How can i return data from 2 days ago at 11:00:00 PM to all of yesterday ending at 11:59:59 PM?
I currently have only yesterdays date query:
SELECT *
FROM table
WHERE code = '00'
AND to_char(RQST_TMSTMP, 'yyyy-mm-dd') = to_char(sysdate-1, 'yyyy-mm-dd')
How about
select *
from table
where code = '00'
and rqst_tmstmp >= trunc(sysdate - 2) + 11/24
and rqst_tmstmp <= trunc(sysdate);
Here's what all those TRUNCs represent (so that you could follow what's going on):
SQL> select sysdate, -- today, right now
2 trunc(sysdate) ts, -- today at midnight
3 trunc(sysdate - 2) ts_2, -- 2 days ago at midnight
4 trunc(sysdate - 2) + 11/24 ts_2_11 -- 2 days ago at midnight + 11 hours
5 from dual;
SYSDATE TS TS_2 TS_2_11
---------------- ---------------- ---------------- ----------------
29.11.2018 17:07 29.11.2018 00:00 27.11.2018 00:00 27.11.2018 11:00
SQL>
If the column is capturing hours & minutes then use,
TO_CHAR(RQST_TMSTMP,'DD-MM-YY HH24:MI')
Related
I have a database structure as below.
period
month
start_date
1
April
2022-04-01
2
May
2022-05-07
3
June
2022-06-04
4
July
2022-07-02
5
August
2022-08-06
6
September
2022-09-03
7
October
2022-10-01
8
November
2022-11-05
9
December
2022-12-03
10
January
2023-01-01
11
February
2023-02-04
12
March
2023-03-04
End date of the year is 2023-03-31.
Based on current_date, how do I select the query to return where the current date falls under Period 6.
My current query as below.
SELECT period FROM table1 as a
WHERE
a.start_date = (SELECT MAX(start_date) FROM table1 as b WHERE
b.start_date <=current_date) and ROWNUM <= 1
Is there anyway to improve the current query which to avoid using subquery?
Today is September 22nd, so - would this do?
Some sample data:
SQL> with test (period, month, start_date) as
2 (select 1, 'april' , date '2022-04-01' from dual union all
3 select 5, 'august' , date '2022-08-06' from dual union all
4 select 6, 'september', date '2022-09-03' from dual union all
5 select 7, 'october' , date '2022-10-01' from dual union all
6 select 10, 'january' , date '2023-01-01' from dual union all
7 select 12, 'march' , date '2023-03-04' from dual
8 ),
Query begins here:
9 temp as
10 (select period, month, start_date,
11 row_number() over (order by start_date desc) rn
12 from test
13 where start_date <= sysdate
14 )
15 select period
16 from temp
17 where rn = 1
18 /
PERIOD
----------
6
SQL>
It still uses a subquery (or a CTE, as in my example), but - as opposed to your approach, it selects from the source table only once, so performance should be improved.
A few more tests: instead of sysdate (line #13), presume that today is September 2nd (which means that it is in period #5):
9 temp as
10 (select period, month, start_date,
11 row_number() over (order by start_date desc) rn
12 from test
13 where start_date <= date '2022-09-02'
14 )
15 select period
16 from temp
17 where rn = 1;
PERIOD
----------
5
SQL>
Or, if today were August 7th:
9 temp as
10 (select period, month, start_date,
11 row_number() over (order by start_date desc) rn
12 from test
13 where start_date <= date '2022-08-07'
14 )
15 select period
16 from temp
17 where rn = 1;
PERIOD
----------
5
SQL>
Your rule for the start_date appears to be:
If the month is January (first month of the calendar year) or April (typically, first month of the financial year) then use the 1st of that month;
Otherwise use the 1st Saturday of the month.
If that is the case then you can calculate the start date of the next month and use the query:
SELECT *
FROM table1
WHERE start_date <= SYSDATE
AND SYSDATE < CASE
WHEN EXTRACT(MONTH FROM ADD_MONTHS(start_date, 1))
IN (1, 4) -- 1st month of calendar or financial year
THEN TRUNC(ADD_MONTHS(start_date, 1), 'MM')
ELSE NEXT_DAY(TRUNC(ADD_MONTHS(start_date, 1), 'MM') - 1, 'SATURDAY')
END
Then, for your sample data:
CREATE TABLE table1 (
period NUMBER(2,0),
month VARCHAR2(9)
GENERATED ALWAYS AS (
CAST(
TO_CHAR(start_date, 'FXMonth', 'NLS_DATE_LANGUAGE=English')
AS VARCHAR2(9)
)
),
start_date DATE
);
INSERT INTO table1 (period, start_date)
SELECT LEVEL,
CASE
WHEN EXTRACT(MONTH FROM ADD_MONTHS(DATE '2022-04-01', LEVEL - 1))
IN (1, 4) -- 1st month of calendar or financial year
THEN ADD_MONTHS(DATE '2022-04-01', LEVEL - 1)
ELSE NEXT_DAY(ADD_MONTHS(DATE '2022-04-01', LEVEL - 1) - 1, 'SATURDAY')
END
FROM DUAL
CONNECT BY LEVEL <= 12;
Outputs:
PERIOD
MONTH
START_DATE
6
September
2022-09-03 00:00:00
fiddle
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.
EX:if yesterday's date is 19th we need to show the records for 19th 12AM to 6 AM and 11 PM to 12 AM.
the query which i tried.Please suggest me.
select * from table_name
where TIMESTAMP between trunc(sysdate-1) and
trunc(sysdate-1) + 6/24
or timestamp between trunc(sysdate-1) + 23/24 and
trunc(sysdate);
You may need:
where yourColumn between trunc(sysdate -1) + 23/24 and trunc (sysdate) + 6/24
How it works: sysdate both contains date and time informations, so trunc(sysdate) only keeps day, with no hours, minutes, ....
If you add a number to a date, that number is meant to represent days, so 1/24 gives one hour and so on.
For clarity:
trunc(sysdate) -> today at 00:00 AM
trunc(sysdate-1) -> yesterday at 00:00 AM
trunc(sysdate-1) + 6/24 -> yesterday at 06:00 AM
trunc(sysdate-1) + 18/24 -> yesterday at 06:00 PM
...
I wanna roll six months dates starting from 2016 to current date, output should be like below
Year Start_Date End_Date
2016 1/1/2016 30/6/2016
2016 1/7/2016 31/12/2016
2017 ... like this for 2017 & 2018
I tried like fetch first date of the year like
SELECT TRUNC(to_date(Date_key, 'YYYYMMDD'), 'YEAR')
FROM Table;
and adding six months from first date, but in this case end date will be problem. Is there a function I can do this without loop?
Something like this?
SQL> alter session set nls_date_format = 'dd.mm.yyyy';
Session altered.
SQL> select extract (year from add_months (date '2016-01-01', (level - 1) * 6)) year,
2 add_months (date '2016-01-01', (level - 1) * 6) start_date,
3 add_months (date '2016-01-01', (level) * 6) - 1 end_date
4 from dual
5 connect by level <= (extract (year from sysdate) - 2016 + 1) * 2;
YEAR START_DATE END_DATE
---------- ---------- ----------
2016 01.01.2016 30.06.2016
2016 01.07.2016 31.12.2016
2017 01.01.2017 30.06.2017
2017 01.07.2017 31.12.2017
2018 01.01.2018 30.06.2018
2018 01.07.2018 31.12.2018
6 rows selected.
SQL>
You can also try this,
SELECT :p_from_year + CEIL(ROWNUM/2)-1,
ADD_MONTHS(TRUNC(TO_DATE(:p_from_year + CEIL(rownum/2)-1, 'YYYY'), 'YYYY'),6-(MOD(rownum, 2)*6)) from_date,
ADD_MONTHS(TRUNC(TO_DATE(:p_from_year + CEIL(rownum/2)-1, 'YYYY'), 'YYYY'),12-(MOD(rownum, 2)*6))-1 to_date
FROM dual
CONNECT BY CEIL(rownum/2)-1 <= (:p_to_year - :p_from_year)
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.