I'm trying to subtract date from Oracle so it even effect the day as well. For example, if
the timestamp is 01/June/2015 00 hours and if I subtract 2 hours, I want to be able to go to to 31/May/2014 22 hours.
I tried
to_char(sysdate-(2/11), 'MM-DD-YYYY HH24')
but it only subtracts the hour; it does not touch the day itself.
Others have commented on the (incorrect) use of 2/11 to specify the desired interval.
I personally however prefer writing things like that using ANSI interval literals which makes reading the query much easier:
sysdate - interval '2' hour
It also has the advantage of being portable, many DBMS support this. Plus I don't have to fire up a calculator to find out how many hours the expression means - I'm pretty bad with mental arithmetics ;)
Try this:
SELECT to_char(sysdate - (2 / 24), 'MM-DD-YYYY HH24') FROM DUAL
To test it using a new date instance:
SELECT to_char(TO_DATE('11/06/2015 00:00','dd/mm/yyyy HH24:MI') - (2 / 24), 'MM-DD-YYYY HH24:MI') FROM DUAL
Output is: 06-10-2015 22:00, which is the previous day.
sysdate-(2/11)
A day consists of 24 hours. So, to subtract 2 hours from a day you need to divide it by 24:
DATE_value - 2/24
Using interval for the same:
DATE_value - interval '2' hour
date - n will subtract n days form given date. In order to subtract hrs you need to convert it into day buy dividing it with 24. In your case it should be to_char(sysdate - (2 + 2/24), 'MM-DD-YYYY HH24'). This will subract 2 days and 2 hrs from sysdate.
you should divide hours by 24 not 11
like this:
select to_char(sysdate - 2/24, 'dd-mon-yyyy HH24') from dual
Related
i have the following problem,
i am trying to find the difference of two dates in hours and minutes, for example
select to_date('13.05.2021 09:30','DD.MM.YYYY HH24:MI')
- to_date('13.05.2021 08:15','DD.MM.YYYY HH24:MI') from dual;
obiously it returns the difference in days, so the output will be 0,
i am expecting somthing like 01:15
Depending on the data type you need for your result...
If an interval day to second will work, then you can do this:
select (date2 - date1) * interval '1' day from dual;
For example:
select (to_date('13.05.2021 09:30','DD.MM.YYYY HH24:MI')
- to_date('13.05.2021 08:15','DD.MM.YYYY HH24:MI'))
* interval '1' day as diff
from dual;
DIFF
-------------------
+00 01:15:00.000000
Give this a try; if you need a different data type for the result, let us know. Note that, stupidly, Oracle doesn't support aggregate functions for intervals; so if you need a sum of such differences, you should apply the aggregation first, and only use this trick to convert to an interval as the last step.
In addition to #mathguy answer in case you need a specific output format:
TO_CHAR does not work with INTERVAL values. Either use EXTRACT
EXTRACT(HOUR FROM ((date2 - date1) DAY TO SECOND)) ||':'|| EXTRACT(MINUTE FROM ((date2 - date1) DAY TO SECOND))
or REGEXP_SUBSTR
REGEXP_SUBSTR((date2 - date1) DAY TO SECOND, ' \d{2}:\d{2}')
You may need some fine-tuning for proper output.
Is it possible to do something like this?
SELECT
TO_CHAR(SYSDATE-1, 'MM-DD-YYYY 08:00:00') "Yesterday",
I'd like to get the sysdate from yesterday at 8am
You would write this as:
trunc(sysdate - 1) + interval '8' hour
Or:
trunc(sysdate) - interval '16' hour
Or you can do date arithmetics with integer values rather than intervals:
trunc(sysdate) - 16/24
Truncate the sysdate. That will remove the time portion of the date.
Then add 8/24 (8 hours of the 24), 8/24 equals to 1/3 - 8 hours is a third of a day.
Then display this date value in the format DD-MON-YYYY HH24:MI:SS.
This all adds up to:
SELECT TO_CHAR(TRUNC(SYSDATE) - 1 + 1/3,'DD-MON-YYYY HH24:MI:SS') FROM DUAL;
Very quick question, if I have a field, for example closed_date, which has a datetime, i.e.
01-JAN-19 09.00.00.000000000
And I would like to only look at cases closed between 09:00am and 10:00am each day, whats the syntax? How do I trunc the field just to time and only pull through cases which were closed each day between the 9am and 10am?
Many thanks
You can look at the hour portion alone; as you have a timestamp you can extract the hour number:
where extract(hour from closed_date) = 9
That will find rows there time is on or after 09:00:00, and before (not including) 10:00:00.
Usually when looking at data in a time period like 'between 9 and 10' you don't actually want to include the upper limit, because if you were also looking at data 'between 10 and 11' then data at exactly 10:00:00 would be included in both, which probably isn't what was intended. So it's common for date/time comparisons to use >= and < instead of between; you could also do this with a string comparison which you might consider clearer:
where to_char(closed_date, 'HH24:MI:SS') >= '09:00:00'
and to_char(closed_date, 'HH24:MI:SS') < '10:00:00'
or slightly more simply
where to_char(closed_date, 'HH24') >= '09'
and to_char(closed_date, 'HH24') < '10'
which in this case, as it's a single hour, is the same as:
where to_char(closed_date, 'HH24') = '09'
but then as you are only looking at the hour part anyway, extracting that as a number simplifies it even more (IMO).
You seem to be looking for a condition to filter a timestamp column based on its hour part. There are various ways to extract date parts from a timestamp, here is a solution that uses TO_CHAR :
TO_CHAR(closed_date, 'hh24') = '09'
This will match on timestamps whose time is higher than or equal to 9 AM and strictly smaller than 10 AM.
You can use the extract function from oracle
SELECT EXTRACT(HOUR FROM timestamp_col_value) AS CURRENT_HOUR
FROM DUAL;
Note that the extract function only works in certain combinations of MONTH/DAY/HOUR and date types. See here for more details.
If you want to extract a hour from a datetime, you need to convert it to a timestamp first, i.e.
SELECT EXTRACT(HOUR FROM CAST(SYSDATE AS TIMESTAMP)) AS CURRENT_HOUR
FROM DUAL;
select regexp_Replace('01-JAN-19 09.00.00.000000000','\d+-\w+-\d+ ') from dual
But if You want to search between hours, try EXTRACT, f.e.:
SELECT *
FROM your_table
WHERE to_char(your_column,
'yyyy-mm-dd hh24:mi:ss.ff a.m.',
'nls_date_language=american') in (9,10)
Updated answer. without nls_Date_language parameter, hour 9 o' clock could be also 9.pm and that is 21.
Need your help to conclude the query to fetch last date time of the sysdate month.
select to_char(last_day(sysdate),'DD-Mon-YYYY HH24:MI:SS') from dual
it gives last date as expected, but I need time as 23:59:00 which is not possible thru above query.
You could use TRUNC on next day i.e. SYSDATE + 1, and then subtract 60 seconds i.e. 60/86400 to get the desired output.
SQL> SELECT to_char((trunc(last_day(sysdate)) +1) - 60/86400,'DD-Mon-YYYY HH24:MI:SS') dt
2 FROM dual;
DT
--------------------
29-Feb-2016 23:59:00
SQL>
You could also use interval '1' minute or interval '60' second instead of 60/86400.
If you just want it for display for some reason you can hard-code the time into the format mask:
select to_char(last_day(sysdate), 'DD-Mon-YYYY "23:59:00"') from dual;
But you probably really want it as a date object, in which case you can add 23 hours and 59 minutes to the truncated (midnight) date, wchi is 1439 of the 1440 minutes in a day:
select to_char(trunc(last_day(sysdate)) + 1439/1440, 'DD-Mon-YYYY HH24:MI:SS')
from dual;
Or you can go to the next day and remove a minute, either with fractional days or with intervals:
select to_char(trunc(last_day(sysdate)) + interval '1' day - interval '1' minute,
'DD-Mon-YYYY HH24:MI:SS') from dual;
Generally if you're working with time periods you want to include up to 23:59:59, which you can also do with any of those methods, but as Damien_The_Unbeliever said in a comment, it's easier to compare against the start of the next period (e.g. < add_months(trunc(sysdate, 'MM'), 1). It's easy to accidentally miss part of a day by not taking the time into account properly, particularly if you actually have a timestamp rather than a date.
By using normal minus '-' function between two timestamps, the answer given from oracle is incorrect.
This is what i want to do:
ALTER SESSION SET NLS_TIMESTAMP_TZ_FORMAT='DD-MON-RR HH24:MI TZR';
Created table:
CREATE TABLE TEST (
StartTime timestamp with time zone
,EndTime timestamp with time zone
,Science varchar2(7)
);
I create the column data type as timestamp with time zone. This is value I have inserted:
INSERT INTO TEST
VALUES('05-OCT-2013 01:00 +08:00'
,'05-OCT-2013 23:00 +06:00'
,'SCIENCE');
INSERT INTO TEST
VALUES('05-OCT-2013 12:00 +08:00'
,'05-OCT-2013 15:00 -12:00'
,'Maths');
Attempted for rounding time:
CREATE VIEW TESTRECRDS AS
SELECT (Extract(hour FROM(ENDTIME- STARTTIME)) || 'Hours' ||
Extract(minute FROM(ENDTIME- STARTTIME))>=60 Then (Extract(hour FROM(ENDTIME- STARTTIME)) + Extract(minute FROM(ENDTIME- STARTTIME))/60 ELSE 0 END || 'Minutes' AS DURATION,
Science
FROM Test;
Now i have two questions regarding on the calculation and rounding off the minutes to nearest hours.
First let's say the endtime is 1535 +0600 and starttime is 01:50 +0800
So when i deduct endtime - starttime:
the formula should be:
2135 - 0950 = 2085 - 0950
= 1135
But if i use my successful attempt answer to calculate, it is not the correct exact answer. The oracle answer would be 15 hours 45 minutes.
In your last CREATE VIEW statement you try to multiply text, which cannot work:
SELECT To_Char(STARTTIME - ENDTIME, 'HH24:MI TZR')*24 AS DURATION
*24 is operating on the text to_char() returns.
You have to multiply the interval before converting to text.
You define the column Science varchar2(6), then you insert 'SCIENCE', a 7-letter word?
I also fixed a syntax error in your INSERT statement: missing '.
About your comment:
"I would like to insert timestamp with timezone during creation of my tables. Can DATE data type do that too?
Read about data types in the manual.
The data type date does not include time zone information.
If by "timezone difference" you mean the difference between the timezone modifiers, use this to calculate:
SELECT EXTRACT(timezone_hour FROM STARTTIME) AS tz_modifier FROM tbl
Keywords here are timezone_hour and is timezone_minute. Read more in the manual.
But be aware that these numbers depend on the daylight saving hours and such shenanigans. Very uncertain territory!
Get it in pretty format - example:
SELECT to_char((EXTRACT (timezone_hour FROM STARTTIME) * 60
+ EXTRACT (timezone_minutes FROM STARTTIME))
* interval '1 min', 'HH:MI')
In PostgreSQL you would have the simpler EXTRACT (timezone FROM STARTTIME), but I don't think Oracle supports that. Can't test now.
Here is a simple demo how you could round minutes to hours:
SELECT EXTRACT(hour FROM (ENDTIME - STARTTIME))
+ CASE WHEN EXTRACT(minute FROM (ENDTIME - STARTTIME)) >= 30 THEN 1 ELSE 0 END
FROM Test;
I'm not sure what number you're trying to calculate, but when you subtract two dates in Oracle, you get the difference between the dates in units of days, not a DATE datatype
SELECT TO_DATE('2011-01-01 09:00', 'yyyy-mm-dd hh24:mi') -
TO_DATE('2011-01-01 08:00', 'yyyy-mm-dd hh24:mi') AS diff
FROM dual
DIFF
----------
.041666667
In this case 8am and 9am are 0.41667 days apart. This is not a date object, this is a scalar number, so formatting it as HH24:MI doesn't make any sense.
To round you will need to do a bit of more math. Try something like:
TO_DATE(ROUND((ENDTIME - STARTTIME) * 96) / 96, 'HH24:MI')
The difference between dates is in days. Multiplying by 96 changes the measure to quarter hours. Round, then convert back to days, and format. It might be better to use a numeric format want to format, in which case you would divide by 4 instead of 96.
Timezone is not particularly relevant to a time difference. You will have to adjust the difference from UTC to that timezone to get the right result with Timezone included.