Is this the best way to determine if an Oracle date is on a weekend?
select * from mytable
where
TO_CHAR (my_date, 'DY', 'NLS_DATE_LANGUAGE=ENGLISH') IN ('SAT', 'SUN');
As of Oracle 11g, yes. The only viable region agnostic alternative that I've seen is as follows:
SELECT *
FROM mytable
WHERE MOD(TO_CHAR(my_date, 'J'), 7) + 1 IN (6, 7);
Not an answer to the question. But some more information. There are many more SQL tricks with date.
to_char(sysdate, 'd') --- day of a week, 1,2,3 .. to 7
to_char(sysdate, 'dd') --- day of a month, 1,2,3 .. to 30 or 31
to_char(sysdate, 'ddd') --- day of a year, 1,2,3 .. to 365 or 366
to_char(sysdate, 'w') --- week of a month, 1,2,3,4 or 5
to_char(sysdate, 'ww') --- week of a year, 1,2,3 .. to 52
More here: to_char function.
MOD(TO_CHAR(my_date, 'J'), 7) + 1 IN (6, 7)
is NOT correct!
The day number of weekend days may be 6 & 7, or 7 and 1, depending on the NLS-settings.
So the PROPER test (irrespective of NLS-settings) is this one mentioned before:
TO_CHAR (my_date, 'DY', 'NLS_DATE_LANGUAGE=ENGLISH') IN ('SAT', 'SUN')
In my opinion the best option is
TO_CHAR (my_date, 'DY', 'NLS_DATE_LANGUAGE=ENGLISH') IN ('SAT', 'SUN')
set NLS_TERRITORY before
alter session set NLS_TERRITORY=SWEDEN;
So you are sure which numbers are weekends
Related
I have a table abc as:
-- start_time |end_time | total_time_taken
-- 27.05.2020 00:52:48 |27.05.2020 02:08:33 |
I want to set the value of total_time_taken as the difference of end_time-start_time. in the format "HH:MM:SS".I searched the similar topic but didnot find the exact answer.
My expected output is like : 01:44:12 (HH:MM:SS)
So,i tried :
SELECT To_Char(end_time,'HH24:MM:SS'),To_Char(start_time,'HH24:MM:SS'),
To_Char(end_time,'HH24:MM:SS')-To_Char(start_time,'HH24:MM:SS') FROM abc;
The datatypes of start_time,end_time,total_time_taken is DATE.Please help me to find the solution.
If you cast those dates as timestamps, you can easily subtract them and see relatively nice result:
SQL> with test (st, et) as
2 (select to_date('27.05.2020 00:52:48', 'dd.mm.yyyy hh24:mi:ss'),
3 to_date('27.05.2020 02:08:33', 'dd.mm.yyyy hh24:mi:ss')
4 from dual
5 )
6 select cast(et as timestamp) - cast(st as timestamp) diff
7 from test;
DIFF
--------------------------------------------------------------------------
+000000000 01:15:45.000000
SQL>
If you want to format it as you wanted (note that mm format mask is for months; mi is for minutes), then you could do some extracting - again from timestamp (won't work for date):
SQL> with test (st, et) as
2 (select to_date('27.05.2020 00:52:48', 'dd.mm.yyyy hh24:mi:ss'),
3 to_date('27.05.2020 02:08:33', 'dd.mm.yyyy hh24:mi:ss')
4 from dual
5 ),
6 diff as
7 (select cast(et as timestamp) - cast(st as timestamp) diff
8 from test
9 )
10 select extract(hour from diff) ||':'||
11 extract(minute from diff) ||':'||
12 extract(second from diff) diff
13 from diff;
DIFF
-------------------------------------------------------------------------
1:15:45
SQL>
You can further make it pretty (e.g. two digits for hours, using LPAD function). Or, you can even write your own function which will actually work on difference of DATE datatype values, do some calculations (using trunc function, subtractions, whatnot), but the above looks pretty elegant if compared to a home-made function.
The answer by Littlefoot is perfectly fine. This answer is just to show there is more than one way to get the result.
First, we can subtract one date from another and get the difference in days, then convert that difference to an interval.
with test (st, et) as
(select to_date('27.05.2020 00:52:48', 'dd.mm.yyyy hh24:mi:ss'),
to_date('27.05.2020 02:08:33', 'dd.mm.yyyy hh24:mi:ss')
from dual
)
select numtodsinterval(et-st, 'day') diff
from test;
Then, since we can't control interval formatting directly, we can add DIFF to an arbitrary date and then use built-in date formatting.
with test (st, et) as
(select to_date('27.05.2020 00:52:48', 'dd.mm.yyyy hh24:mi:ss'),
to_date('27.05.2020 02:08:33', 'dd.mm.yyyy hh24:mi:ss')
from dual
)
select to_char(date '1-1-1' + numtodsinterval(et-st, 'day'), 'hh24:mi:ss') diff
from test;
DIFF
--------
01:15:45
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')
In Oracle, is there a straightforward way to get the first day of the week given a week number?
For example, today's date is 12/4/2012. If I run:
select to_char(sysdate,'WW') from dual;
It returns 49 for the week number.
What I would like to do is somehow return 12/2/2012 for the first day...given week 49 (assuming Sunday as first day of the week).
Any ideas? Thanks in advance for any help!
try this:
select next_day(max(d), 'sun') requested_sun
from (select to_date('01-01-2012', 'dd-mm-yyyy') + (rownum-1) d from dual connect by level <= 366)
where to_char(d, 'ww') = 49-1;
just set your year to_date('01-01-2012' and week number-1 49-1 as applicable.
the sunday in the 49th week of 2008?
SQL> select next_day(max(d), 'sun') requested_sun
2 from (select to_date('01-01-2008', 'dd-mm-yyyy') + (rownum-1) d from dual connect by level <= 366)
3 where to_char(d, 'ww') = 49-1;
REQUESTED
---------
07-DEC-08
and 2012
SQL> select next_day(max(d), 'sun') requested_sun
2 from (select to_date('01-01-2012', 'dd-mm-yyyy') + (rownum-1) d from dual connect by level <= 366)
3 where to_char(d, 'ww') = 49-1;
REQUESTED
---------
02-DEC-12
Try this,
select
next_day(trunc(to_date(in_year,'yyyy'),'yyyy') -1,'Mon') + (7 * (in_week - 1))
from dual;
If you have the date, not just the week number, you can try this:
Get the day number of the week of your date with: to_char(theDate, 'D')
substract that number from your date plus 1, and you'll get the Sunday of that week.
Add 7 and you'll get the date of end of the week(Saturday).
Like this:
SELECT theDate - to_char(theDate, 'D') + 1 as BeginOfWeek,
theDate,
theDate - to_char(theDate, 'D') + 7 as EndOfWeek
FROM TableName
I can't comment on questions yet, so I'll add another one. But this is based on #Dazzals answer.
His solution doesn't work for week one and for ISO-weeks. Also it doesn't work, if the first day of the week is not sunday, which can be controlled by NLS_SETTINGS.
This one does:
SELECT MIN(D)
FROM (SELECT TO_DATE('01-01-2013', 'dd-mm-yyyy') + (ROWNUM-10) D, ROWNUM R
FROM DUAL
CONNECT BY LEVEL <= 376)
WHERE TO_CHAR(D,'IYYYIW') = '201301'
Because we are spanning more than one year, we need to check the year too.
Using the trunc function #Justin used, I think this is what you want:
select trunc(to_date('2012-01-01', 'YYYY-MM-DD') + (49 - 1) * 7, 'WW') from dual;
I ended up doing this:
function getFirstDayOfWeek(y in binary_integer, w in binary_integer) return date
is
td date;
begin
td:=TO_DATE(TO_CHAR(y)||'0101', 'YYYYMMDD');
for c in 0..52
loop
if TO_NUMBER(TO_CHAR(td, 'IW'))=w then
return TRUNC(td, 'IW');
end if;
td:=td+7;
end loop;
return null;
end;
I have a date field
DATE = 10/10/2010
sum = 4 (this are number of years by calculation)
is there a way to add four years to 10/10/2010 and make it
10/10/2014?
Try adding months (12 * number of years) instead. Like this-
add_months(date'2010-10-10', 48)
Use add_months
Example:
SELECT add_months( to_date('10-OCT-2010'), 48 ) FROM DUAL;
Warning
add_months, returns the last day of the resulting month if you input the last day of a month to begin with.
So add_months(to_date('28-feb-2011'),12) will return 29-feb-2012 as a result.
I believe you could use the ADD_MONTHS() function. 4 years is 48 months, so:
add_months(DATE,48)
Here is some information on using the function:
http://www.techonthenet.com/oracle/functions/add_months.php
http://asktom.oracle.com/pls/asktom/f?p=100:11:0::::P11_QUESTION_ID:1157035034361
You can try this:
someDate + interval '4' year
INTERVAL
I am not sure, if I understood Your question correctly, but
select add_months(someDate, numberOfYears * 12) from dual
might do the trick
One more option apart from ADD_MONTHS
SELECT
SYSDATE,
SYSDATE
+ TO_YMINTERVAL ( '1-0' )
FROM
DUAL;
SYSDATE SYSDATE+TO_YMINTERVAL('1-0')
--------- ----------------------------
29-OCT-13 29-OCT-14
1 row selected.
SELECT
SYSDATE,
SYSDATE
+ TO_YMINTERVAL ( '2-0' )
FROM
DUAL;
SYSDATE SYSDATE+TO_YMINTERVAL('2-0')
--------- ----------------------------
29-OCT-13 29-OCT-15
1 row selected.
SELECT
TO_DATE ( '29-FEB-2004',
'DD-MON-YYYY' )
+ TO_YMINTERVAL ( '1-0' )
FROM
DUAL
*
Error at line 4
ORA-01839: date not valid for month specified
But the last one is illegal since there is no 29th day of February in 2005, hence it fails on leap year cases (Feb 29)
Read the documentation for the same
SELECT TO_CHAR(SYSDATE,'YYYY')-2 ANO FROM DUAL
Given 03/09/1982 how can we say it is which week day. In this case it will be Tue.
Is it possible to get in a single query?
SQL> SELECT TO_CHAR(date '1982-03-09', 'DAY') day FROM dual;
DAY
---------
TUESDAY
SQL> SELECT TO_CHAR(date '1982-03-09', 'DY') day FROM dual;
DAY
---
TUE
SQL> SELECT TO_CHAR(date '1982-03-09', 'Dy') day FROM dual;
DAY
---
Tue
(Note that the queries use ANSI date literals, which follow the ISO-8601 date standard and avoid date format ambiguity.)
Nearly ten years late to the party, but I think it's neccessary to enhance #Zohaib 's answer as it's result depend on the language of the client. If you display the week day name to a user, it's absolutely fine, but if your code depends on it, I'd rather control the language like so:
SQL> SELECT TO_CHAR(date '1982-03-09', 'DAY', 'NLS_DATE_LANGUAGE = ENGLISH') day FROM dual;
DAY
---------
TUESDAY
SQL> SELECT TO_CHAR(date '1982-03-09', 'DY', 'NLS_DATE_LANGUAGE = ENGLISH') day FROM dual;
DAY
---
TUE
SQL> SELECT TO_CHAR(date '1982-03-09', 'Dy', 'NLS_DATE_LANGUAGE = ENGLISH') day FROM dual;
DAY
---
Tue
To do this in Oracle sql, I tried like this and it worked for me
SELECT
START_DATE,
CASE WHEN START_DAY = 7 THEN 'SUNDAY'
WHEN START_DAY = 1 THEN 'MONDAY'
WHEN START_DAY = 2 THEN 'TUESDAY'
WHEN START_DAY = 3 THEN 'WEDNESDAY'
WHEN START_DAY = 4 THEN 'THURSDAY'
WHEN START_DAY = 5 THEN 'FRIDAY'
WHEN START_DAY = 6 THEN 'SATURDAY'
END DAY_NAME
FROM
(SELECT
TO_CHAR(T.START_DATE, 'DY') START_DAY,
TO_CHAR(T.START_DATE, 'MM/DD/YYYY') START_DATE
FROM
TABLE T )