I am new with Oracle and still get used to SQL Server.
I do not understand on why the command below would return '1' when clearly sysdate which is today is 29-Apr-15 that clearly before 30-Apr-15
It should return 0. But I do not understand why it returns 1.
Any ideas? Thanks =)
SELECT CASE WHEN sysdate > to_date('30-Apr-15','DD-MON-YYYY') THEN '1' ELSE '0' END
FROM DUAL
you're missing the century from your date
SELECT CASE WHEN sysdate > to_date('30-Apr-2015','DD-MON-YYYY') THEN '1' ELSE '0' END
FROM DUAL
Because you have used 15 instead of 2015. You need to be aware of the Y2K bug. Using 'YY' format for year in a DATE is a bad idea. Always try to use 'YYYY' when converting a string into DATE
Or else, use the RR format to overcome the Y2K issue.
For example,
SQL> alter session set nls_date_format='mm/dd/yyyy';
Session altered.
SQL> SELECT TO_DATE('30-Apr-15','DD-MON-RR') RR,
2 TO_DATE('30-Apr-15','DD-MON-YYYY') YY
3 FROM DUAL;
RR YY
---------- ----------
04/30/2015 04/30/0015
SQL>
Using the same in your query:
SQL> SELECT
2 CASE
3 WHEN sysdate > to_date('30-Apr-15','DD-MON-RR')
4 THEN '1'
5 ELSE '0'
6 END
7 FROM DUAL
8 /
C
-
0
SQL>
Related
Hi everyone I wanna ask u about how I can bring data last 24 hours into bar charts, is there any methods to make it please
I have this table without data
datetime
clientchannel
servicename
service_count
13_02_2022 9:35
*****
notification
2
It is a WHERE clause you need, I presume. Something like this:
select ...
from your_table
where datetime >= sysdate - 1;
Why? Because - when you subtract a number from DATE datatype value in Oracle - it subtracts that many days.
SQL> alter session set nls_date_format = 'dd.mm.yyyy hh24:mi:ss';
Session altered.
SQL> select sysdate right_now,
2 sysdate - 1 yesterday
3 from dual;
RIGHT_NOW YESTERDAY
------------------- -------------------
13.02.2022 11:01:34 12.02.2022 11:01:34
SQL>
If you store date values as strings (which means that DATETIME column is declared as e.g. VARCHAR2(20), and that's really bad idea), then you first have to convert it to a valid date datatype value - use TO_DATE function with appropriate format mask:
where to_date(datetime, 'dd_mm_yyyy hh24:mi') >= sysdate - 1
[EDIT] If you want to go 60 minutes back, then subtract that many minutes:
SQL> select sysdate right_now,
2 sysdate - interval '60' minute an_hour_ago
3 from dual;
RIGHT_NOW AN_HOUR_AGO
------------------- -------------------
14.02.2022 07:09:30 14.02.2022 06:09:30
SQL>
I have afile where i recieve Birthdates and insert them into my Database.
the format is like the following
03-JUN-52
I use the following script to insert the date
update data."PersonBDates" set BIRTHDATE = to_date('13-SEP-47', 'DD-MON-YY');
and i also used
update data."PersonBDates" set BIRTHDATE = to_date('13-SEP-47', 'DD-MON-RR');
but when i check if find it 2074 not 1947.
How to insert this date into my oracle database?
Generally speaking, RR should work, but - not in all cases. You'll have to fix data first because RR will return different values:
for years from 00 to 49 you'll get this century, 20xx, while
50 to 99 will return previous century, 19xx
Here's an example:
SQL> alter session set nls_date_format = 'dd.mm.yyyy';
Session altered.
SQL> select
2 to_date('03-07-52', 'dd-mm-rr') rr1,
3 to_date('03-07-52', 'dd-mm-yy') yy1 ,
4 --
5 to_date('03-07-47', 'dd-mm-rr') rr2,
6 to_date('03-07-47', 'dd-mm-yy') yy2
7 from dual;
RR1 YY1 RR2 YY2
---------- ---------- ---------- ----------
03.07.1952 03.07.2052 03.07.2047 03.07.2047
SQL>
As you can see, both RR and YY format mask for year 47 return 2047.
What to do? Concatenate 19 to all years, e.g.
SQL> with test (col) as
2 (select '03-07-52' from dual union all
3 select '03-07-47' from dual
4 )
5 select col,
6 to_date(substr(col, 1, 6) || '19' || substr(col, -2), 'dd-mm-rrrr') result
7 ---------------- ---------------
8 -- this is "03-07-" "19" the last 2 digits
9 --
10 from test;
COL RESULT
-------- ----------
03-07-52 03.07.1952
03-07-47 03.07.1947
SQL>
[EDIT]
If your current inserting script works OK - which I doubt, regarding error code you mentioned in a comment:
ORA-01858: a non-numeric character was found where a numeric was expected
which means that not all input data have the same, expected & correct format of DD-MON-YY, then a simple way to fix birthdates might be this:
subtract 100 years from all dates whose year is larger than 2000
Here's how:
SQL> create table test (birthdate date);
Table created.
SQL> insert into test
2 select to_date('03-07-52', 'dd-mm-rr') from dual union all
3 select to_date('03-07-47', 'dd-mm-rr') from dual;
2 rows created.
SQL> select * from test;
BIRTHDATE
----------
03.07.1952
03.07.2047
SQL> update test set
2 birthdate = add_months(birthdate, -100 * 12)
3 where extract (year from birthdate) > 2000;
1 row updated.
SQL> select * from test;
BIRTHDATE
----------
03.07.1952
03.07.1947
SQL>
You can modify that, of course, if there's someone who actually was born in 2000 or later.
As of error you got (ORA-01858), well, fixing it depends on how exactly you're entering those values into a table.
if it was a SQL*Loader, invalid values would be rejected and stored into the .bad file and you could fix them and reprocess them later
if it was using an external tables, you could use a where clause and omit invalid rows; for example, use regexp_like
Or, your best option is to make sure that all input values are valid dates. Then any option you choose (I mentioned previously) would work without ORA-xxxxx errors.
Alternate way of concatenating 19 to all years, as Littlefoot suggested.
to_date(regexp_replace('13-SEP-47', '([0-9]+$)', '19\1'), 'DD-MON-YYYY')
I would suggest to implement the solution where 01 is not considered as 1901 but 2001 or something similar (I assume that birthday year is not 1901 for any person in your system).
Case when substr(col, -2) < to_char(sysdate,'YY')
then to_date(col, 'DD-MON-YY')
else to_date(substr(col, 1, 6) || '19' || substr(col, -2), 'dd-mm-rrrr'
End
Cheers!!
I need a list of last date of the previous months for past 13 months; like
2-28-17
3-31-17
4-30-17
5-31-17 and so on until
2-28-18
how can i get this in Oracle Sql
A row generator might help:
SQL> alter session set nls_date_format = 'dd.mm.yyyy';
Session altered.
SQL> select last_day(trunc(add_months(sysdate, -level))) result
2 from dual
3 connect by level <= 13;
RESULT
---------------
28.02.2018
31.01.2018
31.12.2017
30.11.2017
31.10.2017
30.09.2017
31.08.2017
31.07.2017
30.06.2017
31.05.2017
30.04.2017
31.03.2017
28.02.2017
13 rows selected.
SQL>
[EDIT: first day: remove LAST_DAY, truncate the result to 'mm' (month)]
SQL> select trunc(add_months(sysdate, -level), 'mm') result
2 from dual
3 connect by level <= 13;
RESULT
----------------------------------------------------------
01.02.2018
01.01.2018
01.12.2017
01.11.2017
01.10.2017
01.09.2017
01.08.2017
01.07.2017
01.06.2017
01.05.2017
01.04.2017
01.03.2017
01.02.2017
13 rows selected.
SQL>
try this,
SELECT LAST_DAY(ADD_MONTHS(SYSDATE, -1*(14-ROWNUM))) last_day
FROM dual
CONNECT BY rownum < 14;
I want to migrate a table which contains some columns with dates. The issue is my dates are often in dd/mm/yyyyy HH24:MM:YYYY format. But sometimes it appears that the format is only dd/mm/yyyy, or blank.
I guess that's why I'm getting ORA-01830 when I'm trying to migrate the datas.
I tried
CASE
WHEN TO_DATE(MYDATE,'DD/MM/YYYY')
then TO_DATE(MYDATE,'DD/MM/YYYY 00:00:00')
END AS MYDATE
But I'm not sure if it is possible to test the date format (and ofcourse it's not working).
Thank you
TO_DATE cannot test date format, but you can do it. If Lalit's answer would not be enough, try something like
select
case when my_date like '__/__/__' then to_date(my_date, 'dd/mm/yy')
when my_date like '__-__-__' then to_date(my_date, 'dd-mm-yy')
...
end
So you have the data type issue. DATE is stored as string literal. As you have mentioned that the date model has the DD/MM/YYYY part same, just that the time portion is either missing for some rows or the entire value is NULL.
For example, let's say your table have the values like -
SQL> WITH dates AS(
2 SELECT 1 num, '29/12/2014 16:38:57' dt FROM dual UNION ALL
3 SELECT 2, '29/12/2014' FROM dual UNION ALL
4 SELECT 3, NULL FROM dual
5 )
6 SELECT num, dt
7 FROM dates
8 /
NUM DT
---------- -------------------
1 29/12/2014 16:38:57
2 29/12/2014
3
SQL>
TO_DATE with proper format model should do the trick.
Let's stick to a format model first.
SQL> alter session set nls_date_format='dd/mm/yyyy hh24:mi:ss';
Session altered.
Now, let's use TO_DATE to explicitly convert the string literal to date.
SQL> WITH dates AS(
2 SELECT 1 num, '29/12/2014 16:38:57' dt FROM dual UNION ALL
3 SELECT 2, '29/12/2014' FROM dual UNION ALL
4 SELECT 3, NULL FROM dual
5 )
6 SELECT num, to_date(dt, 'dd/mm/yyyy hh24:mi:ss') dt
7 FROM dates
8 /
NUM DT
---------- -------------------
1 29/12/2014 16:38:57
2 29/12/2014 00:00:00
3
SQL>
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;