I am trying to add a day with a specific date using add_months in oracle database.
I wrote this line:
SELECT ADD_MONTHS('01-JAN-2018', MONTHS_BETWEEN('02-JAN-2018', '01-JAN-2018')) FROM DUAL;
this returns:
01-JAN-18
Why doesn't it return 02-JAN-18?? Can I add one day to the date using this function?
Why doesn't it return 02-JAN-18??
According to MONTHS_BETWEEN documentation,
The MONTHS_BETWEEN function calculates the number of months between
two dates. When the two dates have the same day component or are both
the last day of the month, then the return value is a whole number.
Otherwise, the return value includes a fraction that considers the
difference in the days based on a 31-day month
So,
select MONTHS_BETWEEN('02-JAN-2018', '01-JAN-2018') FROM DUAL ;
yields
.0322580645161290322580645161290322580645
ADD_MONTHS returns the date date plus integer months.
So, .0322.. is considered as integer 0 and your query is equivalent to
SELECT ADD_MONTHS('01-JAN-2018', 0) FROM DUAL;
In order to add 1 months, simply take the difference of two dates.
SELECT ADD_MONTHS(DATE '2018-01-01', DATE '2018-01-02' - DATE '2018-01-01') FROM DUAL;
Or better, add an INTERVAL of 1 month
SELECT DATE '2018-01-01' + INTERVAL '1' MONTH FROM DUAL;
To answer your question, add 1 day, simply use
SELECT DATE '2018-01-01' + 1 FROM DUAL;
Related
I have the following query that gets the week of a date:
SELECT pdm.serie, rta.matricula_ant, TO_CHAR (fecha, 'ww') semana,
SUM (rta.kms_acumulados) kms,
COUNT
(DISTINCT (CASE
WHEN v.secuencia BETWEEN rta.sec_origen AND rta.sec_destino
THEN v.cod_inc
ELSE '0'
END
)
)
- 1 numincidencias
FROM (SELECT ms.tren, ms.fecha_origen_tren, ms.secuencia, ri.cod_inc
FROM r_incidencias ri, mer_sitra ms
WHERE ri.cod_serv = ms.tren
AND ri.fecha_origen_tren = ms.fecha_origen_tren
AND ri.cod_tipoin IN (SELECT cod_tipo_iincidencia
FROM v_tipos_incidencias
WHERE grupo = '45')
AND ri.punto_desde = ms.cod_estacion) v,
r_trenes_asignar rta,
r_maquinas rm,
planificador.pl_dh_material pdm
WHERE rta.fecha BETWEEN TO_DATE ('21/09/2018', 'dd/mm/yyyy') AND TO_DATE ('21/09/2018',
'dd/mm/yyyy'
)
AND rta.serie >= 4000
AND rta.matricula_ant IS NOT NULL
AND rm.matricula_maq = rta.matricula_ant
AND rm.cod_serie = pdm.id_material
AND rta.grafico BETWEEN pdm.desde AND pdm.hasta
AND v.tren(+) = rta.tren
AND v.fecha_origen_tren(+) = rta.fecha
GROUP BY pdm.serie, rta.matricula_ant, TO_CHAR (fecha, 'ww')
ORDER BY pdm.serie, rta.matricula_ant, TO_CHAR (fecha, 'ww')
For example week 1
I want to display
week 1 : 1 january - 7 january
How can I get this?
Oracle offers the TRUNC(datestamp, format) function to manipulate dates this way. You may use a variety of format strings to get the first day of a quarter, year, or even the top of the hour.
Given a particular datestamp value, Oracle returns midnight on the first day of the present week with this expression:
TRUNC(datestamp,'DY')
You can add days to a datestamp. Therefore this expression gives you midnight on the last day of the week
TRUNC(datestamp,'DY') + 6
A WHERE-clause selector for all rows in the present week might be this.
WHERE datestamp >= TRUNC(SYSDATE,'DY')
AND datestamp < TRUNC(SYSDATE,'DY') + 7
Notice that the end of the range is just before (<) midnight on the first day of the next week. You need that because you may have datestamps after midnight on the last day of the week. (Beware using BETWEEN for datestamp ranges.)
And,
SELECT TO_CHAR(TRUNC(SYSDATE,'DY'),'YYYY-MM-DD'),
TO_CHAR(TRUNC(SYSDATE,'DY')+6,'YYYY-MM-DD')
FROM DUAL;
displays the first and last dates of the present week in ISO-like format.
Date arithmetic is cool. It's worth your trouble to study the date-arithmetic functions in your DBMS at least once a year.
While writing few queries I needed to return only those rows that have date column set in this year (2017) , that's not my problem I know how to write this query in couple of diffrent ways, but I came across something strange and unexpected for me. Can anyone explain why Oracle db 11.2 is behaving this way?
select sysdate from dual
returns:
2017/12/05 09:22:27
select to_date(2017,'YYYY'),trunc(sysdate,'YYYY') from dual
returns :
2017/12/01 00:00:00 2017/01/01 00:00:00
select to_date(2017,'YYYY'),trunc(sysdate,'YYYY') from dual
where trunc(sysdate,'YYYY') = to_date(2017,'YYYY')
no rows returned
Why does to_date(2017,'YYYY') returns 2017/12/01, will it return 2017/01/01 next month? Why does it work that way? I would expect it to always return 2017/01/01 no matter the current month (if month part is indeed changing depending on sysdate).
In Oracle, TO_DATE will assume that:
If you do not specify the year then it is the current year;
If you do not specify the month then it is the current month;
If you do not specify the day then it is the first day of the month;
If you do not specify the hours then it is the midnight hour (0);
If you do not specify the minutes then it is 0 minutes past the hour; and
If you do not specify the seconds then it is 0 seconds into the minute.
You are specifying only the year (2017) so it will be:
Zero minutes and seconds past midnight of the first day of the current month of the year you specify (2017).
If you want the first day of the year then specify the month (and preferably the rest of the date):
select to_date( '201701','YYYYMM'),
trunc(sysdate,'YYYY')
from dual
where trunc(sysdate,'YYYY') = to_date( '201701','YYYYMM' )
Or use a date literal:
select DATE '2017-01-01',
trunc(sysdate,'YYYY')
from dual
where trunc(sysdate,'YYYY') = DATE '2017-01-01'
Given a field in Oracle that contains dates, how would you calculate what the week ending date is using Sun thru Sat as your week. For example, if the date is 1/26/2015 (which is a Monday), the query should return 1/31/2015 (which is a Saturday. If the date is 1/31/2015, then the query should return 1/31/2015.
Given any particular date / time value, this expression will return midnight of the preceding Sunday.
TRUNC(whatever_time,'DAY')
So, you can do stuff like this:
SELECT TRUNC(whatever_time,'DAY') week_starting,
TRUNC(whatever_time,'DAY') + 6 week_ending,
SUM(sales)
FROM table
GROUP BY TRUNC(whatever_time,'DAY')
and you'll get what you need.
Notice that TRUNC(whatever_time,'DAY') honors the Oracle session initialization parameter called “NLS_TERRITORY”. For example, in Europe Monday is considered the first business day of the week. Try this.
ALTER SESSION SET NLS_TERRITORY=GERMANY;
SELECT TRUNC( DATE '2013-12-31', 'DAY'),
TRUNC( DATE '2014-01-03', 'DAY')
FROM DUAL;
A complete writeup of this is here: http://www.plumislandmedia.net/sql-time-processing/using-sql-report-time-intervals-oracle/
Oracle 11g
here is a quick one hopefully.
Below is part of a script that gets date only from from the next month
first day of next month to last day. But today 29th feb it thrown an error of
ORA-01839: date not valid for month specified
M.MS_DATE between trunc(sysdate + interval '1' month,'MM') and last_day(sysdate + interval '1' month)
Is there a way round this. Many thanks
I have seen this as well and I consider this a bug in Oracle.
The workaround is to use add_months() instead :
between trunc(add_months(sysdate,1),'MM') and last_day(add_months(sysdate,1));
I would probably use add_months() as a_horse_with_no_name suggests, but just as an alternative if you want to use intervals, you can move the point you do the truncation in the first expression, and include the same truncation in the second expression:
select trunc(sysdate, 'MM') + interval '1' month as first_day,
last_day(trunc(sysdate, 'MM') + interval '1' month) as last_day
from dual;
FIRST_DAY LAST_DAY
---------- ----------
2015-02-01 2015-02-28
This works because all months have a first day, so you don't trip over the documented caveat.
When interval calculations return a datetime value, the result must be an actual datetime value or the database returns an error
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.