actual_diff)/(60*60*24)
diff_in_days
from (select systimestamp- movementdate as actual_diff from t);
Good question. IMHO Oracle's timestamp arithmetic is quite ugly. If you need the difference in days, you can convert the TIMESTAMP to a DATE. Depending on your needs, you can also drop the minutes and hours with TRUNC. Once you have DATEs, you can simply substract them to get the difference in days:
CREATE TABLE t (movementdate TIMESTAMP);
INSERT INTO t VALUES (TIMESTAMP '2014-07-21 03:23:02.0');
INSERT INTO t VALUES (TIMESTAMP '2014-07-22 10:54:02.0');
SELECT t.movementdate, DATE '2014-07-23' - trunc(t.movementdate) as daydiff FROM t;
MOVEMENTDATE DAYDIFF
---------------------------- ----
21.07.2014 03:23:02,000000000 2
22.07.2014 10:54:02,000000000 1
I have a table A which contains a Date type attribute. I want to write a query to select the date in another table B with value one month after the value in A.Any one know how to do it in oracle?
uhm... This was the first hit on google:
http://psoug.org/reference/date_func.html
It seems you're looking for the "add_months" function.
You need to use the ADD_MONTHS function in Oracle.
http://www.techonthenet.com/oracle/functions/add_months.php
Additional info: If you want to use this function with today's date you can use ADD_MONTHS(SYSDATE, 1) to get one month from now.
The question is to select a date_field from table b where date_field of table b is one month ahead of a date_field in table a.
An additional requirement must be taken into consideration which is currently unspecified in the question. Are we interested in whole months (days of month not taken into consideration) or do we want to include the days which might disqualify dates that are one month ahead but only by a couple of days (example: a=2011-04-30 and b=2011-05-01, b is 1 month ahead but only by 1 day).
In the first case, we must truncate both dates to their year and month values:
SELECT TRUNC( TO_DATE('2011-04-22','yyyy-mm-dd'), 'mm') as trunc_date
FROM dual;
gives:
trunc_date
----------
2011-04-01
In the second case we don't have to modify the dates.
At least two approaches can be used to solve the initial problem:
First one revolves around adding one month to the date_field in table a and finding a row in table b with a matching date.
SELECT b.date_field
FROM tab_a as a
,tab_b as b
WHERE ADD_MONTHS( TRUNC( a.date_field, 'mm' ), 1) = TRUNC( b.date_field, 'mm' )
;
Note the truncated dates. Leaving this out will require a perfect day to day match between dates.
The second approaches is based on calculating the difference in months between two dates and picking a calculation that gives a 1 month difference.
SELECT b.date_field
FROM tab_a as a
,tab_b as b
WHERE months_between( TRUNC( b.date_field, 'mm') , TRUNC(a.date_field, 'mm') ) = 1
The order of the fields in months_between is important here. In the provided example:
for b.date_field one month ahead of a.date_field the value is 1
for b.date_field one month before a.date_field the value is -1 (negative one)
Reversing the order will also reverse the results.
Hope this answers your question.
I need to add 30 minutes to values in a Oracle date column. I do this in my SELECT statement by specifying
to_char(date_and_time + (.000694 * 31)
which works fine most of the time. But not when the time is on the AM/PM border. For example, adding 30 minutes to 12:30 [which is PM] returns 1:00 which is AM. The answer I expect is 13:00. What's the correct way to do this?
In addition to being able to add a number of days to a date, you can use interval data types assuming you are on Oracle 9i or later, which can be somewhat easier to read,
SQL> ed
Wrote file afiedt.buf
SELECT sysdate, sysdate + interval '30' minute FROM dual
SQL> /
SYSDATE SYSDATE+INTERVAL'30'
-------------------- --------------------
02-NOV-2008 16:21:40 02-NOV-2008 16:51:40
All of the other answers are basically right but I don't think anyone's directly answered your original question.
Assuming that "date_and_time" in your example is a column with type DATE or TIMESTAMP, I think you just need to change this:
to_char(date_and_time + (.000694 * 31))
to this:
to_char(date_and_time + (.000694 * 31), 'DD-MON-YYYY HH24:MI')
It sounds like your default date format uses the "HH" code for the hour, not "HH24".
Also, I think your constant term is both confusing and imprecise. I guess what you did is calculate that (.000694) is about the value of a minute, and you are multiplying it by the number of minutes you want to add (31 in the example, although you said 30 in the text).
I would also start with a day and divide it into the units you want within your code. In this case, (1/48) would be 30 minutes; or if you wanted to break it up for clarity, you could write ( (1/24) * (1/2) ).
This would avoid rounding errors (except for those inherent in floating point which should be meaningless here) and is clearer, at least to me.
UPDATE "TABLE"
SET DATE_FIELD = CURRENT_TIMESTAMP + interval '48' minute
WHERE (...)
Where interval is one of
YEAR
MONTH
DAY
HOUR
MINUTE
SECOND
from http://www.orafaq.com/faq/how_does_one_add_a_day_hour_minute_second_to_a_date_value
The SYSDATE pseudo-column shows the current system date and time. Adding 1 to SYSDATE will advance the date by 1 day. Use fractions to add hours, minutes or seconds to the date
SQL> select sysdate, sysdate+1/24, sysdate +1/1440, sysdate + 1/86400 from dual;
SYSDATE SYSDATE+1/24 SYSDATE+1/1440 SYSDATE+1/86400
-------------------- -------------------- -------------------- --------------------
03-Jul-2002 08:32:12 03-Jul-2002 09:32:12 03-Jul-2002 08:33:12 03-Jul-2002 08:32:13
I prefer using an interval literal for this, because interval '30' minute or interval '5' second is a lot easier to read then 30 / (24 * 60) or 5 / (24 * 60 * 69)
e.g.
some_date + interval '2' hour
some_date + interval '30' minute
some_date + interval '5' second
some_date + interval '2' day
You can also combine several units into one expression:
some_date + interval '2 3:06' day to minute
Adds 2 days, 3 hours and 6 minutes to the date value
The above is also standard SQL and also works in several other DBMS.
More details in the manual: https://docs.oracle.com/database/121/SQLRF/sql_elements003.htm#SQLRF00221
If the data type of the field is date or timestamp, Oracle should always give the correct result if you add the correct number given in number of days (or a the correct fraction of a day in your case). So if you are trying to bump the value in 30 minutes, you should use :
select field + 0.5/24 from table;
Based on the information you provided, I believe this is what you tried to do and I am quite sure it works.
Can we not use this
SELECT date_and_time + INTERVAL '20:00' MINUTE TO SECOND FROM dual;
I am new to this domain.
like that very easily
i added 10 minutes to system date and always in preference use the Db server functions not custom one .
select to_char(sysdate + NUMTODSINTERVAL(10,'MINUTE'),'DD/MM/YYYY HH24:MI:SS') from dual;
Be sure that Oracle understands that the starting time is PM, and to specify the HH24 format mask for the final output.
SELECT to_char((to_date('12:40 PM', 'HH:MI AM') + (1/24/60) * 30), 'HH24:MI') as time
FROM dual
TIME
---------
13:10
Note: the 'AM' in the HH:MI is just the placeholder for the AM/PM meridian indicator. Could be also 'PM'
Oracle now has new built in functions to do this:
select systimestamp START_TIME, systimestamp + NUMTODSINTERVAL(30, 'minute') end_time from dual
Based on what you're asking for, you want the HH24:MI format for to_char.
To edit Date in oracle you can try
select to_char(<columnName> + 5 / 24 + 30 / (24 * 60),
'DD/MM/RRRR hh:mi AM') AS <logicalName> from <tableName>
SELECT to_char(sysdate + (1/24/60) * 30, 'dd/mm/yy HH24:MI am') from dual;
simply you can use this with various date format....