I worked this statement out
SELECT to_date('30.06.2016', 'dd.mm.yyyy') - (LEVEL-1) DATUM
FROM DUAL
CONNECT BY LEVEL <= 366;
which gives me all dates from 30.06.2016 till 366 days in the past.
So far so good.
What I need to add is that to_date('30.06.2016') is more flexible..
What I mean I always want it to use the last day of June in sysdate + 1 year.
In this case we have 2015 at the moment - so we have 30.06.2016.
If we had 2016 I need it to use 30.06.2017.
If we had 2017 I need it to use 30.06.2018.
..
..
Thanks for your help.
EDIT Solution:
SELECT last_day(add_months(to_date('01.06.' || to_char(sysdate, 'YYYY'), 'dd.mm.yyyy'),12)) - (LEVEL-1) DATUM
FROM DUAL
CONNECT BY LEVEL <= 366
If you want 366 days worth of dates:
SELECT TRUNC( SYSDATE, 'YEAR' ) + INTERVAL '18' MONTH - LEVEL AS DATUM
FROM DUAL
CONNECT BY LEVEL <= 366;
Or if you want a year's worth (365 days or 366 days in a leap year) of dates (1st July this year to 30th June next year):
SELECT TRUNC( SYSDATE, 'YEAR' ) + INTERVAL '18' MONTH - LEVEL AS DATUM
FROM DUAL
CONNECT BY TRUNC( SYSDATE, 'YEAR' ) + INTERVAL '18' MONTH - LEVEL >= TRUNC( SYSDATE, 'YEAR' ) + INTERVAL '6' MONTH;
Is your same code, but get from sysdate the year, using to_char:
select to_date('30.06.'||(to_char(sysdate,'yyyy')+1),'dd.mm.yyyy') from dual;
Here's the steps:
Truncate sysdate to the year, using Trunc().
Add 18 months, using Add_Months().
Subtract one day.
Related
How can I get data of the current week in Oracle? Assume that the week goes from Sunday to Saturday
SELECT * FROM TABLENAME
WHERE DATEFIELD IS {WITHIN CURRENT WEEK THAT GOES FROM 12AM ON SUNDAY TO 11:59PM ON SATURDAY}
Use Oracle trunc(sysdate,'IW') returns the first day of the week and add 6 to get the end of week.
-- not tested
SELECT * FROM TABLENAME
WHERE DATEFIELD >= trunc ( sysdate, 'iw' )
AND DATEFIELD < trunc ( sysdate, 'iw' ) + 5
I want to get the start and end days of every week between two dates. The dates's format is dd/mm/yyy hh24:mi:ss. I need the weeks in the format dd/mm/yyyy hh24:mi:ss because I have to calculate the days and hours between the start and end day of the week with the times
I wrote this statement
WITH
date_range AS (
SELECT
pdm.des_comercial serie,
pdm.id_material codserie,
ri.id_accion intervencion,
TO_CHAR(NVL(ri.fecha_salida_rev, SYSDATE), 'dd/mm/RRRR') fecha1,
to_char((CASE
WHEN ri.fecha_salida_rev > TO_DATE('18/06/2019', 'dd/mm/yyyy') THEN TO_DATE('18/06/2019', 'dd/mm/yyyy')
WHEN ri.fecha_salida_Rev IS NULL THEN TO_DATE('18/06/2019', 'dd/mm/yyyy')
ELSE ri.fecha_salida_Rev
END),'dd/mm/yyyy hh24:mi:ss') fechasalida,
to_char((CASE
WHEN ri.fecha_entrada_rev < TO_DATE('01/06/2019', 'dd/mm/yyyy') THEN TO_DATE('01/06/2019', 'dd/mm/yyyy')
ELSE ri.fecha_entrada_Rev
END),'dd/mm/yyyy hh24:mi:ss') fechaentrada
,
ri.cod_taller_rev,
ri.COD_MATRICULA,
ri.fecha_entrada_rev start_date,
ri.fecha_salida_rev end_date
FROM
r_intervencion ri,
planificador.pl_dh_material pdm
WHERE
ri.id_accion = ri.amortizada_por
AND ri.causa_entrada = 1
AND ri.tipo_accion = 1
AND pdm.id_material = ri.cod_serie
AND pdm.hasta = 99999999
AND ri.ID_ACCION = 'IM4'
AND ri.fecha_salida_rev BETWEEN TO_DATE('01/06/2019', 'dd/mm/yyyy') AND TO_DATE('18/06/2019', 'dd/mm/yyyy')
),
semanas AS (
SELECT LEVEL "Week"
,to_char(to_date(start_date,'dd/mm/yyyy hh24:mi:ss') + (7 * (LEVEL - 1)),'IW') startweek
,to_char(to_date(start_date ,'dd/mm/yyyy hh24:mi:ss')+ (7 * (LEVEL - 1)),'IW') + 6 endweek
,TO_CHAR(start_date + (7 * (LEVEL - 1)),'IW') "Iso Week",
serie,
codserie,
intervencion,
cod_taller_rev,
cod_matricula,
fechaentrada,
fechasalida,
start_date,
end_date
FROM date_range
CONNECT BY LEVEL <= (to_char(To_date(end_date,'dd/mm/yyyy hh24:mi:ss'),'IW') - to_char(To_date(start_date,'dd/mm/yyyy hh24:mi:ss'),'IW')) / 7 + 1
)
SELECT startweek,
endweek,
to_date(endweek,'dd/mm/yyyy hh24:mi:ss') - to_date(startweek,'dd/mm/yyyy hh24:mi:ss') dias,
serie,
codserie,
intervencion,
cod_taller_rev,
cod_matricula,
start_Date,
end_date,
fechaentrada,
fechasalida,
rd.descripcion
FROM semanas,r_depositos rd
WHERE cod_taller_rev = rd.cod_deposito
When I execute it, I get
Query execution failed
SQL Error [1840] [22008]: ORA-01840: ORA-01840: input value not long enough for date format
The error is in
,to_char(to_date(start_date,'dd/mm/yyyy hh24:mi:ss') + (7 * (LEVEL - 1)),'IW') startweek
,to_char(to_date(start_date ,'dd/mm/yyyy hh24:mi:ss')+ (7 * (LEVEL - 1)),'IW') + 6 endweek
How can I get the startweek and endweek with the format dd/mm/yyyy hh24:mi:ss
EDITED
start_date end_date
20/05/2019 20:00:00 05/06/2019 08:00:00
weeks
20/05/2019 20:00:00 26/05/2019 -> 6 days and xxx hours
27/05/2019 02/06/2019 -> 7 days
03/06/2019 05/06/2019 08:00:00 -> 3 days and xxx hours
I need to calculate the difference in days and hours for each week.
For example between 20/05/2019 20:00:00 and 26/05/2019
and last one between 03/06/2019 and 05/06/2019 08:00:00
My issue is with the calculation
to_date(endweek,'dd/mm/yyyy hh24:mi:ss') - to_date(startweek,'dd/mm/yyyy hh24:mi:ss') dias,
endweek and startweek have to have dd/mm/yyyy hh24:mi:ss
My issue is with the calculation
to_date(endweek,'dd/mm/yyyy hh24:mi:ss') - to_date(startweek,'dd/mm/yyyy hh24:mi:ss') dias,
endweek and startweek have to have dd/mm/yyyy hh24:mi:ss
Oracle dates are stored in an internal format which you generally don't need to worry about. Your application or client formats the date as a string, based on its own setting or your session NLS settings.
When you do something like:
to_date(endweek,'dd/mm/yyyy hh24:mi:ss')
you're really doing:
to_date(to_char(endweek),'dd/mm/yyyy hh24:mi:ss')
and as there is no explicit format mask specified for the implicit to_char() call it used your current session's NLS settings. Depending on the setting it might error; or might corrupt the value - e.g. mixing YY and YYYY masks can lose the century, converting 2019 to 0019. (Given the error you are getting, your NLS settings might be unusual?).
At best you're converting the date value to a string and back to exactly the same date value, which is pointless. You aren't changing the format of the datem because it doesn't have one. The intermediate string does, but you aren't using that, and you can't for calculations (at least without converting back to a date as you are, which again is pointless.)
Oracle has other functions to manipulate date values, including trunc(), so I think you might want something like this - showing the difference in three ways, though there are others and you can format the last one however you want:
with date_range (start_date, end_date) as (
-- dummy data from your example
select to_date('20/05/2019 20:00:00', 'DD/MM/YYYY HH24;MI:SS') as start_date,
to_date('05/06/2019 08:00:00', 'DD/MM/YYYY HH24;MI:SS') as end_date
from dual
),
semanas as (
select level as week,
start_date,
end_date,
greatest(trunc(start_date + (7 * (level - 1)), 'IW'), start_date) as start_week,
least(trunc(start_date + (7 * level), 'IW'), end_date) as end_week
from date_range
connect by level <= (trunc(end_date, 'IW') - trunc(start_date, 'IW')) / 7 + 1
)
select week,
to_char(start_week, 'IW') as iso_week,
to_char(start_week, 'DD/MM/YYYY HH24:MI:SS') as start_week,
to_char(end_week, 'DD/MM/YYYY HH24:MI:SS') as end_week,
end_week - start_week as diff_num,
numtodsinterval(end_week - start_week, 'DAY') as diff_interval,
to_char(date '1999-12-31' + (end_week - start_week), 'FMDD "days" HH24 "hours"') as diff_words
from semanas;
WEEK IS START_WEEK END_WEEK DIFF_NUM DIFF_INTERVAL DIFF_WORDS
---------- -- ------------------- ------------------- ---------- ------------------- ----------------
1 21 20/05/2019 20:00:00 27/05/2019 00:00:00 6.16666667 +06 04:00:00.000000 6 days 4 hours
2 22 27/05/2019 00:00:00 03/06/2019 00:00:00 7 +07 00:00:00.000000 7 days 0 hours
3 23 03/06/2019 00:00:00 05/06/2019 08:00:00 2.33333333 +02 08:00:00.000000 2 days 8 hours
As currently written the connect by only works properly if the date_range CTE generates a single value; if you actually get multiple rows back from your real query then you'll have to do a bit more work, or switch to recursive CTEs, or cross join/apply, depending on your Oracle version.
Your endweek calculation is
to_char(to_date(start_date ,'dd/mm/yyyy hh24:mi:ss')+ (7 * (LEVEL - 1)),'IW') + 6 endweek
This attempts to add the number 6 to a character string. I suspect that what you wanted was
to_char(to_date(start_date ,'dd/mm/yyyy hh24:mi:ss') + (7 * (LEVEL - 1) + 6),'IW') endweek
Here I've moved the + 6 so you're adding 6 to the date value, rather than to a character string.
In Oracle we get week number from following query:
select to_char(TO_DATE(SYSDATE,'DD-MM-YY'),'IW') from dual
I want to get date range of given week number, for example for week no:1 date range is 01-01-2017 to 08-01-2017.
is there any way to get the date range for given week number?
"week no:1 date range is 01-01-2017 to 08-01-2017"
No it isn't. You're confusing 'IW' (which runs MON - SUN) with 'WW' which runs from the first day of the year:
SQL> with dts as (
2 select date '2017-01-01' + (level-1) as dt
3 from dual
4 connect by level <= 8
5 )
6 select dt
7 , to_char(dt, 'DY') as dy_dt
8 , to_char(dt, 'IW') as iw_dt
9 , to_char(dt, 'WW') as ww_dt
10 from dts
11 order by 1;
DT DY_DT IW WW
--------- ------------ -- --
01-JAN-17 SUN 52 01
02-JAN-17 MON 01 01
03-JAN-17 TUE 01 01
04-JAN-17 WED 01 01
05-JAN-17 THU 01 01
06-JAN-17 FRI 01 01
07-JAN-17 SAT 01 01
08-JAN-17 SUN 01 02
8 rows selected.
SQL>
However, it's easy enough to generate a range for the the IW week number. You need to multiple the IW number by 7 which you can convert to a date with the day of year mask. Then you can use next_day() function to get the previous Monday and the next Sunday relative to that date:
SQL> with tgt as (
2 select to_date( &iw *7, 'DDD') as dt from dual
3 )
4 select next_day(dt-8, 'mon') as start_date
5 , next_day(dt, 'sun') as end_date
6* from tgt;
Enter value for iw: 23
old 2: select to_date( &iw *7, 'DDD') as dt from dual
new 2: select to_date( 23 *7, 'DDD') as dt from dual
START_DAT END_DATE
--------- ---------
05-JUN-17 11-JUN-17
SQL>
Obvious this solution uses my NLS Settings (English): you may need to tweak the solution if you use different settings.
These kinds of problems are easy to solve with calendar tables.
The following query builds on the assumption (ISO 8601) that the 4th of January is present in the first week in a year. Therefore I can generate a valid date in the first week of any year by constructing the 4th of January like: to_date(year || '-01-04', 'yyyy-mm-dd'). Oracle will tell me the day of week (sun=1, sat=7) for any date using to_char(date, 'D'). The 4th of JAN 2017 happens to be a wednesday (day 4). Subtracting 3 days will give me the first day (sunday) of the first week of the year.
Now it is easy to find the start day in any given week in the year by simply adding 7 days for each week (not counting the first week).
with weeks as(
select 2017 as year, 39 as week from dual union all
select 2017 as year, 40 as week from dual union all
select 2018 as year, 35 as week from dual
)
select a.*
,to_date(year || '-01-04', 'yyyy-mm-dd') - to_number(to_char(to_date(year || '-01-04', 'yyyy-mm-dd'), 'D')) + 1 + (7 * (week-1)) as start_day
,to_date(year || '-01-04', 'yyyy-mm-dd') + 7 - to_number(to_char(to_date(year || '-01-04', 'yyyy-mm-dd'), 'D')) + (7 * (week-1)) as end_day
from weeks a;
Edit: These are the "convert" expressions you need to convert from week to date range. Note that 2017 and 39 are variable...
start date = to_date(2017 || '-01-04', 'yyyy-mm-dd') - to_number(to_char(to_date(2017 || '-01-04', 'yyyy-mm-dd'), 'D')) + 1 + (7 * (39-1))
end date = to_date(2017 || '-01-04', 'yyyy-mm-dd') + 7 - to_number(to_char(to_date(2017 || '-01-04', 'yyyy-mm-dd'), 'D')) + (7 * (39-1))
Here's a query to list all ISO weeks from 2001 to 2099
SELECT TO_CHAR(TRUNC(dt, 'IW') + 6, 'IYYY-IW') AS week,
TRUNC(dt, 'IW') AS start_date,
TRUNC(dt, 'IW') + 6 AS end_date
FROM (SELECT DATE '2001-01-01' + ((LEVEL - 1) * 7) dt
FROM DUAL
CONNECT BY LEVEL <= 5165);
For the first and last week of year this query needs some CASE logic, but for other weeks works good. This solution use current NLS settings.
select to_char( start_of_week, 'day dd.mm.yyyy' ) start_of_week,
to_char( start_of_week + 6, 'day dd.mm.yyyy' ) end_of_week
from
(
select trunc( date '2017-01-01' + 38*7 , 'day') start_of_week
from dual
)
1) date '2017-01-01' - in what year we look for weeks
or it may be trunc (sysdate, 'YEAR') to take first day of current year
2) date '2017-01-01' + 38*7 - jump to 38th week
3) trunc ( ... , 'day' ) - gives date of first day of the week
https://docs.oracle.com/cd/B19306_01/server.102/b14200/functions201.htm
https://docs.oracle.com/cd/B19306_01/server.102/b14200/functions230.htm
I use this function:
FUNCTION ISOWeekDate(WEEK INTEGER, YEAR INTEGER) RETURN DATE DETERMINISTIC IS
res DATE;
BEGIN
IF WEEK > 53 OR WEEK < 1 THEN
RAISE VALUE_ERROR;
END IF;
res := NEXT_DAY(TO_DATE( YEAR || '0104', 'YYYYMMDD' ) - 7, 'MONDAY') + ( WEEK - 1 ) * 7;
IF TO_CHAR(res, 'fmIYYY') = YEAR THEN
RETURN res;
ELSE
RAISE VALUE_ERROR;
END IF;
END ISOWeekDate;
Please note, according to my comment it is ambiguous if you only provide a week number without a year. The function returns the first day of given ISO Week.
Hi I want to run a query which extracts data from 7:00 am in the morning to 3:00 pm yesterday.
I run this query in afternoon.
Below query extracts data from time I run the query. I want the data from 7:00 am and backwards.
select *
from NOTES
where creation_date >= sysdate- interval '17:15' hour to minute
I would appreciate any help.
Try this in Oracle:
SELECT *
FROM NOTES
WHERE creation_date BETWEEN to_timestamp(to_char(sysdate) || ' 00:00:00', 'dd-Mon-yy hh24:mi:ss') - interval '9' hour
AND to_timestamp(to_char(sysdate) || ' 00:00:00', 'dd-Mon-yy hh24:mi:ss') + interval '7' hour;
Try this in MySQL:
SELECT *
FROM NOTES
WHERE creation_date >= (concat(current_date(), ' 00:00:00') - interval '09:00' hour)
AND creation_date <= (concat(current_date(), ' 00:00:00') + interval '07:00' hour);
Try this as optimal
For Oracle
SELECT *
FROM NOTES
WHERE creation_date BETWEEN trunc(sysdate) - INTERVAL '9' HOUR
AND trunc(curdate) + INTERVAL '7' HOUR;
For MySQL
SELECT *
FROM NOTES
WHERE creation_date BETWEEN curdate() - INTERVAL 9 HOUR
AND curdate() + INTERVAL 7 HOUR;
If you need detail info - just ask.
I need a query where i can get last 12 month's end date from the present system date in oracle .
Below is the sample query i am using in Oracle
select
trunc(add_months(sysdate,level-1),'MM') first_day,
last_day(add_months(sysdate,level-1)) last_day
from dual
connect by level<=12;
Try this please:
SELECT TRUNC (ADD_MONTHS (SYSDATE, -(LEVEL - 1)), 'MM') FIRST_DAY,
LAST_DAY (ADD_MONTHS (SYSDATE, -(LEVEL - 1))) LAST_DAY
FROM DUAL
CONNECT BY LEVEL <= 12;
select
trunc(add_months(sysdate-numtoyminterval(1, 'YEAR'),level-1),'MM') first_day,
last_day(add_months(sysdate-numtoyminterval(1, 'YEAR'),level-1)) last_day
from dual
connect by level<=12;
Instead to start now i.e. sysdate, the start will be one year ago: sysdate-numtoyminterval(1, 'YEAR')